A165141 The least positive integer that can be written in exactly n ways as the sum of a square, a pentagonal number and a hexagonal number.

3, 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, 1277, 1381, 1857, 1556, 1507, 1657, 1832, 1732, 2336, 2299, 2007, 2677, 2326, 2117, 2591, 2502, 2516, 2592, 3106, 3557

Offset: 1

Author

Zhi-Wei Sun, Sep 05 2009

Keywords

• nice
• nonn

Comments

On Sep 04 2009, Zhi-Wei Sun conjectured that the sequence A160324 contains every positive integer, i.e., for each positive integer n there exists a positive integer s which can be written in exactly n ways as the sum of a square, a pentagonal number and a hexagonal number. Based on this conjecture we create the current sequence. It seems that 0.9 < a(n)/n^2 < 1.6 for n > 33. Zhi-Wei Sun conjectured that a(n)/n^2 has a limit c with 1.1 < c < 1.2. On Sun's request, his friend Qing-Hu Hou produced a list of a(n) for n = 1..913 (see the b-file).

Examples

			For n=5 the a(5)=16 solutions are 0^2+1+15 = 1^2+0+15 = 2^2+12+0 = 3^2+1+6 = 4^2+0+0 = 16.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..2658(terms 1..913 from Quig-Hu Hou)
• F. Ge and Z. W. Sun, On some universal sums of generalized polygonal numbers, preprint, arXiv:0906.2450 [math.NT], 2009-2016.
• M. B. Nathanson, A short proof of Cauchy's polygonal number theorem, Proc. Amer. Math. Soc. 99(1987), 22-24.
• Zhi-Wei Sun, A challenging conjecture on sums of polygonal number (a message to Number Theory List), 2009.
• Zhi-Wei Sun, Polygonal numbers, primes and ternary quadratic forms (a talk given at a number theory conference), 2009.
• Zhi-Wei Sun, Mixed Sums of Primes and Other Terms (a webpage).
• Zhi-Wei Sun, On universal sums of polygonal numbers, preprint, arXiv:0905.0635 [math.NT], 2009-2015.

Crossrefs

Cf. A160324, A000290, A000326, A000384, A160325, A160326.

Programs

Mathematica

SQ[x_] := x>-1 && IntegerPart[Sqrt[x]]^2==x;
RN[n_] := RN[n] = Sum[If[SQ[n - (3 y^2 - y)/2 - (2 z^2 - z)], 1, 0], {y, 0, (1 + Sqrt[1 + 24 n])/6}, {z, 0, (1 + Sqrt[8 (n - (3 y^2 - y)/2) + 1])/4}] (* iterators modified by Robert G. Wilson v, Apr 05 2025 *);
Do[Do[If[RN[m]==n,Print[n," ", m];Goto[aa]],{m,1,1000000}];Label[aa];Continue,{n,1,100}]
f = Compile[{{n, Integer}}, Block[{cnt = 0, pentind, hex, hexind = Floor[(1 + Sqrt[8n +1])/4]}, While[ hexind > -1, hex = hexind (2 hexind - 1); pentind = Floor[(1 + Sqrt[1 + 24 (n - hex)])/6]; While[pentind > -1, If[ Floor[ Sqrt[ n - hex - pentind (3 pentind - 1)/2]]^2 == n - hex - pentind (3 pentind - 1)/2, cnt++]; pentind--]; hexind--]; cnt]]; t[] := 0; k = 1; While[k < 100001, If[ t[f[k]] == 0, t[f[k]] = k]; k++]; t /@ Range@ 250 (* Robert G. Wilson v, Apr 04 2025 *)

Formula

a(n) = min{m>0: m=x^2+(3y^2-y)/2+(2z^2-z) has exactly n solutions with x,y,z=0,1,2,...}.