A257768 Numbers m such that for some power k, m is the sum of d + d^k as d runs through the digits of m.

12, 18, 30, 90, 666, 870, 960, 1998, 7816, 42648, 119394, 302034, 360522, 1741752, 12051036, 909341082, 931186956, 1136424308, 1145082306, 8390370196, 49388550660, 52927388760, 100552730520, 41845367362266, 51671446297908, 245917854035004, 607628544623816, 858683110606660, 4023730658941192

Offset: 1

Author

Pieter Post, May 07 2015

Keywords

• base
• nonn

Comments

The power k of most terms in this sequence is equal to or one more or one less than the number of digits in the term. One exception is 302034: 302034 = 3^9 + 0^9 + 2^9 + 0^9 + 3^9 + 4^9 + 3+0+2+0+3+4.

Examples

			666 = (6+6+6) + (6^3 + 6^3 + 6^3).
7816 = (7+8+1+6) + (7^4 + 8^4 + 1^4 + 6^4).
360522 = (3+6+0+5+2+2) + (3^7 + 6^7 + 0^7 + 5^7 + 2^7 + 2^7).
		

Crossrefs

Cf. A115518, A130680, A111434.

Programs

Maple

mmax:= 10:  # to get all terms < 10^mmax
Res:= NULL:
score:= (c,p) -> add(c[i+1]*(i+i^p), i=0..9):
for m from 2 to mmax do
comps:= convert(map(`-`,combinat:-composition(10+m,10),[1$10]),list):
for c in comps do
  cL:= [seq(i$c[i+1], i=0..9)];
  if max(c[3..-1]) = 0 then slim:= 0 else slim:= 10^m fi;
  for p from 1 do
    s:= score(c,p);
    L:= sort(convert(s,base,10));
    if L = cL then Res:= Res,s; break fi;
    if s >= slim then break fi;
  od:
od:
od:
sort([Res]); # Robert Israel, May 08 2015

Python

# WARNING: this prints numbers in the sequence, but not in increasing order.
def moda(n,a):
    kk = 0
    while n > 0:
        kk= kk+(n%10)**a
        n = n//10
    return kk
def sod(n):
    kk = 0
    while n > 0:
        kk += n % 10
        n = n//10
    return kk
for a in range (1, 10):
    for c in range (10, 10**6):
        if c == moda(c,a)+sod(c):
            print(c, end=",")

Extensions

a(14)-a(29) from Giovanni Resta, May 08 2015