A280472 Number of ways to write n as the sum of an octagonal number (A000567), a second octagonal number (A045944), and a strict partition number (A000009).

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

Offset

1,2

Comments

Conjecture: (i) a(n) > 0 for all n > 0.

(ii) lim_n a(n)/(log n)^2 = 1/Pi^2.

On the author's request, Prof. Qing-Hu Hou at Tianjin Univ. has verified part (i) of the above conjecture for n up to 10^9.

See also A280455 for a similar conjecture of the author involving the partition function.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28--Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.

Example

a(1) = 1 since 1 = 0*(3*0-2) + 0*(3*0+2) + A000009(2).
a(50) = 2 since 50 = 4*(3*4-2) + 1*(3*1+2) + A000009(7) = 4*(3*4-2) + 0*(3*0+2) + A000009(10).
a(1399) = 1 since 1399 = 1*(3*1-2) + 18*(3*18+2) + A000009(32).

Mathematica

Oct[n_]:=Oct[n]=IntegerQ[Sqrt[3n+1]]&&Mod[Sqrt[3n+1], 3]==1;
q[n_]:=q[n]=PartitionsQ[n];
Do[r=0; m=2; Label[bb]; If[q[m]>n, Goto[cc]]; Do[If[Oct[n-q[m]-x(3x-2)], r=r+1], {x, 0, (Sqrt[3(n-q[m])+1]+1)/3}]; m=m+If[m<3, 2, 1]; Goto[bb]; Label[cc]; Print[n, " ", r]; Continue, {n, 1, 80}]

Crossrefs

Cf. A000009, A000567, A045944, A280386, A280455.

Sequence in context: A330324 A341132 A291309 * A068063 A087181 A034973

Adjacent sequences: A280469 A280470 A280471 * A280473 A280474 A280475

Keyword

nonn

Author

Zhi-Wei Sun, Jan 04 2017

Status

approved