%I #17 Dec 02 2015 11:57:38
%S 1,2,3,4,5,6,7,8,9,37,48,415,231591,3829377463694454,
%T 56407086228259246207394322684
%N Numbers n such that the sum of the digits of n to some power divided by the sum of the digits equal n.
%C The first nine terms are trivial, but then the terms become very rare. It appears that this sequence is finite.
%e 37 = (3^3+7^3)/(3+7).
%e 231591 = (2^7+3^7+1^7+5^7+9^7+1^7)/(2+3+1+5+9+1).
%o (Python)
%o def moda(n,a):
%o ....kk = 0
%o ....while n > 0:
%o ........kk= kk+(n%10)**a
%o ........n =int(n//10)
%o ....return kk
%o def sod(n):
%o ....kk = 0
%o ....while n > 0:
%o ........kk= kk+(n%10)
%o ........n =int(n//10)
%o ....return kk
%o for a in range (1, 10):
%o ....for c in range (1, 10**6):
%o ........if c*sod(c)==moda(c, a):
%o ............print (a,c, moda(c,a),sod(c))
%Y Cf. A061209, A115518, A111434, A114135, A130680, A257784, A257768.
%K nonn,base,more
%O 1,2
%A _Pieter Post_, May 08 2015
%E a(14) from _Giovanni Resta_, May 09 2015
%E a(15) from _Chai Wah Wu_, Nov 30 2015