

A254522


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


1



0, 1, 1, 2, 3, 16, 9, 16, 85, 256, 93, 512, 315, 4096, 5461, 2048, 3855, 65536, 13797, 131072, 349525, 1048576, 182361, 1048576, 3355443, 16777216, 22369621, 33554432, 9256395, 268435456, 34636833, 67108864, 1431655765, 4294967296, 17179869183, 8589934592, 1857283155, 68719476736, 91625968981
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OFFSET

1,4


COMMENTS

An autosequence of the first kind is a sequence which main diagonal is A000004.
Difference table of a(n)/A093803(n):
0, 1, 1, 2, 3, 16/3, ...
1, 0, 1, 1, 7/3, 11/3, ...
1, 1, 0, 4/3, 4/3, 10/3, ...
2, 1, 4/3, 0, 2, 2, ...
3, 7/3, 4/3, 2, 0, 16/5, ...
16/3, 11/3, 10/3, 2, 16/5, 0, ...
etc.
This is an autosequence of the first kind.
Its first (or second) upper diagonal is A075101(n)/(2*A000265(n)).
From Robert Israel, Apr 03 2017: (Start)
If p is a prime == 5 (mod 8), then a(5*p) = (2^(5*p1)1)/5 and a(5*p+3) = 2^(5*p) = 10*a(5*p)+2. This explains pairs such as
a(25) = 3355443
a(28) = 33554432
and
a(65) = 3689348814741910323
a(68) = 36893488147419103232. (End)


LINKS

Robert Israel, Table of n, a(n) for n = 1..3321


MAPLE

seq(numer((2^n1+(1)^n)/(2*n)), n=1..50); # Robert Israel, Feb 01 2015


MATHEMATICA

Table[Numerator[(2^n  1 + (1)^n)/(2*n)], {n, 39}] (* Michael De Vlieger, Feb 01 2015 *)


CROSSREFS

Cf. A000004, A075101, A093803, A099430.
Sequence in context: A066841 A266211 A074270 * A007120 A092973 A126007
Adjacent sequences: A254519 A254520 A254521 * A254523 A254524 A254525


KEYWORD

nonn


AUTHOR

Paul Curtz, Jan 31 2015


EXTENSIONS

a(25) corrected by Robert Israel, Apr 03 2017


STATUS

approved



