%I
%S 1,6,11,104,116,129,218,363,408,440,481,534,566,568,590,638,646,684,
%T 718,807,895,979,999,1003,1007,1137,1251,1282,1557,1935,2197,2367,
%U 2571,2582,2808,2855,3132,3283,3336,3578,3737,3891,3946,3980,4172,4484,4886,5158
%N Numbers n such that the sum of divisors of n is a pentagonal number.
%C This is to A006532 (numbers n such that sum of divisors is a square) as A000326 Pentagonal numbers is to A000290 squares, and as A045746 Numbers n such that the sum of divisors of n is a triangular number, is to A000217 triangular numbers. 6 and 11 are the first two numbers whose sum of divisors is the same pentagonal number. 104 and 116 are the second two numbers whose sum of divisors is the same pentagonal number. 363 and 481 are the third two numbers whose sum of divisors is the same pentagonal number. 408, 440, 534, 568, 590, 638, 646, 718, 807, 895, 979, 1003, and 1007 are the first thirteen numbers whose sum of divisors is the same pentagonal number.
%C The subsequence of values which are themselves pentagonal numbers begins a(1) = 1, a(15) = 590, 395403426.
%H Harvey P. Dale, <a href="/A180929/b180929.txt">Table of n, a(n) for n = 1..1000</a>
%F A000203(a(n)) is in A000326. sigma(a(n) = k*(3*k1)/2 for some nonnegative integer k. Sum of divisors of a(n) is a pentagonal number. sigma_1(a(n)) is a pentagonal number.
%e a(1) = 1 because the sum of divisors of 1 is the pentagonal number 1.
%e a(2) = 6 because the sum of divisors of 6 is the pentagonal number 12.
%e a(3) = 11 because the sum of divisors of 11 is the pentagonal number 12.
%e a(4) = 104 because the sum of divisors of 104 is the pentagonal number 210.
%e a(5) = 116 because the sum of divisors of 116 is the pentagonal number 210.
%e a(6) = 129 because the sum of divisors of 129 is the pentagonal number 176.
%e a(7) = 218 because the sum of divisors of 218 is the pentagonal number 330.
%p Contribution from _R. J. Mathar_, Sep 26 2010: (Start)
%p isA000326 := proc(n) if not issqr(24*n+1) then false; else sqrt(24*n+1)+1 ; (% mod 6) = 0 ; end if; end proc:
%p for n from 1 to 5000 do if isA000326(numtheory[sigma](n)) then printf("%d,",n) ; end if; end do: (End)
%t pnos=Table[(n (3n1))/2,{n,500}]; okQ[n_]:=Module[{ds=DivisorSigma[1,n]},MemberQ[pnos,ds]] Select[Range[5000],okQ] [From _Harvey P. Dale_, Sep 26 2010]
%t Select[Range[5200],IntegerQ[(Sqrt[1+24DivisorSigma[1,#]]+1)/6]&] (* _Harvey P. Dale_, Jun 14 2013 *)
%Y Cf. A000203, A000217, A000290, A000326, A006532, A045746.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Sep 25 2010
%E Corrected and extended by _R. J. Mathar_ and _Harvey P. Dale_, Sep 26 2010
