|
%I #11 May 02 2018 09:27:51
%S 1,5,6,35,42,50,57,60,64,72,81,85,102,121,124,182,188,201,232,260,261,
%T 267,308,312,351,440,452,495,519,528,594,645,649,688,735,741,774,784,
%U 805,854,861,875,882,901,966,1025,1027,1045,1050,1081,1105,1112,1119
%N Numbers whose sum of prime factors (counted with multiplicity) is a pentagonal number (A000326).
%C This is to pentagonal numbers A000326 as A000290 squares are to A212831 numbers whose sum of prime factors is a square (counted with multiplicity) and as A000217 triangular numbers are to A212849 Numbers whose sum of prime factors (counted with multiplicity) is a triangular number.
%H G. C. Greubel, <a href="/A212918/b212918.txt">Table of n, a(n) for n = 1..5000</a>
%F {k such that A001414(k) = sopfr(k) is in A000326(j) = j*(3*j-1)/2 for some integer j}.
%e a(3) = 35 because sopfr(35) = sum of prime factors of 35 = 5 + 7 = 12, and 12 is the 3rd pentagonal number.
%t pentagonalQ[n_] := IntegerQ[(1 + Sqrt[1 + 24*n])/6]; pfs[n_] := Module[{p, e}, {p, e} = Transpose[FactorInteger[n]]; Dot[p, e]]; Select[Range[1500], pentagonalQ[pfs[#]] &] (* _T. D. Noe_, May 31 2012 *)
%o (PARI) sopfr(n) = my(f=factor(n)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]); \\ A001414
%o isok(n) = ispolygonal(sopfr(n), 5); \\ _Michel Marcus_, May 02 2018
%Y Cf. A000326, A001414, A212831, A212849.
%K nonn,easy
%O 1,2
%A _Jonathan Vos Post_, May 31 2012
%E Corrected by _T. D. Noe_, May 31 2012
|
