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Numbers incremented by their digit product produce a cube.
1

%I #29 Oct 16 2024 14:52:42

%S 0,4,85,168,184,212,368,549,681,919,999,1000,1283,1593,2181,3664,4096,

%T 4717,6811,7497,8000,9919,10648,12143,15275,15425,21664,21728,21824,

%U 27000,39304,42427,42811,47629,50653,63424,64000,74088,79507,84416,103823,110592

%N Numbers incremented by their digit product produce a cube.

%e 681 + 6*8*1 = 729 = 9^3.

%t Select[Range[0,100000], IntegerQ[(# + Times @@ IntegerDigits[#])^(1/3)] &] (* _T. D. Noe_, Sep 16 2013 *)

%o (Python)

%o def DP(n):

%o p = 1

%o for i in str(n):

%o p *= int(i)

%o return p

%o for n in range(10**4):

%o k = 0

%o P = n + DP(n)

%o while P >= k**3:

%o if P == k**3:

%o print(n, end=', ')

%o break

%o k += 1

%o # Simplified by _Derek Orr_, Mar 12 2015

%o (PARI) for(n=0, 10^5, d=digits(n); P=n+prod(i=1, #d, d[i]); if(ispower(P, 3), print1(n, ", "))) \\ _Derek Orr_, Mar 12 2015

%Y Cf. A007954, A066567, A229184.

%K nonn,easy,base,changed

%O 1,2

%A _Derek Orr_, Sep 15 2013

%E Prepended a(1) = 0 from _Derek Orr_, Mar 12 2015