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A306592
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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).
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1
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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, 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, 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
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OFFSET
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0,4
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COMMENTS
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Conjecture: a(n) > 0 for any nonnegative integer n.
I'd like to call this the 1-2-3-4-5 conjecture. I have verified it for all n = 0..4*10^5.
It seems that a(n) = 1 only for n = 0,1,2.
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LINKS
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EXAMPLE
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Let f(x) = x*(3x-1)/2. Then
a(2) = 1 with 2 = f(0)^2 + 2*f(1)^2 + 3*f(0)^2 + 4*f(0)^2 + 5*f(0)^2,
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,
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.
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MATHEMATICA
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f[x_]:=f[x]=(x(3x-1)/2)^2;
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[n_]:=PQ[n]=SQ[n]&&SQ[24Sqrt[n]+1];
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]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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