A277011 -id:A277011 - cp's OEIS frontend

A277012 Factorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

0, 1, 2, 5, 6, 13, 4, 9, 10, 21, 22, 45, 12, 25, 26, 53, 54, 109, 28, 57, 58, 117, 118, 237, 8, 17, 18, 37, 38, 77, 20, 41, 42, 85, 86, 173, 44, 89, 90, 181, 182, 365, 92, 185, 186, 373, 374, 749, 24, 49, 50, 101, 102, 205, 52, 105, 106, 213, 214, 429, 108, 217, 218, 437, 438, 877, 220, 441, 442, 885, 886, 1773, 56, 113, 114, 229, 230, 461, 116, 233

Offset: 0

Antti Karttunen, Sep 25 2016

			9 = "111" in factorial base (3! + 2! + 1! = 9) is converted to three 1-bits with separating zeros between, in binary as "10101" = A007088(21), thus a(9) = 21.
91 = "3301" in factorial base (91 = 3*4! + 3*3! + 1!) is converted to binary number "1110111001" = A007088(953), thus a(91) = 953. Between the rightmost 1-runs the other zero comes from the factorial base representation, while the other zero is an extra separating zero inserted after each run of 1-bits apart from the rightmost 1-run. The single zero between the two leftmost 1-runs is similarly used to separate the two "unary representations" of 3's.

Cf. A000035, A000079, A000120, A000225, A007088, A007623, A034968, A060130, A069010, A156552, A227153, A227349.
Cf. A277008 (terms sorted into ascending order).
Cf. A277011 (a left inverse).
Differs from analogous A277022 for the first time at n=24, where a(24) = 8, while A277022(24) = 60.

(define (A277012 n) (let loop ((n n) (z 0) (i 2) (j 0)) (if (zero? n) z (let ((d (remainder n i))) (loop (quotient n i) (+ z (* (A000225 d) (A000079 j))) (+ 1 i) (+ 1 j d))))))

a(n) = A156552(A276076(n)).
Other identities. For all n >= 0:
A277011(a(n)) = n.
A005940(1+a(n)) = A276076(n).
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A000120(a(n)) = A034968(n).
A069010(a(n)) = A060130(n).
A227349(a(n)) = A227153(n).

A277007 Number of maximal runs of 1-bits (in binary expansion of n) such that the length of run > 1 + the total number of zeros anywhere right of that run.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0

Offset: 0

Antti Karttunen, Sep 25 2016

			For n=3, "11" in binary, the only maximal run of 1-bits is of length 2, and 2 > 0+1 (where 0 is the total number of zeros to the right of it), thus a(3) = 1.
For n=59, "111011" in binary, both the length of run "11" at the least significant end exceeds the limit (see case n=3 above), and also the length of run "111" exceeds 1 + the total number of 0's to the right of it, thus a(59) = 1+1 = 2.
For n=60, "111100" in binary, the length of only run of 1's is 4, and 4 > 2+1, thus a(60) = 1.
For n=118, "1110110" in binary, the length of rightmost run of 1-bits is 2, but that is not > 1+1 (one more than the number of 0-bits right to it). Also, the length of the leftmost run of 1-bits is 3, but that is not > 1+2, thus a(118) = 0.
For n=246, "11110110" in binary, the rightmost run of 1-bits does not contribute, but the leftmost run of 1-bits has now length 4, which is more than 1+2 (where 2 is the total number of 0-bits right of it), thus a(246) = 1.

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n

Cf. A277008 (positions of zeros), A277009 (of nonzeros).
Cf. A005940, A277011, A277012, A276077.
Differs from the similar entry A277017 for the first time at n=60, where a(60)=1, while A277017(60)=0.

;; Standalone iterative implementation:
(define (A277007 n) (let loop ((e 0) (n n) (z 0) (r 0)) (cond ((zero? n) (+ e (if (> r (+ 1 z)) 1 0))) ((even? n) (loop (+ e (if (> r (+ 1 z)) 1 0)) (/ n 2) (+ 1 z) 0)) (else (loop e (/ (- n 1) 2) z (+ 1 r))))))

a(n) = A276077(A005940(1+n)).

cp's OEIS Frontend

A277012 Factorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

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