A054390 Number of ways of writing n as a sum of powers of 3, each power being used at most three times.

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

Offset: 0

Author

John W. Layman, May 09 2000

Keywords

• nonn
• look

Comments

Let M be an infinite matrix with (1, 1, 1, 1, 0, 0, 0, ...) in each column shifted down thrice from the previous column (for k>0). Then A054390 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Apr 14 2010

Conjecture: Number of ways of partitioning n into distinct parts of A038754. - R. J. Mathar, Mar 01 2023

Examples

			a(33) = 4 because we have 33 = 27+3+3 = 27+3+1+1+1 = 9+9+9+3+3 = 9+9+9+3+1+1+1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000
• Karl Dilcher, Larry Ericksen, Polynomials Characterizing Hyper b-ary Representations, J. Int. Seq., Vol. 21 (2018), Article 18.4.3.
• Timothy B. Flowers, Extending a Recent Result on Hyper m-ary Partition Sequences, Journal of Integer Sequences, Vol. 20 (2017), #17.6.7.

Crossrefs

Cf. A002487.

Programs

Maple

a[0]:=1: a[1]:=1: a[2]:=1: for n from 1 to 35 do a[3*n]:=a[n]+a[n-1]: a[3*n+1]:=a[n]: a[3*n+2]:=a[n] od: A:=[seq(a[n],n=0..104)]; # Emeric Deutsch, Apr 02 2006
g:=product((1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))),j=0..10): gser:=series(g,x=0,125): seq(coeff(gser,x,n),n=0..104); # Emeric Deutsch, Apr 02 2006
# third Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
       add(`if`(n-j*3^i<0, 0, b(n-j*3^i, i-1)), j=0..3)))
    end:
a:= n-> b(n, ilog[3](n)):
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 21 2012

Mathematica

a[0]=1; a[1]=1; a[2]=1; For[n=1, n <= 35, n++, a[3*n] = a[n] + a[n-1]; a[3*n+1] = a[n]; a[3*n+2] = a[n]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Dec 20 2016, after Emeric Deutsch *)

Formula

a(0)=1, a(1)=1, a(2)=1 and, for n>0, a(3n)=a(n)+a(n-1), a(3n+1)=a(n), a(3n+2)=a(n).

G.f.: Product_{j >= 0} (1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))). - Emeric Deutsch, Apr 02 2006

G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3) * A(x^3). - Ilya Gutkovskiy, Jul 09 2019