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A306592 Number of ways to write n as u^2 + 2*v^2 + 3*x^2 + 4*y^2 + 5*z^2, where u,v,x,y,z are generalized pentagonal numbers (A001318).

%I A306592 #8 Feb 26 2019 09:41:10
%S A306592 1,1,1,2,3,3,4,4,5,6,4,5,7,6,4,5,7,6,6,5,8,8,6,6,10,10,5,8,9,8,8,5,8,
%T A306592 9,5,5,10,7,4,7,6,6,6,3,5,7,3,2,7,6,4,6,4,4,9,4,5,8,6,5,6,7,4,7,2,4,8,
%U A306592 5,3,6,7,5,6,7,6,9,3,5,9,5,4,8,6,7,7,5,5,10,4,3,6,4,4,5,2,3,6,4,5,10,5
%N A306592 Number of ways to write n as u^2 + 2*v^2 + 3*x^2 + 4*y^2 + 5*z^2, where u,v,x,y,z are generalized pentagonal numbers (A001318).
%C A306592 Conjecture: a(n) > 0 for any nonnegative integer n.
%C A306592 I'd like to call this the 1-2-3-4-5 conjecture. I have verified it for all n = 0..4*10^5.
%C A306592 It seems that a(n) = 1 only for n = 0,1,2.
%H A306592 Zhi-Wei Sun, <a href="/A306592/b306592.txt">Table of n, a(n) for n = 0..10000</a>
%e A306592 Let f(x) = x*(3x-1)/2. Then
%e A306592 a(2) = 1 with 2 = f(0)^2 + 2*f(1)^2 + 3*f(0)^2 + 4*f(0)^2 + 5*f(0)^2,
%e A306592 a(415) = 2 with 415 = f(-1)^2 + 2*f(3)^2 + 3*f(1)^2 + 4*f(2)^2 + 5*f(-1)^2 = f(3)^2 + 2*f(0)^2 + 3*f(2)^2 + 4*f(-2)^2 + 5*f(0)^2,
%e A306592 a(427) = 2 with 427 = f(-3)^2 + 2*f(2)^2 + 3*f(-2)^2 + 4*f(0)^2 + 5*f(1)^2 = f(-3)^2 + 2*f(1)^2 + 3*f(2)^2 + 4*f(0)^2 + 5*f(2)^2.
%t A306592 f[x_]:=f[x]=(x(3x-1)/2)^2;
%t A306592 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t A306592 PQ[n_]:=PQ[n]=SQ[n]&&SQ[24Sqrt[n]+1];
%t A306592 tab={};Do[r=0;Do[If[PQ[n-5f[x]-4f[y]-3f[z]-2f[w]],r=r+1],{x,-Floor[(Sqrt[24Sqrt[n/5]+1]-1)/6],(Sqrt[24Sqrt[n/5]+1]+1)/6},{y,-Floor[(Sqrt[24Sqrt[(n-5f[x])/4]+1]-1)/6],(Sqrt[24Sqrt[(n-5f[x])/4]+1]+1)/6},{z,-Floor[(Sqrt[24Sqrt[(n-5f[x]-4f[y])/3]+1]-1)/6],(Sqrt[24Sqrt[(n-5f[x]-4f[y])/3]+1]+1)/6},{w,-Floor[(Sqrt[24Sqrt[(n-5f[x]-4f[y]-3f[z])/2]+1]-1)/6],(Sqrt[24Sqrt[(n-5f[x]-4f[y]-3f[z])/2]+1]+1)/6}];tab=Append[tab,r],{n,0,100}];Print[tab]
%Y A306592 Cf. A000290, A001318.
%K A306592 nonn
%O A306592 0,4
%A A306592 _Zhi-Wei Sun_, Feb 26 2019