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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A270616 Number of ordered ways to write n as the sum of a positive square, the square of a triangular number, and a generalized pentagonal number (A001318).

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%I A270616 #12 Jul 21 2023 06:53:10
%S A270616 1,2,2,2,2,3,2,1,3,4,4,3,2,3,3,4,6,4,3,3,2,3,3,3,6,4,5,4,1,4,4,5,2,1,
%T A270616 3,5,6,5,6,5,5,5,2,5,6,3,5,3,5,6,6,10,4,2,3,4,5,4,5,7,6,5,4,4,6,6,7,2,
%U A270616 3,3,6,6,5,6,5,6,5,3,4,8
%N A270616 Number of ordered ways to write n as the sum of a positive square, the square of a triangular number, and a generalized pentagonal number (A001318).
%C A270616 Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 29, 34, 5949, 10913.
%C A270616 See also A270566 and A270594 for more similar conjectures.
%C A270616 By the author's work in Sci. China Math. 58(2015), any natural number can be written as the sum of a triangular number, a square and a generalized pentagonal number.
%H A270616 Zhi-Wei Sun, <a href="/A270616/b270616.txt">Table of n, a(n) for n = 1..10000</a>
%H A270616 Zhi-Wei Sun, <a href="http://dx.doi.org/10.4064/aa127-2-1">Mixed sums of squares and triangular numbers</a>, Acta Arith. 127 (2007), 103-113.
%H A270616 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.
%H A270616 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1502.03056">On universal sums ax^2+by^2+f(z), aT_x+bT_y+f(z) and zT_x+by^2+f(z)</a>, preprint, arXiv:1502.03056 [math.NT], 2015.
%e A270616 a(1) = 1 since 1 = 1^2 + (0*1/2)^2 + 0*(3*0+1)/2.
%e A270616 a(8) = 1 since 8 = 1^2 + (0*1/2)^2 + 2*(3*2+1)/2.
%e A270616 a(29) = 1 since 29 = 4^2 + (1*2/2)^2 + 3*(3*3-1)/2.
%e A270616 a(34) = 1 since 34 = 5^2 + (2*3/2)^2 + 0*(3*0+1)/2.
%e A270616 a(5949) = 1 since 5949 = 47^2 + (10*11/2)^2 + 22*(3*22-1)/2.
%e A270616 a(10913) = 1 since 10913 = 23^2 + (2*3/2)^2 +83*(3*83+1)/2.
%t A270616 pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
%t A270616 Do[r=0;Do[If[pQ[n-x^2-(y(y+1)/2)^2],r=r+1],{x,1,Sqrt[n]},{y,0,(Sqrt[8*Sqrt[n-x^2]+1]-1)/2}];Print[n," ",r];Continue,{n,1,80}]
%Y A270616 Cf. A000217, A000290, A001318, A262813, A262815, A262816, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516, A270533, A270559, A270566, A270594.
%K A270616 nonn
%O A270616 1,2
%A A270616 _Zhi-Wei Sun_, Mar 20 2016