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%I #11 Jul 21 2023 04:32:32
%S 1,2,2,2,2,3,4,3,1,3,5,3,2,2,5,5,3,3,5,5,3,6,6,3,3,8,6,5,5,3,7,5,5,3,
%T 4,4,8,9,3,5,7,6,3,5,5,7,5,3,4,5,6,6,9,4,5,7,7,5,4,4,7,6,1,5,5,7,7,7,
%U 1,6,10,8,6,3,4,3,6,4,6,9,5,7,9,3,5,8,9,8,3,3,11,10,6,6,8,12,5,6,4,7
%N Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a pentagonal number.
%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 63, 69, 489, 714, 1089.
%C (ii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is twice a square or twice a pentagonal number.
%C (iii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - 2*x^4 - y*(y+1)/2 is a square or a pentagonal number.
%C See also A262941 and A262945 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A262944/b262944.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(9) = 1 since 9 = 1^4 + 2*3/2 + p_5(2).
%e a(63) = 1 since 63 = 0^4 + 7*8/2 + p_5(5).
%e a(69) = 1 since 69 = 2^4 + 7*8/2 + 5^2.
%e a(489) = 1 since 489 = 3^4 + 12*13/2 + p_5(15).
%e a(714) = 1 since 714 = 4^4 + 18*19/2 + p_5(14).
%e a(1089) = 1 since 1089 = 4^4 + 38*39/2 + p_5(8).
%t SQ[n_]:=IntegerQ[Sqrt[n]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1,6]==0)
%t Do[r=0;Do[If[SQ[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, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Oct 05 2015
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