A362843 -id:A362843 - cp's OEIS frontend

A032799 Numbers k such that k equals the sum of its digits raised to the consecutive powers (1,2,3,...).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798, 12157692622039623539

Offset: 1

Patrick De Geest, May 15 1998

Lemma: The sequence is finite with all terms in the sequence having at most 22 digits. Proof: Let n be an m-digit natural number in the sequence for some m. Then 10^(m-1)<=n and n<=9+9^2+...9^m = 9(9^m-1)/8<(9^(m+1))/8. Thus 10^(m-1)<(9^(m+1))/8. Taking logarithms of both sides and solving yields m<22.97 QED. Note proof is identical to that for A208130. [Francis J. McDonnell, Apr 14 2012]

Sometimes referred to as disarium numbers. - Dumitru Damian, Jul 22 2024

			89 = 8^1 + 9^2.
175 = 1^1 + 7^2 + 5^3.
2427 = 2^1 + 4^2 + 2^3 + 7^4.
2646798 = 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 175, p. 55, Ellipses, Paris 2008.
Ken Follett, Code to Zero, Dutton, a Penguin Group, NY 2000, p. 84.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 37.
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, London, 1986, Entry 175.

Rosetta Code, Disarium numbers.
Tim Cross, Problem 2002.1, Student Problems, The Mathematical Gazette, Vol. 86, No. 505 (2002), p. 152.
Shyam Sunder Gupta, Digital Invariants and Narcissistic Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 21, 513-526.
Eric Weisstein's World of Mathematics, Narcissistic Number.

Cf. A005188, A208130, A362843.

N:= 10: # to get solutions of up to N digits
Branch:= proc(level,sofar)
  option remember;
  local Res, x, x0, lb, ub, y;
  Res:= NULL;
  if perm[level] = 1 then x0:= 1 else x0:= 0 fi;
  for x from x0 to 9 do
    lb:= sofar + b[x,perm[level]] + scmin[level];
    ub:= sofar + b[x,perm[level]] + scmax[level];
    if lb <= 0 and ub >= 0 then
       if level = n then Res:= Res, [x]
       else
         for y in Branch(level+1,sofar+b[x,perm[level]]) do
            Res:= Res, [x, op(y)]
         od
        fi
     fi
   od;
   [Res]
end:
count:= 0:
for n from 1 to N do
  printf("Looking for %d digit solutions\n",n);
  forget(Branch);
  for j from 1 to n do
    for x from 0 to 9 do
      b[x,j]:= x^j - x*10^(n-j)
    od
  od:
  for j from 1 to n do
    smin[j]:= min(seq(b[x,j],x=0..9));
    smax[j]:= max(seq(b[x,j],x=0..9));
  od:
  perm:= sort([seq(smax[j]-smin[j],j=1..n)],`>`,output=permutation):
  for j from 1 to n do
    scmin[j]:= add(smin[perm[i]],i=j+1..n);
    scmax[j]:= add(smax[perm[i]],i=j+1..n);
  end;
  for X in Branch(1,0) do
    xx:= add(X[i]*10^(n-perm[i]),i=1..n);
    count:= count+1;
    A[count]:= xx;
    print(xx);
  od
od:
seq(A[i],i=1..count); # Robert Israel, Aug 07 2014

f[n_] := Plus @@ (IntegerDigits[n]^Range[ Floor[ Log[10, n] + 1]]); Select[ Range[10^7], f[ # ] == # &] (* Robert G. Wilson v, May 04 2005 *)
Join[{0},Select[Range[10^7],Total[IntegerDigits[#]^Range[ IntegerLength[ #]]] == #&]] (* Harvey P. Dale, Oct 13 2015 *)
sdcpQ[n_]:=n==Inner[Power,IntegerDigits[n],Range[IntegerLength[n]],Plus]; Join[{0},Select[Range[27*10^5],sdcpQ]] (* Harvey P. Dale, May 30 2020 *)

for(n=1,10^22,d=digits(n);s=sum(i=1,#d,d[i]^i);if(s==n,print1(n,", "))) \\ Derek Orr, Aug 07 2014

Corrected by Macsy Zhang (macsy(AT)21cn.com), Feb 17 2002

A377012 Numbers k whose digits can be split into substrings so that the sum of these substrings raised to consecutive powers (1, 2, 3, ...) is the number k itself.

Original entry on oeis.org

89, 135, 175, 518, 598, 1306, 1370, 1371, 1676, 2045, 2427, 3055, 5755, 5918, 9899, 11053, 24429, 88297, 135010, 234322, 255050, 255051, 360030, 360031, 494703, 512780, 517380, 568217, 767368, 779243, 785920, 785921, 788834, 819116, 931316, 986562, 998999, 1000100

Offset: 1

Francesco Di Matteo, Oct 28 2024

At least two substrings are required and each substring must have at least one digit.
A subsequence is A032799 (except for 1-digit numbers).
Unlike A032799, which is finite, this sequence is infinite because; e.g., the pattern 89, 9899, 998999, ... can always be split into two equal-length substrings that generate a term as (10^i - 2)^1 + (10^i - 1)^2 = 10^i*(10^i - 2) + (10^i - 1) for all i > 0.
Leading zeros in strings are allowed here. The first such term generated by the author's program is 1010053 = 1^1 + (01005)^2 + 3^3. - Michael S. Branicky, Nov 30 2024
Another pattern is deduced from a(38) = 1000100 = 100^1 + 0^2 + 100^3 with formula (10^i)^1 + 0^2 + ... + 0^n + ([0...0]10^i)^(n+1) = (10^i)^(n+1) + 10^i with i > 1 and n > 1. - Francesco Di Matteo, Jan 15 2025

			175 = 1^1 + 7^2 + 5^3 is a term.
5755 = 5^1 + 75^2 + 5^3 is a term.
88297 = 88^1 + 297^2 is a term.
234322 = 23^1 + 4^2 + 3^3 + 22^4 is a term.

Cf. A005188, A208130, A347189, A362843.

import itertools
analys = range(1, 7) # increase this if you want
for limite in analys:
    numbers = range(pow(10,limite-1),pow(10,limite))
    r = range(1, limite+1)
    disp_temp = []
    for s in r:
        disp = list(itertools.product(r, repeat=s+1))
        disp_temp.extend(disp)
    disp_ok = [d for d in disp_temp if sum(d)==limite]
    for numero in numbers:
        str_numero = str(numero)
        for combo in disp_ok:
            k = limite
            totale = 0
            for c in range(len(combo), 0, -1):
                partenza = k-combo[c-1]
                porzione = str_numero[partenza:k]
                if c == 1:
                    totale = totale + int(porzione)
                else:
                    totale = totale + pow(int(porzione),c)
                k = k - combo[c-1]
            if totale == numero:
                print(numero)

cp's OEIS Frontend

A032799 Numbers k such that k equals the sum of its digits raised to the consecutive powers (1,2,3,...).

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