Numbers with equal rises and falls
When a number is written in base 10, adjacent digits may "rise" or "fall"
as the number is read (usually from left to right).
OEIS Sequence A296712 describes numbers whose digit sequence in base 10 have equal "rises" and "falls".
Definition:
Given the digits of the number are written as a series d:
- A rise is an index i such that d(i) < d(i+1)
- A fall is an index i such that d(i) > d(i+1).
Examples:
- The number 726169 has 3 rises and 2 falls, so it is not in the sequence.
- The number 83548 has 2 rises and 2 falls, so it is in the sequence.
- Task
Print the first 200 numbers in the sequence. Show that the 10 millionth (10,000,000th) number in the sequence is 41909002.
See also: OEIS:A296712 the Oeis entry.
8080 Assembly
<lang 8080asm>puts: equ 9 ; CP/M calls putch: equ 2 org 100h ;;; Print first 200 numbers lxi d,first mvi c,puts call 5 mvi b,200 ; 200 numbers f200: push b call next ; Get next number call pnum ; Print the number pop b ; Restore counter dcr b ; Are we there yet? jnz f200 ; If not, next number ;;; Find 10,000,000th number lxi d,tenmil mvi c,puts call 5 f1e7: call next ; Keep generating numbers until ten million reached jnz f1e7 ; Then print the number ;;; Print the current number pnum: lxi d,num pscan: dcx d ; Scan for zero ldax d ana a jnz pscan mvi c,puts ; Once found, print string jmp 5 ;;; Increment number until rises and falls are equal next: lxi h,num incdgt: mov a,m ; Get digit ana a ; If 0, then initialize jz grow inr a ; Otherwise, increment mov m,a ; Store back cpi '9'+1 ; Rollover? jnz idone ; If not, we're done mvi m,'0' ; If so, set digit to 0 dcx h ; And increment previous digit jmp incdgt grow: mvi m,'1' idone: lxi h,num ; Find rises and falls mvi b,0 ; B = rises - falls mov c,m ; C = right digit in comparison pair: dcx h mov a,m ; A = left digit in comparison ana a ; When zero, done jz check cmp c ; Compare left digit to right digit jc fall ; A<C = fall jnz rise ; A>C = rise nxdgt: mov c,a ; C is now left digit jmp pair ; Check next pair fall: dcr b ; Fall: decrement B jmp nxdgt rise: inr b ; Rise: increment B jmp nxdgt check: mov a,b ; If B=0 then rises and falls are equal ana a jnz next ; Otherwise, increment number and try again lxi h,ctr ; But if so, decrement the counter to 10 million mov a,m ; First byte sui 1 mov m,a inx h ; Second byte mov a,m sbb b ; B=0 here mov m,a inx h ; Third byte mov a,m sbb b mov m,a dcx h ; OR them together to see if the number is zero ora m dcx h ora m ret ;;; Strings first: db 'The first 200 numbers are:',13,10,'$' tenmil: db 13,10,10,'The 10,000,000th number is: $' ;;; Current number (stored as ASCII) db 0,0,0,0,0,0,0,0 num: db '0 $' ;;; 24-bit counter to keep track of ten million ctr: db 80h,96h,98h ; 1e7 = 989680h</lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is: 41909002
8086 Assembly
