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

0,3

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

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]

Prog

(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))))))

Crossrefs

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

Sequence in context: A163507 A003963 A003960 * A124223 A377027 A379730

Adjacent sequences: A243496 A243497 A243498 * A243500 A243501 A243502

Keyword

nonn,look

Author

Antti Karttunen, Jun 28 2014

Status

approved