cp's OEIS Frontend

A254661 Number of ways to write n as the sum of a triangular number, an even square and a second pentagonal number.

%I A254661 #10 Jul 02 2015 04:02:46
%S A254661 1,1,1,2,1,2,2,3,2,1,3,1,3,1,2,2,3,4,2,4,1,5,3,2,2,3,4,2,3,3,3,3,4,3,
%T A254661 3,1,5,3,3,4,4,4,3,5,5,4,5,5,2,2,2,6,5,2,4,3,2,6,3,6,2,5,5,4,5,3,7,5,
%U A254661 4,1,4,6,8,3,5,1,6,6,5,6,4,6,6,4,4,7,3,5,2,5,2,5,5,7,6,2,7,6,4,4,5
%N A254661 Number of ways to write n as the sum of a triangular number, an even square and a second pentagonal number.
%C A254661 Conjecture: (i) a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 2, 4, 9, 11, 13, 20, 35, 69, 75, 188.
%C A254661 (ii) For each a = 2,3, any nonnegative integer n can be written as x(x+1)/2 + a*y^2 + z*(3*z+1)/2 with x,y,z nonnegative integers.
%C A254661 Compare part (i) of this conjecture with the conjecture in A160325.
%H A254661 Zhi-Wei Sun, <a href="/A254661/b254661.txt">Table of n, a(n) for n = 0..10000</a>
%H A254661 Zhi-Wei Sun, <a href="http://arxiv.org/abs/0905.0635">On universal sums of polygonal numbers</a>, arXiv:0905.0635 [math.NT], 2009-2015.
%e A254661 a(20) = 1 since 20 = 1*2/2 + 2^2 + 3*(3*3+1)/2.
%e A254661 a(35) = 1 since 35 = 7*8/2 + 0^2 + 2*(3*2+1)/2.
%e A254661 a(69) = 1 since 69 = 2*3/2 + 8^2 + 1*(3*1+1)/2.
%e A254661 a(75) = 1 since 75 = 9*10/2 + 2^2 + 4*(3*4+1)/2.
%e A254661 a(188) = 1 since 188 = 1*2/2 + 0^2 + 11*(3*11+1)/2.
%t A254661 TQ[n_]:=IntegerQ[Sqrt[8n+1]]
%t A254661 Do[r=0;Do[If[TQ[n-4y^2-z(3z+1)/2],r=r+1],{y,0,Sqrt[n/4]},{z,0,(Sqrt[24(n-4y^2)+1]-1)/6}];
%t A254661 Print[n," ",r];Label[aa];Continue,{n,0,100}]
%Y A254661 Cf. A000217, A000290, A016742, A005449, A160325.
%K A254661 nonn
%O A254661 0,4
%A A254661 _Zhi-Wei Sun_, Feb 04 2015