A306727 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals: A(n,k) is the number of partitions of 3*n into powers of 3 less than or equal to 3^k.

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 4, 1, 1, 2, 3, 5, 5, 1, 1, 2, 3, 5, 7, 6, 1, 1, 2, 3, 5, 7, 9, 7, 1, 1, 2, 3, 5, 7, 9, 12, 8, 1, 1, 2, 3, 5, 7, 9, 12, 15, 9, 1, 1, 2, 3, 5, 7, 9, 12, 15, 18, 10, 1, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 11, 1, 1, 2, 3, 5, 7, 9, 12, 15, 18, 23, 26, 12, 1

Offset: 0

Author

Serguei Zolotov, Mar 06 2019

Keywords

• nonn
• tabl

Comments

Column sequences converge to A005704.

Examples

			A(3,3) = 5, because there are 5 partitions of 3*3=9 into powers of 3 less than or equal to 3^3=9: [9], [3,3,3], [3,3,1,1,1], [3,1,1,1,1,1,1], [1,1,1,1,1,1,1,1,1].
Square array A(n,k) begins:
  1, 1, 1, 1, 1, 1,  ...
  1, 2, 2, 2, 2, 2,  ...
  1, 3, 3, 3, 3, 3,  ...
  1, 4, 5, 5, 5, 5,  ...
  1, 5, 7, 7, 7, 7,  ...
  1, 6, 9, 9, 9, 9,  ...
		

Links

• Serguei Zolotov, Table of n, a(n) for n = 0..10584

Crossrefs

Main diagonal gives A005704.

A181322 gives array for base p=2.

Programs

Mathematica

nmax = 12;
f[k_] := f[k] = 1/(1-x) 1/Product[1-x^(3^j), {j, 0, k-1}] + O[x]^(nmax+1) // CoefficientList[#, x]&;
A[n_, k_] := f[k][[n+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 20 2019 *)

Python

def aseq(p, x, k):
    # generic algorithm for any p - power base, p=3 for this sequence
    if x < 0:
        return 0
    if x < p:
        return 1
    # coefficients
    arr = [0]*(x+1)
    arr[0] = 1
    m = p**k
    while m > 0:
        for i in range(m, x+1, m):
            arr[i] += arr[i-m]
        m //= p
    return arr[x]
def A(n, k):
    p = 3
    return aseq(p, p*n, k)
# A(n, k), 5 = A(3, 3) = aseq(3, 3*3, 3)
# Serguei Zolotov, Mar 13 2019

Formula

G.f. of column k: 1/(1-x) * 1/Product_{j=0..k-1} (1 - x^(3^j)).