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%I #4 Mar 30 2012 16:50:47
%S 3,15,7,341,819,255,9709,2047,475107,31,9699291,41943,5461,8388607,
%T 3556769739,31675383749,65498251203,575525617597,34359738367,511,
%U 549755813887,182518930210733,2047,1627389855,113715890591104923,2251799813685247,963770320257286037
%N Let p be the n-th 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.
%C Karpenkov asks how often is it the case that if p is the n-th 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)?
%D O. N. Karpenkov, On examples of difference operators ..., Funct. Anal. Other Math., 1 (2006), 175-180. [The function q(n)]
%H N. J. A. Sloane, <a href="/A135546/b135546.txt">Table of n, a(n) for n = 2..1000</a>
%p (First load the b-file for A014664 as the array b1.)
%p a := proc(i) local p,g; p:=ithprime(i); g:=b1[i-1]; if g mod 2 = 0 then p*(2^(g/2)-1) else 2^g-1; fi; end;
%Y Cf. A038553, A014664.
%K nonn
%O 2,1
%A _N. J. A. Sloane_, Feb 24 2008
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