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A275824 Number of ways to write n as pen(x) + pen(y) + pen (z) + 2*pen(w), where x,y,z,w are nonnegative integers with x <= y <= z, and pen(k) denotes the pentagonal number k*(3*k-1)/2. 1

%I #11 Dec 06 2018 12:03:50

%S 1,1,2,2,1,2,1,2,1,1,2,2,3,3,2,3,2,3,1,1,2,1,3,2,6,4,3,4,2,3,2,2,3,1,

%T 5,4,4,5,4,4,1,3,3,1,3,5,7,6,4,5,3,5,3,3,4,4,5,4,6,7,3,5,4,4,3,2,6,3,

%U 6,5,5,7,7,7,3,6,7,5,4,4,9

%N Number of ways to write n as pen(x) + pen(y) + pen (z) + 2*pen(w), where x,y,z,w are nonnegative integers with x <= y <= z, and pen(k) denotes the pentagonal number k*(3*k-1)/2.

%C Conjecture: (i) a(n) > 0 for all n = 0,1,2,...., and a(n) = 1 only for n = 0, 1, 4, 6, 8, 9, 18, 19, 21, 33, 40, 43.

%C (ii) For every m = 5,6,7,... and each integer d > 1 not divisible by 4, any sufficiently large integer can be written as x + y + z + d*w (or x + y + 2*z + d*w) with x,y,z,w m-gonal numbers.

%C Part (i) of the conjecture is stronger than the classical result that any natural number is the sum of five pentagonal numbers.

%H Zhi-Wei Sun, <a href="/A275824/b275824.txt">Table of n, a(n) for n = 0..10000</a>

%H Xiang-Zi Meng and Zhi-Wei Sun, <a href="http://arxiv.org/abs/1608.02022">Sums of four polygonal numbers with coefficients</a>, arXiv:1608.02022 [math.NT], 2016.

%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/170o.pdf">A result similar to Lagrange's theorem</a>, J. Number Theory 162(2016), 190-211.

%e a(4) = 1 since 4 = 0*(3*0-1)/2 + 1*(3*1-1)/2 + 1*(3*1-1)/2 + 1*(3*1-1).

%e a(18) = 1 since 18 = 1*(3*1-1)/2 + 2*(3*2-1)/2 + 3*(3*3-1)/2 + 0*(3*0-1).

%e a(19) = 1 since 19 = 0*(3*0-1)/2 + 2*(3*2-1)/2 + 3*(3*3-1)/2 + 1*(3*1-1).

%e a(21) = 1 since 21 = 1*(3*1-1)/2 + 2*(3*2-1)/2 + 2*(3*2-1)/2 + 2*(3*2-1).

%e a(33) = 1 since 33 = 0*(3*0-1)/2 + 1*(3*1-1)/2 + 4*(3*4-1)/2 + 2*(3*2-1).

%e a(40) = 1 since 40 = 0*(3*0-1)/2 + 2*(3*2-1)/2 + 5*(3*5-1)/2 + 0*(3*0-1).

%e a(43) = 1 since n = 1*(3*1-1)/2 + 2*(3*2-1)/2 + 5*(3*5-1)/2 + 1*(3*1-1).

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]

%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-x*(3x-1)-y(3y-1)/2-z(3z-1)/2],r=r+1],{x,0,(Sqrt[12n+1]+1)/6},{y,0,(Sqrt[8(n-x*(3x-1))+1]+1)/6},{z,y,(Sqrt[12(n-x*(3x-1)-y(3y-1)/2)+1]+1)/6}];Print[n," ",r];Continue,{n,0,80}]

%Y Cf. A000326, A256106.

%K nonn

%O 0,3

%A _Zhi-Wei Sun_, Aug 10 2016

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