A254617 Least positive integer m such that A254574(m) = n.

49, 1, 3, 20, 8, 15, 30, 22, 58, 48, 93, 78, 92, 148, 113, 127, 155, 198, 197, 323, 268, 272, 288, 345, 358, 338, 555, 568, 443, 498, 612, 547, 653, 730, 708, 687, 722, 778, 1002, 897, 1107, 1030, 1112, 1205, 1535, 1343, 1458, 1093, 1203, 1588, 1548, 1822, 1623, 2162, 2208, 1577, 1497, 1948, 2228, 2473

Offset: 1

Author

Zhi-Wei Sun, Feb 03 2015

Keywords

• nonn

Comments

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

(ii) No term a(n) with n>2 is congruent to 1 or -1 modulo 5.

See also the comments in A254595 for a similar conjecture.

Examples

			a(4) = 20 since 20 is the first positive integer m with A254574(m) = 4. Note that 20 = 0*1/2 + 3*(3*3+1)/2 + 2*(3*2-1)/2 = 1*2/2 + 2*(3*2+1)/2 + 3*(3*3-1)/2 = 3*4/2 + 1*(3*1+1)/2 + 3*(3*3-1)/2 = 5*6/2 + 0*(3*0+1)/2 + 2*(3*2-1)/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, A000326, A005449, A254573, A254574, A254595.

Programs

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[Do[m=0;Label[aa];m=m+1;r=0;Do[If[TQ[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}]