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Numbers of the form (4^k - 1)/3 whose greatest prime divisor is of the form 2^q - 1 or 2^q + 1.
1

%I #19 Mar 02 2020 09:38:48

%S 5,21,85,341,5461,21845,22369621,89478485,1431655765,5726623061,

%T 91625968981,1501199875790165,1537228672809129301,

%U 98382635059784275285,1690200800304305868662270940501,1772303994379887830538409413707126101

%N Numbers of the form (4^k - 1)/3 whose greatest prime divisor is of the form 2^q - 1 or 2^q + 1.

%C The binary expansion of (4^k-1)/3 has no consecutive equal binary digits.

%C 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

%H Amiram Eldar, <a href="/A187063/b187063.txt">Table of n, a(n) for n = 1..25</a>

%e (4^6-1)/3 = 1365 = 3 * 5 * 7 * 13 is not in the sequence because 13 is not of the form 2^q +/- 1 ;

%e (4^16-1)/3 = 1431655765 = 5 * 17 * 257 * 65537 and 65537 = 2^16 + 1.

%p with(numtheory):

%p a:= proc(n) option remember; local k, t, d, h;

%p for k from 1+ `if`(n=1, 0, ilog[4](a(n-1)*3+1))

%p do t:= (4^k-1)/3;

%p d:= max(factorset(t)[]);

%p for h in [d+1, d-1] do

%p if 2^ilog[2](h)=h then RETURN(t) fi

%p od

%p od

%p end:

%p seq(a(n), n=1..17); # _Alois P. Heinz_, Mar 04 2011

%t 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 *)

%Y Cf. A002450 ((4^n-1)/3), A274906.

%K nonn

%O 1,1

%A _Michel Lagneau_, Mar 03 2011