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

1,2

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.

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.

Example

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.

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}]

Crossrefs

Cf. A000326, A002378, A005449, A254573, A254617.

Sequence in context: A121837 A160439 A245287 * A137453 A063381 A019078

Adjacent sequences: A254592 A254593 A254594 * A254596 A254597 A254598

Keyword

nonn

Author

Zhi-Wei Sun, Feb 02 2015

Status

approved