

A135546


Let p be the nth prime and let g be the order of 2 mod p (see A014664). Then if g is even, a(n) = p*(2^(g/2)  1), otherwise a(n) = 2^g  1.


2



3, 15, 7, 341, 819, 255, 9709, 2047, 475107, 31, 9699291, 41943, 5461, 8388607, 3556769739, 31675383749, 65498251203, 575525617597, 34359738367, 511, 549755813887, 182518930210733, 2047, 1627389855, 113715890591104923, 2251799813685247, 963770320257286037
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OFFSET

2,1


COMMENTS

Karpenkov asks how often is it the case that if p is the nth prime (n >= 2) then A038553(p) = a(n)? The first failure is at p = 37. Is it true that a(n) is always divisible by A038553(p)?


REFERENCES

O. N. Karpenkov, On examples of difference operators ..., Funct. Anal. Other Math., 1 (2006), 175180. [The function q(n)]


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 2..1000


MAPLE

(First load the bfile for A014664 as the array b1.)
a := proc(i) local p, g; p:=ithprime(i); g:=b1[i1]; if g mod 2 = 0 then p*(2^(g/2)1) else 2^g1; fi; end;


CROSSREFS

Cf. A038553, A014664.
Sequence in context: A066832 A102777 A102531 * A138006 A335696 A256557
Adjacent sequences: A135543 A135544 A135545 * A135547 A135548 A135549


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Feb 24 2008


STATUS

approved



