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Number of ways to express n as the sum of a square, a pentagonal number and a hexagonal number.
12

%I #21 Sep 10 2018 13:51:40

%S 1,3,3,1,1,3,4,3,1,2,4,3,2,2,2,4,5,4,2,2,3,3,5,3,3,2,3,5,4,5,2,5,5,2,

%T 2,1,6,8,5,2,3,5,4,3,4,5,3,3,2,5,7,7,5,4,7,4,4,3,4,4,3,6,3,2,5,5,9,7,

%U 3,3,6,9,5,3,1,8,7,6,2,5,6,3,10,4,3,3,8,7,5,4,1,4,10,7,5,4,8,6,2,8,6,10,7,5

%N Number of ways to express n as the sum of a square, a pentagonal number and a hexagonal number.

%C In April 2009, _Zhi-Wei Sun_ conjectured that a(n)>0 for every n=0,1,2,3,.... Note that pentagonal numbers and hexagonal numbers are more sparse than squares and that there are infinitely many positive integers which cannot be written as the sum of three squares.

%C On Aug 12 2009, _Zhi-Wei Sun_ made the following general conjecture on diagonal representations by polygonal numbers: For each integer m>2, any natural number n can be written in the form p_{m+1}(x_1)+...+p_{2m}(x_m) with x_1,...,x_m nonnegative integers, where p_k(x)=(k-2)x(x-1)/2+x (x=0,1,2,...) are k-gonal numbers. Sun has verified this with m=3 for n up to 10^6, and with m=4,5,6,7,8,9,10 for n up to 5*10^5. - _Zhi-Wei Sun_, Aug 15 2009

%C On Aug 21 2009, _Zhi-Wei Sun_ formulated the following strong version for his conjecture on diagonal representations by polygonal numbers: For any integer m>2, each natural number n can be expressed as p_{m+1}(x_1)+p_{m+2}(x_2)+p_{m+3}(x_3)+r with x_1,x_2,x_3 nonnegative integers and r an integer among 0,...,m-3. For m=3 and m=4,5,6,7,8,9,10, Sun has verified this conjecture for n up to 10^6 and 5*10^5 respectively. Sun also guessed that for each m=3,4,... all sufficiently large integers have the form p_{m+1}(x_1)+p_{m+2}(x_2)+p_{m+3}(x_3) with x_1,x_2,x_3 nonnegative integers. For example, it seems that 387904 is the largest integer not in the form p_{20}(x_1)+p_{21}(x_2)+p_{22}(x_3). - _Zhi-Wei Sun_, Aug 21 2009

%C On Sep 04 2009, _Zhi-Wei Sun_ conjectured that the sequence contains every positive integer. For n=1,2,3,... let s(n) denote the least nonnegative integer m such that a(m)=n. Here is the list of s(1),...,s(30): 0, 9, 1, 6, 16, 36, 50, 37, 66, 82, 167, 121, 162, 236, 226, 276, 302, 446, 478, 532, 457, 586, 677, 521, 666, 852, 976, 877, 1006, 1046. - _Zhi-Wei Sun_, Sep 04 2009

%H Zhi-Wei Sun, <a href="/A160324/b160324.txt">Table of n, a(n) for n = 0..50000</a>

%H M. B. Nathanson, <a href="http://dx.doi.org/10.1090/S0002-9939-1987-0866422-3">A short proof of Cauchy's polygonal number theorem</a>, Proc. Amer. Math. Soc. 99(1987), 22-24.

%H G. Pall, <a href="http://www.jstor.org/stable/2371077">Large positive integers are sums of four or five values of a quadratic function</a>, Amer. J. Math. 54(1932), 66-78.

%H Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;48c9be36.0905">Various new conjectures involving polygonal numbers and primes</a> (a message to Number Theory List), May 2009.

%H Zhi-Wei Sun, <a href="http://math.nju.edu.cn/~zwsun/MSPT.htm">Mixed Sums of Primes and Other Terms (a webpage)</a>.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/0905.0635">On universal sums of polygonal numbers, preprint</a>, arXiv:0905.0635 [math.NT], 2009.

%F a(n)=|{<x,y,z>: x,y,z=0,1,2,... & x^2+(3y^2-y)/2+(2z^2-z)=n}|

%e For n=10 the a(10)=4 solutions are 4+0+6, 4+5+1, 9+0+1, 9+1+0.

%t SQ[x_]:=x>-1&&IntegerPart[Sqrt[x]]^2==x RN[n_]:=Sum[If[SQ[n-(3y^2-y)/2-(2z^2-z)],1,0], {y,0,Sqrt[n]},{z,0,Sqrt[Max[0,n-(3y^2-y)/2]]}] Do[Print[n," ", RN[n]],{n,0,50000}]

%t planeFiguratePi[n_, r_] := Floor[((r -4) + Sqrt[(r -4)^2 + 8n (r -2)])/(2 (r -2))]; z[r_] := PolygonalNumber[r, Range[0, planeFiguratePi[mx, r]]]; mx = 105; Join[{1}, Take[Transpose[ Tally[ Sort[ Plus @@@ FlattenAt[ Tuples[{z[4], z[5], z[6]}], 2]]]][[2]], {2, mx}]] (* _Robert G. Wilson v_, May 22 2017 *)

%Y Cf. A000290, A000326, A000384.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, May 08 2009