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A278161 Run length transform of A008619 (floor(n/2)+1). 4
1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 3, 3, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 3, 3, 3, 4, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 3, 3, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 4, 4, 4, 6, 2, 2, 2, 4, 2, 2, 4, 4, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

LINKS

Antti Karttunen, Table of n, a(n) for n = 0..16383

Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Index entries for sequences related to binary expansion of n

FORMULA

a(n) = A046951(A005940(1+n)), a(A156552(n)) = A046951(n).

a(n) = Sum_{k=0..n} ((binomial(n+3k,6k)*binomial(n,k)) mod 2).  - Chai Wah Wu, Nov 19 2019

EXAMPLE

n=111 is "1101111" in binary, which has two runs of 1-bits: the other has length 2, and the other has length 4, thus we take the product A008619(2)*A008619(4) = (floor(2/2)+1) * (floor(4/2)+1) = 2*3, which is the result, so a(111) = 6.

MATHEMATICA

f[n_] := Floor[n/2] + 1; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 120}] (* Jean-Fran├žois Alcover, Jul 11 2017 *)

PROG

(Scheme)

(define (A278161 n) (fold-left (lambda (a r) (* a (A008619 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))

(define (A008619 n) (+ 1 (/ (- n (modulo n 2)) 2)))

;; See A227349 for the required other functions.

CROSSREFS

Cf. A005940, A008619, A046951, A156552.

Cf. A106737, A227349 for other run length transforms, and also A278222.

Sequence in context: A079882 A317335 A014709 * A069258 A273134 A126207

Adjacent sequences:  A278158 A278159 A278160 * A278162 A278163 A278164

KEYWORD

nonn,base

AUTHOR

Antti Karttunen, Nov 14 2016

STATUS

approved

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Last modified April 10 14:07 EDT 2021. Contains 342845 sequences. (Running on oeis4.)