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%I #10 Jul 21 2023 04:28:16
%S 1,2,3,3,2,3,3,3,2,1,4,4,3,2,3,5,4,3,3,3,4,5,5,4,3,5,6,5,5,3,6,4,4,4,
%T 1,4,5,7,6,2,6,3,3,3,5,8,5,4,3,5,4,4,4,5,5,5,7,4,3,3,7,3,3,2,2,8,5,6,
%U 2,3,5,7,6,2,1,4,4,3,6,7,6,3,5,4,3,2,6,6,6,4,6,8
%N Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a pentagonal number (A000326) or twice a pentagonal number.
%C Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 10, 35, 75, 134, 415, 515, 1465, 2365, 3515, 4140.
%H Zhi-Wei Sun, <a href="/A262955/b262955.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/s11425-015-4994-4">On universal sums of polygonal numbers</a>, Sci. China Math. 58(2015), no. 7, 1367-1396.
%e a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
%e a(10) = 1 since 10 = 0^4 + 4*5/2 + p_5(0).
%e a(35) = 1 since 35 = 1^4 + 4*5/2 + 2*p_5(3).
%e a(75) = 1 since 75 = 2^4 + 5*6/2 + 2*p_5(4).
%e a(134) = 1 since 134 = 2^4 + 1*2/2 + p_5(9).
%e a(415) = 1 since 415 = 0^4 + 21*22/2 + 2*p_5(8).
%e a(515) = 1 since 515 = 0^4 + 6*7/2 + 2*p_5(13).
%e a(1465) = 1 since 1465 = 5^4 + 35*36/2 + p_5(12).
%e a(2365) = 1 since 2365 = 5^4 + 8*9/2 + 2*p_5(24).
%e a(3515) = 1 since 3515 = 5^4 + 51*52/2 + 2*p_5(23).
%e a(4140) = 1 since 4140 = 1^4 + 90*91/2 + 2*p_5(4).
%t PenQ[n_]:=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0)
%t PQ[n_]:=PenQ[n]||PenQ[n/2]
%t Do[r=0;Do[If[PQ[n-x^4-y(y+1)/2],r=r+1],{x,0,n^(1/4)},{y,1,(Sqrt[8(n-x^4)+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]
%Y Cf. A000217, A000326, A000583, A262941, A262944, A262945, A262954.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Oct 05 2015
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