A254595 Least positive integer m such that m can be written as x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 in exactly n ways, where x,y,z are nonnegative integers.

1, 2, 9, 7, 14, 37, 64, 68, 57, 119, 112, 168, 194, 147, 267, 259, 222, 477, 427, 404, 519, 652, 567, 497, 512, 749, 722, 719, 952, 1209, 904, 1139, 1267, 1184, 1069, 1737, 1594, 1667, 1734, 2077, 1799, 1659, 1729, 1814, 1762, 1862, 2577, 2444, 2997, 2072, 2457, 2842, 3029, 3249, 3094, 3589, 3999, 4208, 3479, 3232

Offset: 1

Author

Zhi-Wei Sun, Feb 02 2015

Keywords

• nonn

Comments

Conjecture: (i) a(n) exists for any n > 0. Moreover, n^2 is the main term of a(n) as n tends to the infinity.

(ii) a(n) is never divisible by 5. Also, for any n > 1 the term a(n) is not congruent to 1 modulo 5.

Examples

			a(3) = 9 since 9 is the first positive integer m with A254573(m)=3. Note that 9 = 2*3 +1*(3*1+1)/2 + 1*(3*1-1)/2 = 1*2 + 1*(3*1+1)/2 + 2*(3*2-1)/2 = 1*2 + 2*(3*2+1)/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. A000326, A002378, A005449, A254573, A254617.

Programs

Mathematica

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