A072939 Define a sequence c depending on n as follows: c(1)=1 and c(2)=n; c(k+2) = (c(k+1) + c(k))/2 if c(k+1) and c(k) have the same parity; otherwise c(k+2) = abs(c(k+1) - 2*c(k)); sequence gives values of n such that lim_{k->oo} c(k) = infinity.

3, 7, 9, 11, 15, 19, 23, 25, 27, 31, 33, 35, 39, 41, 43, 47, 51, 55, 57, 59, 63, 67, 71, 73, 75, 79, 83, 87, 89, 91, 95, 97, 99, 103, 105, 107, 111, 115, 119, 121, 123, 127, 129, 131, 135, 137, 139, 143, 147, 151, 153, 155, 159, 161, 163, 167, 169, 171, 175, 179

Offset

1,1

Comments

If c(2) is even then c(k) = 1 for k >= 2*c(2), hence there is no even value in the sequence. If n is in the sequence, there exist an integer k(n) and an integer m(n) such that, for any k >= k(n), c(2k) - c(2k-1) = 2*m(n) and c(2k+1) - c(2k) = -m(n). Sometimes m(n) = (n-1)/2 but not always. If B(n) = a(n+1) - a(n) then B(n) = 2 or 4, but B(n) does not seem to follow any pattern.

Conjecture: a(n) = A036554(n)+1. - Vladeta Jovovic, Apr 01 2003

a(n) = A036554(n)+1 = A079523(n)+2. - Ralf Stephan, Jun 09 2003

Conjecture: this sequence gives the positions of 0's in the limiting 0-word of the morphism 0->11, 1->10, A285384. - Clark Kimberling, Apr 26 2017

Conjecture: This also gives the positions of the 1's in A328979. - N. J. A. Sloane, Nov 05 2019

Links

Table of n, a(n) for n=1..60.

Formula

Conjecture : lim_{n->oo} a(n)/n = 3.

Example

41 is in the sequence: if c(2)=41, then it follows that c(3)=21, c(4)=31, c(5)=26, c(6)=36, c(7)=31, c(8)=41, c(9)=36, ...; for k >= 2, c(2k) - c(2k-1) = 10 and c(2k+1) - c(2k) = -5, which implies that c(k) -> infinity.

Prog

(Python)
from itertools import count, islice
def A072939_gen(startvalue=2): return filter(lambda n:(~(n-1)&(n-2)).bit_length()&1, count(max(startvalue, 2))) # generator of terms >= startvalue
A072939_list = list(islice(A072939_gen(), 30)) # Chai Wah Wu, Jul 05 2022
(Python)
def A072939(n):
    def bisection(f, kmin=0, kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1		
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c, s = n+x, bin(x)[2:]
        l = len(s)
        for i in range(l&1, l, 2):
            c -= int(s[i])+int('0'+s[:i], 2)
        return c
    return bisection(f, n, n)+1 # Chai Wah Wu, Jan 29 2025

Crossrefs

Cf. A036554, A079523, A285384, A328979.

Sequence in context: A047529 A359567 A125667 * A171947 A287914 A291348

Adjacent sequences: A072936 A072937 A072938 * A072940 A072941 A072942

Keyword

nonn

Author

Benoit Cloitre, Aug 12 2002

Status

approved