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%I #10 Apr 11 2016 21:16:45
%S 1,2,1,2,1,1,2,1,2,1,3,3,3,3,3,3,4,2,6,4,2,1,1,8,4,5,2,2,7,10,9,2,3,4,
%T 5,6,6,5,2,7,11,11,4,1,5,8,13,8,6,5,3,8,8,12,7,3,8,18,16,12,2,7,10,15,
%U 11,10,4,4,11,15,22
%N Number of ordered ways to write n as pen(u) + pen(v) + pen(x) + pen(y) + pen(z) with u,v,x,y,z nonnegative integers such that u + 2*v + 4*x + 5*y + 6*z is a pentagonal number, where pen(k) denotes the pentagonal number k*(3k-1)/2.
%C Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 2, 4, 5, 7, 9, 21, 22, 43. Also, every n = 0,1,2,... can be written as pen(u) + pen(v) + pen(x) + pen(y) + pen(z) with u,v,x,y,z nonnegative integers such that 3*u + 5*v + 11*x + 16*y + 19*z is also a pentagonal number.
%C (ii) Any integer n > 43 can be written as the sum of five pentagonal numbers u, v, x, y and z such that u + 2*v + 5*x + 7*y + 10*z is also a pentagonal number. Also, each integer n > 10 can be written as the sum of five pentagonal numbers u, v, x, y and z such that u + 2*v + 5*x + 7*y + 10*z is a square.
%C (iii) Any natural number n can be written as u^2 + v^2 + x^2 + y^2 + z^2 with u^2 + 2*v^2 + 3*x^2 + 4*y^2 + 5*z^2 a square, where u, v, x, y and z are integers.
%C As conjectured by Fermat and proved by Cauchy, each natural number can be written as the sum of five pentagonal numbers.
%C See also A271510, A271513, A271518 and A271644 for some similar conjectures refining Lagrange's four-square theorem.
%H Zhi-Wei Sun, <a href="/A271608/b271608.txt">Table of n, a(n) for n = 0..1500</a>
%H Z.-W. Sun, <a href="http://math.scichina.com:8081/sciAe/EN/abstract/abstract517007.shtml">On universal sums of polygonal numbers</a>, Sci. China Math. 58(2015), no. 7, 1367-1396.
%e a(7) = 1 since 7 = 5 + 0 + 1 + 0 + 1 = pen(2) + pen(0) + pen(1) + pen(0) + pen(1) with 2 + 2*0 + 4*1 + 5*0 + 6*1 = 12 = pen(3).
%e a(9) = 1 since 9 = 1 + 1 + 5 + 1 + 1 = pen(1) + pen(1) + pen(2) + pen(1) + pen(1) with 1 + 2*1 + 4*2 + 5*1 + 6*1 = 22 = pen(4).
%e a(22) = 1 since 22 = 0 + 0 + 5 + 12 + 5 = pen(0) + pen(0) + pen(2) + pen(3) + pen(2) with 0 + 2*0 + 4*2 + 5*3 + 6*2 = 35 = pen(5).
%e a(43) = 1 since 43 = 5 + 1 + 35 + 1 + 1 = pen(2) + pen(1) + pen(5) + pen(1) + pen(1) with 2 + 2*1 + 4*5 + 5*1 + 6*1 = 35 = pen(5).
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
%t pen[x_]:=pen[x]=x*(3x-1)/2
%t pQ[n_]:=pQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0)
%t Do[r=0;Do[If[pQ[n-pen[x]-pen[y]-pen[z]-pen[w]]&&pQ[x+2y+4z+5w+6*Floor[(Sqrt[24(n-pen[x]-pen[y]-pen[z]-pen[w])+1]+1)/6]],r=r+1],{x,0,(Sqrt[24n+1]+1)/6},{y,0,(Sqrt[24(n-pen[x])+1]+1)/6},{z,0,(Sqrt[24(n-pen[x]-pen[y])+1]+1)/6},{w,0,(Sqrt[24(n-pen[x]-pen[y]-pen[z])+1]+1)/6}];Print[n," ",r];Label[aa];Continue,{n,0,70}]
%Y Cf. A000290, A000326, A271510, A271513, A271518, A271644.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Apr 10 2016
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