A067170 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.

2, 3, 5, 7, 250, 735, 2500, 25000, 250000, 1858560, 2500000, 18585600, 25000000, 91990080, 185856000, 242121642, 250000000, 919900800, 1081088775, 1390120992, 1768635648, 1858560000, 2500000000, 5435938431, 7245987840, 9199008000, 9475854336, 17996666688, 18585600000, 24214634829, 25000000000

Offset

1,1

Comments

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

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

Links

David A. Corneth, Table of n, a(n) for n = 1..11790 (first 260 terms from Giovanni Resta, terms < 10^34)

Example

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.

Mathematica

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[ # ] &]

Prog

(PARI) sd(n) = my(d=digits(n)); sum(k=1, #d, d[k]^3); \\ A055012
sp(n) = my(f=factor(n)); sum(k=1, #f~, f[k, 1]^3); \\ A005064
isok(n) = sp(n) == sd(n); \\ Michel Marcus, Sep 28 2019

Crossrefs

Cf. A005064, A006753, A055012, A067184.

Sequence in context: A257483 A178371 A067184 * A090719 A088297 A180015

Adjacent sequences: A067167 A067168 A067169 * A067171 A067172 A067173

Keyword

nonn,base

Author

Joseph L. Pe, Feb 18 2002

Extensions

a(10)-a(14) from Amiram Eldar, Sep 28 2019

a(15)-a(18) from Michel Marcus, Sep 28 2019

a(20)-a(29) from David A. Corneth, Sep 28 2019

Missing a(19) from Giovanni Resta, Sep 28 2019

Status

approved