OFFSET
1,1
COMMENTS
The binary expansion of (4^k-1)/3 has no consecutive equal binary digits.
The corresponding values of k are 2, 3, 4, 5, 7, 8, 13, 14, 16, 17, 19, 26, 31, 34, 51, 61, 62, 89, 107, 122, 127, 178, 214, 254, 521, ... - Amiram Eldar, Mar 02 2020
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..25
EXAMPLE
(4^6-1)/3 = 1365 = 3 * 5 * 7 * 13 is not in the sequence because 13 is not of the form 2^q +/- 1 ;
(4^16-1)/3 = 1431655765 = 5 * 17 * 257 * 65537 and 65537 = 2^16 + 1.
MAPLE
with(numtheory):
a:= proc(n) option remember; local k, t, d, h;
for k from 1+ `if`(n=1, 0, ilog[4](a(n-1)*3+1))
do t:= (4^k-1)/3;
d:= max(factorset(t)[]);
for h in [d+1, d-1] do
if 2^ilog[2](h)=h then RETURN(t) fi
od
od
end:
seq(a(n), n=1..17); # Alois P. Heinz, Mar 04 2011
MATHEMATICA
okQ[n_] := Module[{p = FactorInteger[n][[-1, 1]]}, IntegerQ[Log[2, p + 1]] || IntegerQ[Log[2, p - 1]]]; t = Table[(4^n-1)/3, {n, 2, 50}]; Select[t, okQ] (* T. D. Noe, Mar 04 2011 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 03 2011
STATUS
approved