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A254522 Numerators of (2^n - 1 + (-1)^n)/(2*n), n > 0.

%I A254522 #41 Apr 04 2017 03:19:42
%S A254522 0,1,1,2,3,16,9,16,85,256,93,512,315,4096,5461,2048,3855,65536,13797,
%T A254522 131072,349525,1048576,182361,1048576,3355443,16777216,22369621,
%U A254522 33554432,9256395,268435456,34636833,67108864,1431655765,4294967296,17179869183,8589934592,1857283155,68719476736,91625968981
%N A254522 Numerators of (2^n - 1 + (-1)^n)/(2*n), n > 0.
%C A254522 An autosequence of the first kind is a sequence which main diagonal is A000004.
%C A254522 Difference table of a(n)/A093803(n):
%C A254522      0,     1,    1,   2,    3, 16/3, ...
%C A254522      1,     0,    1,   1,  7/3, 11/3, ...
%C A254522     -1,     1,    0, 4/3,  4/3, 10/3, ...
%C A254522      2,    -1,  4/3,   0,    2,    2, ...
%C A254522     -3,   7/3, -4/3,   2,    0, 16/5, ...
%C A254522   16/3, -11/3, 10/3,  -2, 16/5,    0, ...
%C A254522 etc.
%C A254522 This is an autosequence of the first kind.
%C A254522 Its first (or second) upper diagonal is A075101(n)/(2*A000265(n)).
%C A254522 From _Robert Israel_, Apr 03 2017: (Start)
%C A254522 If p is a prime == 5 (mod 8), then a(5*p) = (2^(5*p-1)-1)/5 and a(5*p+3) = 2^(5*p) = 10*a(5*p)+2. This explains pairs such as
%C A254522    a(25) = 3355443
%C A254522    a(28) = 33554432
%C A254522 and
%C A254522    a(65) = 3689348814741910323
%C A254522    a(68) = 36893488147419103232. (End)
%H A254522 Robert Israel, <a href="/A254522/b254522.txt">Table of n, a(n) for n = 1..3321</a>
%p A254522 seq(numer((2^n-1+(-1)^n)/(2*n)), n=1..50); # _Robert Israel_, Feb 01 2015
%t A254522 Table[Numerator[(2^n - 1 + (-1)^n)/(2*n)], {n, 39}] (* _Michael De Vlieger_, Feb 01 2015 *)
%Y A254522 Cf. A000004, A075101, A093803, A099430.
%K A254522 nonn
%O A254522 1,4
%A A254522 _Paul Curtz_, Jan 31 2015
%E A254522 a(25) corrected by _Robert Israel_, Apr 03 2017