Disarium numbers

Disarium numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A Disarium number is an integer where the sum of each digit raised to the power of its position in the number, is equal to the number.

E.G.

135 is a Disarium number:

1¹ + 3² + 5³ == 1 + 9 + 125 == 135

There are a finite number of Disarium numbers.

Task

• Find and display the first 18 Disarium numbers.

Stretch

• Find and display all 20 Disarium numbers.

See also

• Geeks for Geeks - Disarium numbers
• OEIS:A032799 - Numbers n such that n equals the sum of its digits raised to the consecutive powers (1,2,3,...)
• Related task: Narcissistic decimal number
• Related task: Own digits power sum Which seems to be the same task as Narcissistic decimal number...

Factor

Works with: Factor version 0.99 2021-06-02

<lang factor>USING: io kernel lists lists.lazy math.ranges math.text.utils math.vectors prettyprint sequences ;

disarium? ( n -- ? )

   dup 1 digit-groups dup length 1 [a,b] v^ sum = ;

disarium ( -- list ) 0 lfrom [ disarium? ] lfilter ;

19 disarium ltake [ pprint bl ] leach nl</lang>

Output:

0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427 2646798 

Julia

<lang julia>isdisarium(n) = sum(last(p)^first(p) for p in enumerate(reverse(digits(n)))) == n

function disariums(numberwanted)

   n, ret = 0, Int[]
   while length(ret) < numberwanted
       isdisarium(n) && push!(ret, n)
       n += 1
   end
   return ret

end

println(disariums(19)) @time disariums(19)

</lang>

Output:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]
  0.555962 seconds (5.29 M allocations: 562.335 MiB, 10.79% gc time)

Raku

Not an efficient algorithm. First 18 in less than 1/4 second. 19th in around 45 seconds. Pretty much unusable for the 20th. <lang perl6>my $disarium = (^∞).hyper.map: { $_ if $_ == sum .polymod(10 xx *).reverse Z** 1..* };

put $disarium[^18]; put $disarium[18];</lang>

Output:

0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427
2646798

Wren

Library: Wren-math

Well, for once, we can brute force the first 19 Disarium numbers without resorting to BigInt or GMP. Surprisingly, much quicker than Raku at 3.35 seconds.

Taking a rain check on the 20th number which is 20 digits long (too big for Wren's 53 bit integers) to see whether more efficient methods emerge. <lang ecmascript>import "./math" for Int

var limit = 19 var count = 0 var disarium = [] var n = 0 while (count < limit) {

   var sum = 0
   var digits = Int.digits(n)
   for (i in 0...digits.count) sum = sum + digits[i].pow(i+1)
   if (sum == n) {
       disarium.add(n)
       count = count + 1
   }
   n = n + 1

} System.print("The first 19 Disarium numbers are:") System.print(disarium)</lang>

Output:

The first 19 Disarium numbers are:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798]