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%I #14 Mar 11 2019 17:21:12
%S 1,2,2,3,4,3,3,3,3,6,5,4,7,5,2,5,5,6,6,3,4,5,1,2,5,6,5,5,6,3,3,2,2,6,
%T 5,3,9,7,3,6,4,5,8,3,6,7,2,4,7,6,8,9,9,6,6,5,3,9,9,6,11,7,4,10,5,6,13,
%U 5,5,9,2,5,7,7,7,10,7,3,5,3,3,6,6,5,11,6,6,9,4,6,11,3,5,9,2,5,4,4,7,11
%N Number of ways to write n as w^2 + T(x)^2 + Pen(y)^2 + 2*Pen(z)^2, where w,x,y,z are integers with w > 0 and x >= 0, and T(m) = m*(m+1)/2 and Pen(m) = m*(3m-1)/2.
%C Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as s^2 + t^2 + u^2 + 2*v^2, where s is a positive integer, t is a triangular number, u and v are generalized pentagonal numbers.
%C I have verified a(n) > 0 for all n = 1..5*10^6.
%C It is well known that any positive odd integer can be written as x^2 + y^2 + 2*z^2 with x,y,z integers.
%C See also A306614 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A306606/b306606.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.
%e a(1) = 1 with 1 = 1^2 + T(0)^2 + Pen(0)^2 + 2*Pen(0)^2.
%e a(23) = 1 with 23 = 4^2 + T(1)^2 + Pen(-1)^2 + 2*Pen(1)^2.
%e a(335) = 1 with 335 = 18^2 + T(2)^2 + Pen(0)^2 + 2*Pen(1)^2.
%e a(3695) = 1 with 3695 = 53^2 + T(7)^2 + Pen(-1)^2 + 2*Pen(-2)^2.
%t t[x_]:=t[x]=x(x+1)/2; p[x_]:=p[x]=x(3x-1)/2;
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t tab={};Do[r=0;Do[If[SQ[n-t[x]^2-p[y]^2-2*p[z]^2],r=r+1],{x,0,(Sqrt[8*Sqrt[n-1]+1]-1)/2},{y,-Floor[(Sqrt[24*Sqrt[n-1-t[x]^2]+1]-1)/6],(Sqrt[24*Sqrt[n-1-t[x]^2]+1]+1)/6},
%t {z,-Floor[(Sqrt[24*Sqrt[(n-1-t[x]^2-p[y]^2)/2]+1]-1)/6],(Sqrt[24*Sqrt[(n-1-t[x]^2-p[y]^2)/2]+1]+1)/6}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000217, A000290, A001318, A275344, A306614.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Feb 28 2019
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