A254661 -id:A254661 - cp's OEIS frontend

A254680 Least positive integer m with A254661(m) = n.

1, 3, 7, 17, 21, 51, 66, 72, 157, 147, 121, 136, 246, 297, 332, 367, 402, 506, 547, 577, 677, 796, 892, 731, 926, 1216, 1116, 976, 1181, 1402, 1556, 1416, 1507, 1496, 2287, 1622, 1977, 2112, 1942, 2131, 2017, 2882, 2767, 2501, 3162, 3671, 3097, 3187, 3047, 3762, 3867, 2952, 4356, 4111, 4826, 5112, 5211, 4811, 4686, 5461

Offset: 1

Zhi-Wei Sun, Feb 05 2015

nonn

Conjecture: a(n) exists for any n > 0. Moreover, no term is divisible by 5, and no term with n>2 is congruent to 3 modulo 5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..170
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Cf. A000217, A000290, A005449, A160325, A254661, A254677.

a(3) = 7 since 7 is the first positive integer that can be written as x*(x+1)/2 + (2y)^2 + z*(3*z+1)/2 (with x,y,z nonnegative integers) in exactly 3 ways. In fact, 7 = 0*1/2 + 0^2 +2*(3*2+1)/2 = 1*2/2 + 2^2 +1*(3*1+1)/2 = 2*3/2 + 2^2 + 0*(3*0+1)/2.

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}]

A254668 Number of ways to write n as the sum of a square, a second pentagonal number, and a hexagonal number.

Original entry on oeis.org

1, 2, 2, 2, 2, 1, 2, 3, 3, 3, 2, 2, 3, 1, 1, 3, 5, 6, 2, 3, 1, 2, 4, 2, 4, 3, 4, 3, 3, 2, 4, 7, 4, 4, 2, 2, 4, 3, 3, 4, 3, 5, 5, 3, 6, 3, 5, 4, 2, 4, 4, 6, 5, 3, 2, 6, 5, 7, 4, 3, 2, 4, 4, 4, 7, 3, 8, 4, 5, 3, 5, 6, 8, 3, 2, 3, 4, 9, 2, 8, 3, 7, 7, 4, 5, 5, 4, 4, 4, 6, 5, 4, 6, 7, 9, 2, 8, 4, 3, 4, 3

Offset: 0

Zhi-Wei Sun, Feb 04 2015

nonn

Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 5, 13, 14, 20, 112, 125.
Compare this conjecture with the conjecture in A160324.
The conjecture that a(n) > 0 for all n = 0,1,2,... appeared in Conjecture 1.2(ii) of the author's JNT paper in the links. - Zhi-Wei Sun, Oct 03 2016

			a(20) = 1 since 20 = 2^2 + 3*(3*3+1)/2 + 1*(2*1-1).
a(112) = 1 since 112 = 7^2 + 6*(3*6+1)/2 + 2*(2*2-1).
a(125) = 1 since 125 = 5^2 + 8*(3*8+1)/2 + 0*(2*0-1).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.
Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Cf. A000290, A000384, A005449, A160324, A160325, A254661.

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-y(3y+1)/2-z(2z-1)],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[8(n-y(3y+1)/2)+1]+1)/4}];
Print[n," ",r];Continue,{n,0,100}]

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A254680 Least positive integer m with A254661(m) = n.

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