<lang asm>puts: equ 9 ; MS-DOS print string cpu 8086 bits 16 org 100h section .text mov bp,98h ; BP:DI = 989680h = ten million mov di,9680h ;;; Print first 200 numbers mov dx,first ; Print message mov ah,puts int 21h n200: call next ; Get next number call pnum ; Print the number cmp di,95B8h ; Have we had 200 yet? ja n200 ; If not, print next number ;;; Print the 10 millionth number mov dx,tenmil ; Print message mov ah,puts int 21h n1e7: call next ; Get next number jnz n1e7 ; Until we have the 10 millionth ;;; Print the current number pnum: std ; Read backwards xchg si,di ; Keep DI safe mov di,num mov cx,9 xor al,al ; Find the first zero repnz scasb inc di ; Go to first digit inc di xchg si,di ; Put DI back mov dx,si ; Call DOS to print the number mov ah,puts int 21h ret ;;; Increment number until rises and falls are equal next: std ; Read number backwards .inc: mov bx,num .iloop: mov al,[bx] ; Get digit test al,al ; If uninitialized, write a 1 jz .grow inc ax ; Otherwise, increment mov [bx],al ; Write it back cmp al,'9'+1 ; Rollover? jnz .idone ; If not, the increment is done mov [bx],byte '0' ; But if so, this digit should be 0, dec bx ; and the next digit incremented. jmp .iloop .grow: mov [bx],byte '1' ; The number gains an extra digit .idone: xor bl,bl ; BL = rise and fall counter mov si,num lodsb ; Read first digit to compare to .pair: xchg ah,al ; Previous digit to compare lodsb ; Read next digit test al,al ; Done yet? jz .done cmp al,ah ; If not, compare the digits ja .fall ; If they are different, jb .rise ; there is a fall or a rise jmp .pair ; Otherwise, try next pair .fall: dec bl ; Fall: decrement BL jmp .pair .rise: inc bl ; Rise: increment BL jmp .pair .done: test bl,bl ; At the end, check if BL is zero jnz .inc ; If not, try next number sub di,1 ; Decrement the million counter in BP:DI sbb bp,0 mov ax,di ; Test if BP:DI is zero or ax,bp ret section .data ;;; Strings first: db 'The first 200 numbers are:',13,10,'$' tenmil: db 13,10,10,'The 10,000,000th number is: $' ;;; Current number, stored as ASCII db 0,0,0,0,0,0,0,0 num: db '0 $'</lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is: 41909002
ALGOL 68
... with a single counter for rises and falls.
<lang algol68>BEGIN
# returns TRUE if the number of digits in n followed by a higher digit (rises) #
# equals the number of digits followed by a lower digit (falls) #
# FALSE otherwise #
PROC rises equals falls = ( INT n )BOOL:
BEGIN
INT rf := 0;
INT prev := n MOD 10;
INT v := n OVER 10;
WHILE v > 0 DO
INT d = v MOD 10;
IF d < prev THEN
rf +:= 1 # rise #
ELIF d > prev THEN
rf -:= 1 # fall #
FI;
prev := d;
v OVERAB 10
OD;
rf = 0
END; # rises equals falls #
# task tests #
print( ( "The first 200 numbers in the sequence are:", newline ) );
INT count := 0;
INT max count = 10 000 000;
FOR n WHILE count < max count DO
IF rises equals falls( n ) THEN
count +:= 1;
IF count <= 200 THEN
print( ( whole( n, -4 ) ) );
IF count MOD 20 = 0 THEN print( ( newline ) ) FI
ELIF count = max count THEN
print( ( newline, "The 10 millionth number in the sequence is ", whole( n, -8 ), ".", newline ) )
FI
FI
OD
END </lang>
- Output:
The first 200 numbers in the sequence are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10 millionth number in the sequence is 41909002.
APL
<lang APL>risefall←{
⍝ Determine if a number is in the sequence inSeq←0=(+/2(<->)/10(⊥⍣¯1)⊢)
⍝ First 200 numbers ⎕←'The first 200 numbers are:' ⎕←(⊢(/⍨)inSeq¨)⍳404
⍝ 10,000,000th number
⍝ You can't just make a list that big and filter
⍝ it, because that will just get you a WS FULL.
