A125974 Function whose restriction to A014486 induces signature-permutation A125976.

0, 1, 2, 5, 10, 3, 2, 21, 42, 11, 12, 13, 10, 11, 10, 85, 170, 43, 44, 13, 52, 7, 6, 53, 42, 11, 12, 45, 10, 43, 42, 341, 682, 171, 172, 45, 180, 39, 38, 53, 212, 23, 56, 57, 50, 51, 22, 213, 170, 43, 44, 45, 52, 39, 38, 181, 42, 43, 44, 173, 42, 171, 170, 1365, 2730

Offset

0,3

Comments

A125975 gives the terms i, for which a(a(i)) = i. Question: would it be possible to construct a more elegant and natural variant which were an involution for all the natural numbers? (and acting in the same way on the set A125975, or at least on the set A014486.)

Links

Table of n, a(n) for n=0..64.

Prog

(Scheme) (define (A125974 n) (let ((runlens (binexp->runcount1list n))) (let loop ((chosen (reverse! (bisect runlens 0))) (others (reverse! (bisect runlens 1))) (s 0) (b (modulo n 2)) (p 1)) (cond ((and (null? chosen) (null? others)) s) ((and (pair? chosen) (= 1 (car chosen)) (pair? (cdr chosen))) (loop (cdr chosen) others (+ s (* b p)) b (+ p p))) (else (loop others (if (or (null? chosen) (= 1 (car chosen))) '() (cons (- (car chosen) 1) (cdr chosen))) (+ s (* b p)) (- 1 b) (+ p p)))))))
(Scheme) (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))))))) ;; (binexp->runcount1list 25) returns (2 2 1)
(Scheme) (define (bisect lista parity) (let loop ((lista lista) (i 0) (z (list))) (cond ((null? lista) (reverse! z)) ((eq? i parity) (loop (cdr lista) (modulo (1+ i) 2) (cons (car lista) z))) (else (loop (cdr lista) (modulo (1+ i) 2) z)))))
(Python)
def A125974(n):
    if 0 == n:
        return n
    chosen = A000265(n)  # Initially ones, get rid of lsb-0's.
    others = n >> A007814(n + 1)  # Initially zeros, get rid of lsb-1's.
    s = 0  # the resulting sum
    b = n % 2  # n's parity.
    p = 1  # powers of two.
    while (chosen != 0) or (others != 0):
        if (1 == chosen) or (1 == A036987(chosen + 1)):  # Last one or zero at hand.
            chosen = others
            others = 0
            nb = 1 - b
        elif (0 == (chosen % 4)) or (
            3 == (chosen % 4)
        ):  # Source run continues, dest changes.
            tmp = chosen
            chosen = others
            others = tmp >> 1
            nb = 1 - b
        elif 1 == (
            chosen % 4
        ):  # Source run changes, from ones to zeros, skip past zeros.
            chosen = A000265(chosen - 1)
            nb = b
        else:  # Source run changes, from zeros to ones, skip past ones.
            chosen = chosen >> A007814(chosen + 2)
            nb = b
        s += b * p
        p <<= 1
        b = nb
    return s

Crossrefs

Python code uses the following functions: A000265, A007814 and A036987.

Sequence in context: A172411 A264784 A306679 * A218336 A059955 A099796

Adjacent sequences: A125971 A125972 A125973 * A125975 A125976 A125977

Keyword

nonn,base

Author

Antti Karttunen, Jan 02 2007

Status

approved