|
|
A135546
|
|
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.
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
2,1
|
|
COMMENTS
|
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)?
|
|
REFERENCES
|
O. N. Karpenkov, On examples of difference operators ..., Funct. Anal. Other Math., 1 (2006), 175-180. [The function q(n)]
|
|
LINKS
|
|
|
MAPLE
|
(First load the b-file for A014664 as the array b1.)
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;
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|