⍝ Instead it's necessary to loop over them the old-
⍝ fashioned way
⍞←'The 10,000,000th number is: '
⎕←1e7{⍺=0:⍵-1 ⋄ (⍺-inSeq ⍵)∇ ⍵+1}1
}</lang>
- Output:
The first 200 numbers are:
1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102
103 104 105 106 107 108 109 111 120 121 130 131
132 140 141 142 143 150 151 152 153 154 160 161
162 163 164 165 170 171 172 173 174 175 176 180
181 182 183 184 185 186 187 190 191 192 193 194
195 196 197 198 201 202 203 204 205 206 207 208
209 212 213 214 215 216 217 218 219 222 230 231
232 240 241 242 243 250 251 252 253 254 260 261
262 263 264 265 270 271 272 273 274 275 276 280
281 282 283 284 285 286 287 290 291 292 293 294
295 296 297 298 301 302 303 304 305 306 307 308
309 312 313 314 315 316 317 318 319 323 324 325
326 327 328 329 333 340 341 342 343 350 351 352
353 354 360 361 362 363 364 365 370 371 372 373
374 375 376 380 381 382 383 384 385 386 387 390
391 392 393 394 395 396 397 398 401 402 403 404
The 10,000,000th number is:
41909002
C
<lang C>#include <stdio.h>
/* Check whether a number has an equal amount of rises
* and falls */
int riseEqFall(int num) {
int rdigit = num % 10;
int netHeight = 0;
while (num /= 10) {
netHeight += ((num % 10) > rdigit) - ((num % 10) < rdigit);
rdigit = num % 10;
}
return netHeight == 0;
}
/* Get the next member of the sequence, in order,
* starting at 1 */
int nextNum() {
static int num = 0;
do {num++;} while (!riseEqFall(num));
return num;
}
int main(void) {
int total, num;
/* Generate first 200 numbers */
printf("The first 200 numbers are: \n");
for (total = 0; total < 200; total++)
printf("%d ", nextNum());
/* Generate 10,000,000th number */
printf("\n\nThe 10,000,000th number is: ");
for (; total < 10000000; total++) num = nextNum();
printf("%d\n", num);
return 0;
}</lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is: 41909002
F#
<lang fsharp> // A296712. Nigel Galloway: October 9th., 2020 let fN g=let rec fN Ψ n g=match n,Ψ with (0,0)->true |(0,_)->false |_->let i=n%10 in fN (Ψ + (compare i g)) (n/10) i in fN 0 g (g%10) let A296712=seq{1..2147483647}|>Seq.filter fN A296712|>Seq.take 200|>Seq.iter(printf "%d "); printfn"\n" [999999;9999999;99999999]|>List.iter(fun n->printfn "The %dth element is %d" (n+1) (Seq.item n A296712)) </lang>
- Output:
1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 1000000th element is 3284698 The 10000000th element is 41909002 The 100000000th element is 375551037
Factor
<lang factor>USING: grouping io kernel lists lists.lazy math math.extras prettyprint tools.memory.private ;
- rises-and-falls-equal? ( n -- ? )
0 swap 10 /mod swap [ 10 /mod rot over - sgn rotd + spin ] until-zero drop 0 = ;
- OEIS:A296712 ( -- list )
1 lfrom [ rises-and-falls-equal? ] lfilter ;
! Task "The first 200 numbers in OEIS:A296712 are:" print 200 OEIS:A296712 ltake list>array 20 group simple-table. nl
"The 10 millionth number in OEIS:A296712 is " write OEIS:A296712 9,999,999 [ cdr ] times car commas print</lang>
- Output:
The first 200 numbers in OEIS:A296712 are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10 millionth number in OEIS:A296712 is 41,909,002
Forth
<lang forth>: in-seq? ( n -- is N in the sequence? )
0 swap \ height 10 /mod \ digit and rest of number begin dup while \ as long as the number isn't zero... 10 /mod \ get next digit and quotient -rot swap \ retrieve previous digit over - sgn \ see if higher, lower or equal (-1, 0, 1) >r rot r> + \ add to height -rot swap \ quotient on top of stack repeat drop drop \ drop number and last digit 0= \ is height equal to zero?
