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%I #39 Sep 28 2019 22:04:13
%S 2,3,5,7,250,735,2500,25000,250000,1858560,2500000,18585600,25000000,
%T 91990080,185856000,242121642,250000000,919900800,1081088775,
%U 1390120992,1768635648,1858560000,2500000000,5435938431,7245987840,9199008000,9475854336,17996666688,18585600000,24214634829,25000000000
%N Numbers n such that sum of the cubes of the distinct prime factors of n equals the sum of the cubes of the digits of n.
%C If 10*m is a term (e.g. m = 25, 185856, 9199008), then 10^k * m is a term for all k >= 1. Therefore this sequence is infinite. - _Amiram Eldar_, Sep 28 2019
%C The sum of cubes of digits of a k-digit number is at most 729*k. Therefore any term with at most k digits is p-smooth where p is the largest prime < (729*k)^(1/3). - _David A. Corneth_, Sep 28 2019
%H David A. Corneth, <a href="/A067170/b067170.txt">Table of n, a(n) for n = 1..11790</a> (first 260 terms from Giovanni Resta, terms < 10^34)
%e The prime factors of 735 are 3,5,7, the sum of whose cubes = 495 = sum of the cubes of the digits of 735; so 735 is a term of the sequence.
%t f[n_] := Module[{a, l, t, r}, a = FactorInteger[n]; l = Length[a]; t = Table[a[[i]][[1]], {i, 1, l}]; r = Sum[(t[[i]])^3, {i, 1, l}]]; g[n_] := Module[{b, m, s}, b = IntegerDigits[n]; m = Length[b]; s = Sum[(b[[i]])^3, {i, 1, m}]]; Select[Range[2, 10^6], f[ # ] == g[ # ] &]
%o (PARI) sd(n) = my(d=digits(n)); sum(k=1, #d, d[k]^3); \\ A055012
%o sp(n) = my(f=factor(n)); sum(k=1, #f~, f[k,1]^3); \\ A005064
%o isok(n) = sp(n) == sd(n); \\ _Michel Marcus_, Sep 28 2019
%Y Cf. A005064, A006753, A055012, A067184.
%K nonn,base
%O 1,1
%A _Joseph L. Pe_, Feb 18 2002
%E a(10)-a(14) from _Amiram Eldar_, Sep 28 2019
%E a(15)-a(18) from _Michel Marcus_, Sep 28 2019
%E a(20)-a(29) from _David A. Corneth_, Sep 28 2019
%E Missing a(19) from _Giovanni Resta_, Sep 28 2019
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