A254677 Least positive integer m with A160325(m) = n.

2, 1, 5, 19, 15, 22, 37, 92, 71, 156, 136, 222, 206, 211, 257, 292, 506, 402, 577, 521, 632, 789, 682, 796, 742, 1006, 1046, 1192, 1346, 1482, 1312, 1507, 2021, 1522, 2172, 1977, 1962, 2007, 2161, 2479, 2502, 3047, 2761, 2326, 3097, 2876, 3316, 3216, 3421, 3386, 3902, 3652, 4406, 4356, 4587, 4492, 4342, 4917, 4811, 5472

Offset: 1

Author

Zhi-Wei Sun, Feb 05 2015

Keywords

• nonn

Comments

Conjecture: a(n) exists for any n > 0. Moreover, no term a(n) is congruent to 3 modulo 5.

Examples

			a(3) = 5 since 5 is the least positive integer that can be written as x(x+1)/2 + (2y)^2 + z(3z-1)/2 (with x,y,z nonnegative integers) in exactly 3 ways. In fact, 5 = 0*1/2 + 0^2 + 2*(3*2-1)/2 = 0*1/2 + 2^2 + 1*(3*1-1)/2 = 1*2/2 + 2^2 + 0*(3*0-1)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..200
• Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Crossrefs

Cf. A000217, A000290, A000326, A160325.

Programs

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[Do[m=0;Label[aa];m=m+1;r=0;Do[If[TQ[m-4y^2-z(3z-1)/2],r=r+1;If[r>n,Goto[aa]]],{y,0,Sqrt[m/4]},{z,0,(Sqrt[24(m-4y^2)+1]+1)/6}];
If[r==n,Print[n," ", m];Goto[bb],Goto[aa]]];Label[bb];Continue,{n,1,60}]