- next-val ( n -- n: retrieve first element of sequence higher than N )
begin 1+ dup in-seq? until
- two-hundred
begin over 200 < while next-val dup . swap 1+ swap repeat
- ten-million
begin over 10000000 < while next-val swap 1+ swap repeat
0 0 \ top of stack: current index and number ." The first 200 numbers are: " two-hundred cr cr ." The 10,000,000th number is: " ten-million . cr bye</lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is: 41909002
Go
<lang go>package main
import "fmt"
func risesEqualsFalls(n int) bool {
if n < 10 {
return true
}
rises := 0
falls := 0
prev := -1
for n > 0 {
d := n % 10
if prev >= 0 {
if d < prev {
rises = rises + 1
} else if d > prev {
falls = falls + 1
}
}
prev = d
n /= 10
}
return rises == falls
}
func main() {
fmt.Println("The first 200 numbers in the sequence are:")
count := 0
n := 1
for {
if risesEqualsFalls(n) {
count++
if count <= 200 {
fmt.Printf("%3d ", n)
if count%20 == 0 {
fmt.Println()
}
}
if count == 1e7 {
fmt.Println("\nThe 10 millionth number in the sequence is ", n)
break
}
}
n++
}
}</lang>
- Output:
The first 200 numbers in the sequence are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10 millionth number in the sequence is 41909002
Haskell
<lang haskell>import Data.Char
pairs :: [a] -> [(a,a)] pairs (a:b:as) = (a,b):pairs (b:as) pairs _ = []
riseEqFall :: Int -> Bool riseEqFall n = rel (>) digitPairs == rel (<) digitPairs
where rel r = sum . map (fromEnum . uncurry r)
digitPairs = pairs $ map digitToInt $ show n
a296712 :: [Int] a296712 = [n | n <- [1..], riseEqFall n]
main :: IO () main = do putStrLn "The first 200 numbers are: " putStrLn $ unwords $ map show $ take 200 a296712 putStrLn "" putStr "The 10,000,000th number is: " putStrLn $ show $ a296712 !! 9999999 </lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is: 41909002
Julia
<lang julia>using Lazy
function rises_and_falls(n)
if n < 10
return 0, 0
end
lastr, rises, falls = n % 10, 0, 0
while n != 0
n, r = divrem(n, 10)
if r > lastr
falls += 1
elseif r < lastr
rises += 1
end
lastr = r
end
return rises, falls
end
isA296712(x) = ((a, b) = rises_and_falls(x); return a == b)
function genA296712(N, M)
A296712 = filter(isA296712, Lazy.range(1));
arr = take(N, A296712)
j = 0
for i in take(200, A296712)
j += 1
print(lpad(i, 4), j % 20 == 0 ? "\n" : "")
end
for i in take(M, A296712)
j = i
end
println("\nThe $M-th number in sequence A296712 is $j.")
end
genA296712(200, 10_000_000)
</lang>
- Output:
1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10000000-th number in sequence A296712 is 41909002.
Phix
<lang Phix>atom t1 = time()+1 integer count = 0, n = 0 printf(1,"The first 200 numbers are:\n") while true do
n += 1
integer rmf = 0,
l = remainder(n,10),
r = floor(n/10)
while r do
integer p = remainder(r,10)
rmf += compare(l,p)
l = p
r = floor(r/10)
end while
if rmf=0 then
count += 1
if count<=200 then
printf(1,"%3d ",n)
if remainder(count,20)=0 then
printf(1,"\n")
end if
end if
if count == 1e7 then
progress("")
printf(1,"\nThe %,dth number is %,d\n",{count,n})
exit
end if
if time()>t1 then
progress("%,d:%,d\r",{count,n})
t1 = time()+1
end if
end if
end while</lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is 41,909,002
Python
<lang python>import itertools
def riseEqFall(num):
"""Check whether a number belongs to sequence A296712."""
height = 0
d1 = num % 10
num //= 10
while num:
d2 = num % 10
height += (d1<d2) - (d1>d2)
d1 = d2
num //= 10
return height == 0
def sequence(start, fn):
"""Generate a sequence defined by a function"""
num=start-1
while True:
num += 1
while not fn(num): num += 1
yield num
a296712 = sequence(1, riseEqFall)
- Generate the first 200 numbers
print("The first 200 numbers are:") print(*itertools.islice(a296712, 200))
- Generate the 10,000,000th number
print("The 10,000,000th number is:") print(*itertools.islice(a296712, 10000000-200-1, 10000000-200))
- It is necessary to subtract 200 from the index, because 200 numbers
- have already been consumed.
