A243499 Product of parts of integer partitions as enumerated in the table A125106.

1, 1, 2, 1, 3, 2, 4, 1, 4, 3, 6, 2, 9, 4, 8, 1, 5, 4, 8, 3, 12, 6, 12, 2, 16, 9, 18, 4, 27, 8, 16, 1, 6, 5, 10, 4, 15, 8, 16, 3, 20, 12, 24, 6, 36, 12, 24, 2, 25, 16, 32, 9, 48, 18, 36, 4, 64, 27, 54, 8, 81, 16, 32, 1, 7, 6, 12, 5, 18, 10, 20, 4, 24, 15, 30, 8, 45, 16, 32, 3

Offset: 0

Author

Antti Karttunen, Jun 28 2014

Keywords

• nonn
• look

Comments

This sequence and A341392 have the same set of values on intervals from 2^m to 2^(m+1) - 1 for m >= 0. - Mikhail Kurkov, Jun 18 2021 [verification needed]

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8191

Crossrefs

Cf. A125106, A161511 (gives the corresponding sums), A227184, A003963, A243504, A006068, A005940, A163511, A000110, A007814, A023416, A053645, A329369 (similar recurrence), A341392.

Programs

Scheme

(define (A243499 n) (let loop ((n n) (i 1) (p 1)) (cond ((zero? n) p) ((even? n) (loop (/ n 2) (+ i 1) p)) (else (loop (/ (- n 1) 2) i (* p i))))))

Formula

Can also be obtained by mapping with an appropriate permutation from the products of parts of each partition computed for other enumerations similar to A125106:

a(n) = A227184(A006068(n)).

a(n) = A003963(A005940(n+1)).

a(n) = A243504(A163511(n)).

From Mikhail Kurkov, Jul 11 2021: (Start)

a(n) = (1 + A023416(n))*a(A053645(n)) for n > 0 with a(0) = 1.

a(2n+1) = a(n) for n >= 0.

a(2n) = A341392(2*A059894(n)) = a(n - 2^f(n)) + a(2n - 2^f(n)) = (2 + f(n))*a(n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).

Sum_{k=0..2^n - 1} a(k) = A000110(n+1) for n >= 0.

a((4^n - 1)/3) = n! for n >= 0.

a(2^m*(2^n - 1)) = (m+1)^n for n >= 0, m >= 0. (End) [verification needed]