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A286468
Run lengths in the binary expansion of n are one more than the exponents of corresponding primes in the prime factorization of a(n).
4
1, 1, 2, 2, 1, 3, 4, 4, 3, 1, 2, 6, 5, 9, 8, 8, 9, 5, 6, 2, 1, 3, 4, 12, 15, 7, 10, 18, 25, 27, 16, 16, 27, 25, 18, 10, 7, 15, 12, 4, 3, 1, 2, 6, 5, 9, 8, 24, 45, 35, 30, 14, 11, 21, 20, 36, 75, 49, 50, 54, 125, 81, 32, 32, 81, 125, 54, 50, 49, 75, 36, 20, 21, 11, 14, 30, 35, 45, 24, 8, 9, 5, 6, 2, 1, 3, 4, 12, 15, 7, 10, 18, 25, 27, 16, 48, 135, 175, 90, 70
OFFSET
1,3
LINKS
FORMULA
a(1) = 1; and then after, a(4n) = 2*a(2n), a(4n+1) = A003961(a((n-1)/2)), a(4n+2) = A003961(a(n/2)), a(4n+3) = 2*a((n-1)/2).
a(n) = A052126(1+A075157(n)) = A052126(A075159(1+n)).
Other identities. For all n >= 1:
a(A007283(n)) = A007283(n-1).
EXAMPLE
For n = 25, "11001" in binary, the lengths of runs from the right are 1, 2 and 2, thus we form a product prime(1)^(1-1) * prime(2)^(2-1) * prime(3)^(2-1) = 2^0 * 3^1 * 5^1 = 15, and a(26) = 15.
PROG
(PARI) A286468(n) = { my(p=((n+1)%2), i=0, m=1); while(n>0, if(((n%2)==p), m *= prime(i), p = (n%2); i = i+1); n = n\2); m };
(Scheme)
(define (A286468 n) (fold-left (lambda (a r) (* (A003961 a) (A000079 (- r 1)))) 1 (binexp->runcount1list n)))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
;; Or by implementing the given recurrence:
(define (A286468 n) (cond ((= 1 n) n) ((zero? (modulo n 4)) (* 2 (A286468 (/ n 2)))) ((= 1 (modulo n 4)) (A003961 (A286468 (/ (- n 1) 2)))) ((= 2 (modulo n 4)) (A003961 (A286468 (/ n 2)))) ((= 3 (modulo n 4)) (* 2 (A286468 (/ (- n 1) 2))))))
CROSSREFS
Cf. A000975 (positions of ones).
Sequence in context: A113594 A368604 A246425 * A102563 A121496 A276325
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 17 2017
STATUS
approved