</lang>
- Output:
The first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th number is: 41909002
Raku
<lang perl6>use Lingua::EN::Numbers; use Base::Any;
sub rf (int $base = 10, $batch = Any) {
my %batch = batch => $batch if $batch;
lazy flat 1 ..^ $base, ($base .. ∞).hyper(|%batch).map: {
my int ($this, $last) = $_, $_ % $base;
my int ($rise, $fall) = 0, 0;
while $this {
my int $rem = $this % $base;
$this = $this div $base;
if $rem > $last { $fall = $fall + 1 }
elsif $rem < $last { $rise = $rise + 1 }
$last = $rem
}
next unless $rise == $fall;
$_
}
}
my $upto = 200; put "First $upto (base 10):"; .put for rf[^$upto]».fmt("%3d").batch(20);
$upto = 10_000_000; put "\nThe {$upto.&comma}th (base 10): ", comma rf(10, 65536)[$upto - 1];
$upto = ^6; put "\n{tc $upto.map({.exp(10).&ordinal}).join: ', '}, values in some other bases:"; for 2, 'A005408',
3, 'A296691',
4, 'A296694',
5, 'A296697',
6, 'A296700',
7, 'A296703',
8, 'A296706',
9, 'A296709',
10, 'A296712',
11, 'A296744',
12, 'A296747',
13, 'A296750',
14, 'A296753',
15, 'A296756',
16, 'A296759',
20, 'A296762',
60, 'A296765'
-> $base, $oeis {
put "Base {$base.fmt(<%2d>)} (OEIS:$oeis): ",
$upto.map({rf($base, 5000)[.exp(10) - 1].&to-base($base)}).join: ', '
}</lang>
- Output:
First 200 (base 10): 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10,000,000th (base 10): 41,909,002 First, tenth, one hundredth, one thousandth, ten thousandth, one hundred thousandth, values in some other bases: Base 2 (OEIS:A005408): 1, 10011, 11000111, 11111001111, 100111000011111, 110000110100111111 Base 3 (OEIS:A296691): 1, 201, 22112, 10101111, 1100022001, 112010220010 Base 4 (OEIS:A296694): 1, 111, 3333, 221012, 13002120, 1023032003 Base 5 (OEIS:A296697): 1, 102, 1441, 40011, 1431201, 40142240 Base 6 (OEIS:A296700): 1, 55, 512, 20424, 400402, 10405133 Base 7 (OEIS:A296703): 1, 44, 365, 12620, 155554, 2525215 Base 8 (OEIS:A296706): 1, 33, 316, 7466, 60404, 1336621 Base 9 (OEIS:A296709): 1, 22, 275, 5113, 40217, 665836 Base 10 (OEIS:A296712): 1, 11, 252, 3396, 29201, 393329 Base 11 (OEIS:A296744): 1, A, 216, 2240, 21718, 217785 Base 12 (OEIS:A296747): 1, A, 201, 10AA, 19723, B3025 Base 13 (OEIS:A296750): 1, A, 1B8, A0A, 172A7, 823C8 Base 14 (OEIS:A296753): 1, A, 1B5, 8B9, 14B81, 6476D Base 15 (OEIS:A296756): 1, A, 1B2, 7D4, 11BBA, 51013 Base 16 (OEIS:A296759): 1, A, 1A9, 716, 10424, 41E9D Base 20 (OEIS:A296762): 1, A, 196, 523, 8011, 24HE2 Base 60 (OEIS:A296765): 1, A, ff, 1f2, 63Q, g5D
REXX
To do the heavy lifting, this REXX program constructs a table of every two-digit sequence which indicates a
rise (+1), fall (-1), or neither (0).
