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A249150 Number of trailing zeros in the factorial base representation of products of binomial coefficients: a(n) = A230403(A001142(n)). 7
0, 0, 1, 0, 3, 1, 5, 0, 1, 3, 9, 6, 11, 5, 3, 0, 15, 1, 17, 3, 5, 9, 21, 10, 3, 11, 1, 5, 27, 24, 29, 0, 9, 15, 5, 35, 35, 17, 11, 39, 39, 5, 41, 9, 22, 21, 45, 18, 5, 3, 15, 11, 51, 1, 9, 34, 17, 27, 57, 46, 59, 29, 62, 0, 11, 9, 65, 15, 21, 48, 69, 40, 71, 35, 3, 17, 9, 11, 77, 79, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,5

COMMENTS

a(n) = A249151(n)-1. Please see the comments and graph of that sequence.

LINKS

Table of n, a(n) for n=0..80.

FORMULA

a(n) = A230403(A001142(n)).

PROG

(Scheme) (define (A249150 n) (A230403 (A001142 n)))

CROSSREFS

One less than A249151.

Cf. A249423 (values k such that a(k) = k).

Cf. A249425 (record positions).

Cf. A249426 (record values).

Cf. A001142, A187059, A230403.

Sequence in context: A125846 A318382 A143677 * A143467 A143315 A237588

Adjacent sequences:  A249147 A249148 A249149 * A249151 A249152 A249153

KEYWORD

nonn

AUTHOR

Antti Karttunen, Oct 25 2014

STATUS

approved

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Last modified September 15 16:33 EDT 2019. Contains 327078 sequences. (Running on oeis4.)