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%I #20 Jul 21 2023 04:11:21
%S 1,2,2,3,2,2,3,2,3,3,2,2,3,3,3,3,2,2,4,5,2,5,4,5,7,3,1,1,4,4,6,4,1,4,
%T 4,3,5,6,5,6,4,1,1,2,5,4,5,3,3,2,1,5,4,7,9,5,4,2,2,2,5,3,2,5,2,1,3,4,
%U 3,8,4,4,5,6,3,3,3,2,7,6,1,3,3,4,7,4,6,6,7,5,2,3,3
%N Number of ordered pairs (x,y) with x >= 0 and y >= 0 such that n - x^4 - 2*y^2 is a triangular number or a pentagonal number.
%C Conjecture: a(n) > 0 for every n = 0,1,2,..., and a(n) = 1 only for the following 55 values of n: 0, 26, 27, 32, 41, 42, 50, 65, 80, 97, 112, 122, 130, 160, 196, 227, 239, 272, 322, 371, 612, 647, 736, 967, 995, 1007, 1106, 1127, 1205, 1237, 1240, 1262, 1637, 1657, 1757, 2912, 2987, 3062, 3107, 3524, 3647, 3902, 5387, 5587, 5657, 6047, 6107, 11462, 13427, 14717, 15002, 17132, 20462, 30082, 35750.
%C See also A262941, A262944, A262954 and A262955 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A262945/b262945.txt">Table of n, a(n) for n = 0..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(26) = 1 since 26 = 2^4 + 2*0^2 + 4*5/2.
%e a(32) = 1 since 32 = 0^4 + 2*4^2 + 0*1/2.
%e a(41) = 1 since 41 = 1^4 + 2*3^2 + p_5(4), where p_5(n) denotes the pentagonal number n*(3*n-1)/2.
%e a(196) = 1 since 196 = 1^4 + 2*5^2 + p_5(10).
%e a(3524) = 1 since 3524 = 0^4 + 2*22^2 + 71*72/2.
%e a(3647) = 1 since 3647 = 0^4 + 2*34^2 + p_5(30).
%e a(6047) = 1 since 6047 = 5^4 + 2*39^2 + p_5(40).
%e a(6107) = 1 since 6107 = 0^4 + 2*1^2 + 110*111/2.
%e a(11462) = 1 since 11462 = 9^4 + 2*5^2 + 98*99/2.
%e a(13427) = 1 since 13427 = 7^4 + 2*0^2 + 148*149/2.
%e a(14717) = 1 since 14717 = 8^4 + 2*72^2 + 22*23/2.
%e a(15002) = 1 since 15002 = 0^4 + 2*86^2 + 20*21/2.
%e a(17132) = 1 since 17132 = 3^4 + 2*30^2 + p_5(101).
%e a(20462) = 1 since 20462 = 0^4 + 2*26^2 + 195*196/2.
%e a(30082) = 1 since 30082 = 11^4 + 2*63^2 + 122*123/2.
%e a(35750) = 1 since 35750 = 0^4 + 2*44^2 + 252*253/2.
%t SQ[n_]:=IntegerQ[Sqrt[8n+1]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1, 6]==0)
%t Do[r=0;Do[If[SQ[n-x^4-2y^2],r=r+1],{x, 0, n^(1/4)},{y,0,Sqrt[(n-x^4)/2]}];Print[n, " ", r];Continue,{n,0,100}]
%Y Cf. A000217, A000290, A000326, A000583, A262941, A262944, A262954, A262955.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Oct 05 2015
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