|
%I #12 Oct 30 2022 11:01:59
%S 1,2,9,8,17,34,64,129,252,515,1026,2044,4091,8184,16375,32758,65525,
%T 131060,262131,524279,1048566,2097167,4194322,8388590,16777203,
%U 33554450,67108877,134217712,268435473,536870929,1073741807,2147483622,4294967278,8589934615
%N a(n) is the unique k such that A357961(k) = 2^n.
%C 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
%H Rémy Sigrist, <a href="/A357993/a357993.gp.txt">PARI program</a>
%F Empirically: a(n) ~ 2^n.
%e A357961(1026) = 1024 = 2^10, so a(10) = 1026.
%o (PARI) See Links section.
%Y Cf. A357961.
%K nonn,base
%O 0,2
%A _Rémy Sigrist_, Oct 23 2022
|
