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A111434 Numbers k such that the sums of the digits of k, k^2 and k^3 coincide.

%I A111434 #30 Jul 24 2021 02:29:42
%S A111434 0,1,10,100,468,585,1000,4680,5850,5851,5868,10000,28845,46800,58500,
%T A111434 58510,58680,58968,100000,288450,468000,585000,585100,586800,589680,
%U A111434 1000000,2884500,4680000,5850000,5851000,5868000,5896800,10000000
%N A111434 Numbers k such that the sums of the digits of k, k^2 and k^3 coincide.
%C A111434 The sequence is clearly infinite, since we can add trailing zeros. Is the subset of values not ending in 0 infinite too (see A114135)?
%H A111434 David A. Corneth, <a href="/A111434/b111434.txt">Table of n, a(n) for n = 1..1124</a> (using the b-file in A114135).
%e A111434 468 is in the sequence since 468^2 = 219024 and 468^3 = 102503232 and we have 18 = 4+6+8 = 2+1+9+0+2+4 = 1+0+2+5+0+3+2+3+2.
%e A111434 5851 is in the sequence because 5851, 34234201 (= 5851^2) and 200304310051 (=5851^3) all have digital sum 19.
%p A111434 s:=proc(n) local nn: nn:=convert(n,base,10): sum(nn[j],j=1..nops(nn)): end: a:=proc(n) if s(n)=s(n^2) and s(n)=s(n^3) then n else fi end: seq(a(n),n=0..1000000); # _Emeric Deutsch_, May 13 2006
%t A111434 SumOfDig[n_]:=Apply[Plus, IntegerDigits[n]]; Do[s=SumOfDig[n]; If[s==SumOfDig[n^2] && s==SumOfDig[n^3], Print[n]], {n, 10^6}]
%t A111434 Select[Range[0,10000000],Length[Union[Total/@IntegerDigits[{#,#^2,#^3}]]] == 1&] (* _Harvey P. Dale_, Apr 26 2014 *)
%Y A111434 Cf. A011557, A058369, A070276.
%K A111434 base,nonn
%O A111434 1,3
%A A111434 _Giovanni Resta_, Nov 21 2005
%E A111434 b-file Corrected by _David A. Corneth_, Jul 22 2021