login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A357993
a(n) is the unique k such that A357961(k) = 2^n.
2
1, 2, 9, 8, 17, 34, 64, 129, 252, 515, 1026, 2044, 4091, 8184, 16375, 32758, 65525, 131060, 262131, 524279, 1048566, 2097167, 4194322, 8388590, 16777203, 33554450, 67108877, 134217712, 268435473, 536870929, 1073741807, 2147483622, 4294967278, 8589934615
OFFSET
0,2
COMMENTS
Conjecture: if we write a(m) = 2^m + d then d < 2*m for m > 2. The reason for this conjecture: the Hamming weight of a number is smaller than its binary logarithm. If we assume in A357961 a random distribution of Hamming weights with values < log_2(k) for A357961(k), then we may expect for each dyadic interval an increase in displacement by the half of the intervals exponent. If we assume instead of randomness a stronger repeating of any Hamming weight, we would even reduce the gained displacement by this. - Thomas Scheuerle, Oct 24 2022
FORMULA
Empirically: a(n) ~ 2^n.
EXAMPLE
A357961(1026) = 1024 = 2^10, so a(10) = 1026.
PROG
(PARI) See Links section.
CROSSREFS
Cf. A357961.
Sequence in context: A069815 A215025 A162954 * A129194 A300780 A272347
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Oct 23 2022
STATUS
approved