<lang rexx>/*REXX pgm finds and displays N numbers that have an equal number of rises and falls,*/
parse arg n . /*obtain optional argument from the CL.*/
if n== | n=="," then n= 200 /*Not specified? Then use the default.*/
append= n>0 /*a flag that is used to append numbers*/
n= abs(n) /*use the absolute value of N. */
call init /*initialize the rise/fall database. */
do j=1 until #==n; Lm= length(j) - 1
s= 0 /*initialize the sum of rises/falls. */
do k=1 for Lm; t= substr(j,k,2) /*obtain a set of two digs from number.*/
s= s + @.t /*sum the rises and falls in the number*/
end /*k*/
if s\==0 then iterate /*Equal # of rises & falls? Then add it*/
#= # + 1 /*bump the count of numbers found. */
if append then $= $ j /*append to the list of numbers found. */
else $= j /*merely set the last number found. */
end /*j*/
if append then call show /*display a list of N numbers──►term.*/
else say 'the ' commas(n)th(n) " number is: " commas($) /*show Nth #.*/
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ init: @.= 0; do i=1 for 9; _= i' '; @._= 1; _= '0'i; @._= +1; end /*i*/
do i=10 to 99; parse var i a 2 b; if a>b then @.i= -1
else if a<b then @.i= +1
end /*i*/; #= 0; $=; return
/*──────────────────────────────────────────────────────────────────────────────────────*/ show: say 'the first ' commas(#) " numbers are:"; say; w= length( word($, #) )
_=; do o=1 for n; _= _ right( word($, o), w); if o//20\==0 then iterate
say substr(_, 2); _= /*display a line; nullify a new line. */
end /*o*/ /* [↑] display 20 numbers to a line.*/
if _\== then say substr(_, 2); return /*handle any residual numbers in list. */
/*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do c=length(_)-3 to 1 by -3; _=insert(',', _, c); end; return _ th: parse arg th; return word('th st nd rd',1+(th//10)*(th//100%10\==1)*(th//10<4))</lang>
- output when using the default input:
the first 200 numbers are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404
- output when using the input of: -10000000
the 10,000,000th number is: 41,909,002
Wren
<lang ecmascript>import "/fmt" for Fmt
var risesEqualsFalls = Fn.new { |n|
if (n < 10) return true
var rises = 0
var falls = 0
var prev = -1
while (n > 0) {
var d = n%10
if (prev >= 0) {
if (d < prev) {
rises = rises + 1
} else if (d > prev) {
falls = falls + 1
}
}
prev = d
n = (n/10).floor
}
return rises == falls
}
System.print("The first 200 numbers in the sequence are:") var count = 0 var n = 1 while (true) {
if (risesEqualsFalls.call(n)) {
count = count + 1
if (count <= 200) {
Fmt.write("$3d ", n)
if (count % 20 == 0) System.print()
}
if (count == 1e7) {
Fmt.print("\nThe 10 millionth number in the sequence is $,d.", n)
break
}
}
n = n + 1
}</lang>
- Output:
The first 200 numbers in the sequence are: 1 2 3 4 5 6 7 8 9 11 22 33 44 55 66 77 88 99 101 102 103 104 105 106 107 108 109 111 120 121 130 131 132 140 141 142 143 150 151 152 153 154 160 161 162 163 164 165 170 171 172 173 174 175 176 180 181 182 183 184 185 186 187 190 191 192 193 194 195 196 197 198 201 202 203 204 205 206 207 208 209 212 213 214 215 216 217 218 219 222 230 231 232 240 241 242 243 250 251 252 253 254 260 261 262 263 264 265 270 271 272 273 274 275 276 280 281 282 283 284 285 286 287 290 291 292 293 294 295 296 297 298 301 302 303 304 305 306 307 308 309 312 313 314 315 316 317 318 319 323 324 325 326 327 328 329 333 340 341 342 343 350 351 352 353 354 360 361 362 363 364 365 370 371 372 373 374 375 376 380 381 382 383 384 385 386 387 390 391 392 393 394 395 396 397 398 401 402 403 404 The 10 millionth number in the sequence is 41,909,002.