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A325803 Nonzero terms of Product_{k=0..floor(log_2(n))} (1 + A004718(floor(n/(2^k)))). 2
1, 2, 6, -6, 24, -18, -48, 120, 18, -72, -192, 48, -360, 720, 54, 144, -360, 384, -960, 144, -1800, 720, -2880, 5040, -54, 216, 576, -144, 1080, -2160, 1536, -384, 2880, -5760, -144, 576, 5400, -10800, 2880, -720, -17280, 8640, -25200, 40320, -162, -432, 1080 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Number of positive terms in the original sequence on the interval 2^m <= n < 2^(m+1) for m > 0 equals f(m-1,2,1) (and f(m-1,4,3) for negative) with f(m,g,h) = binomial(m, floor(m/2) + floor((m+g)/4) - floor((m+h)/4)), so total number of nonzero terms equals binomial(m, floor(m/2)).

LINKS

Mikhail Kurkov, Table of n, a(n) for n = 1..13495

FORMULA

Sum_{n=0..2^m-1} Product_{k=0..floor(log_2(n))} (1 + A004718(floor(n/(2^k)))) = 3^m, m >= 0.

More generally, if we define a(n,k) = (-1)^(n+1)*a(floor(n/k),k) + n mod k, a(0,k) = 0, so Sum_{n=0..k^m-1} Product_{i=0..floor(log_k(n))} (1 + a(floor(n/(k^i)),k)) = binomial(k+1, 2)^m for any k = 2p, p > 0.

PROG

(PARI) b(n) = if(n==0, 0, (-1)^(n+1)*b(n\2) + n%2); \\ A004718

f(n) = if(n==0, 1, prod(k=0, logint(n, 2), 1+b(n\2^k)));

lista(nn) = for (n=0, nn, if (f(n), print1(f(n), ", "))); \\ Michel Marcus, May 26 2019

CROSSREFS

Cf. A000120, A004718, A284005, A325804.

Sequence in context: A100634 A242527 A304680 * A130865 A327630 A282170

Adjacent sequences:  A325800 A325801 A325802 * A325804 A325805 A325806

KEYWORD

sign

AUTHOR

Mikhail Kurkov, May 22 2019

STATUS

approved

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Last modified November 11 22:31 EST 2019. Contains 329046 sequences. (Running on oeis4.)