A373309 Number of binary partitions of n into three kinds of parts.

1, 3, 9, 19, 42, 78, 146, 246, 420, 668, 1068, 1620, 2470, 3618, 5310, 7546, 10746, 14910, 20706, 28134, 38262, 51090, 68238, 89706, 117964, 153012, 198468, 254332, 325914, 413214, 523778, 657606, 825444, 1027292, 1278060, 1577748, 1947062, 2386002, 2922702, 3557162, 4327644

Offset

0,2

Comments

Submitted at the request of Joerg Arndt.

Links

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 521 terms from Paul D. Hanna)

Formula

G.f. A(x) = Sum_{n>=0} a(n) * x^n satisfies the following formulas.

(1) A(x) = 1 / Product_{n>=0} (1 - x^(2^n))^3, from Joerg Arndt, Fri Jun 21 2024.

(2) A(x) = Product_{n>=0} (1 + x^(2^n))^(3*(n+1)), deduced from a formula by Joerg Arndt in A018819.

(3) A(x) = A(x^2) / (1-x)^3.

(4) Convolution cube of A018819, which is the number of partitions of n into powers of 2.

Example

G.f.: A(x) = 1 + 3*x + 9*x^2 + 19*x^3 + 42*x^4 + 78*x^5 + 146*x^6 + 246*x^7 + 420*x^8 + 668*x^9 + 1068*x^10 + 1620*x^11 + 2470*x^12 + ...
where A(x) = 1/((1-x)^3*(1-x^2)^3*(1-x^4)^3* ... * (1 - x^(2^k))^3 * ...).

Mathematica

nmax = 40; CoefficientList[Series[1/Product[(1 - x^(2^k))^3, {k, 0, Log[2, nmax] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 25 2024 *)

Prog

(PARI) {a(n) = my(A = 1/prod(k=0, #binary(n), (1 - x^(2^k) +x*O(x^n))^3 )); polcoeff(A, n)}
for(n=0, 40, print1(a(n), ", "))

Crossrefs

Cf. A018819, A171238, A373308.

Sequence in context: A279673 A146662 A145947 * A153084 A285927 A147371

Adjacent sequences: A373306 A373307 A373308 * A373310 A373311 A373312

Keyword

nonn

Author

Paul D. Hanna, Jun 24 2024

Status

approved