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.

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

Offset: 2

Author

N. J. A. Sloane, Feb 24 2008

Keywords

• nonn

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)?

Links

• N. J. A. Sloane, Table of n, a(n) for n = 2..1000
• O. N. Karpenkov, On examples of difference operators for {0,1}-valued functions over finite sets, Funct. Anal. Other Math. 1 (2006), 175-180. [The function q(n)]
• O. N. Karpenkov, On examples of difference operators for {0,1}-valued functions over finite sets, arXiv:math/0611940 [math.CO], 2006.

Crossrefs

Cf. A038553, A014664.

Programs

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;

Mathematica

g[n_]:=MultiplicativeOrder[2, Prime[n]];a[n_]:=If[EvenQ[g[n]],Prime[n]*(2^(g[n]/2)-1),2^g[n]-1];Table[a[n],{n,2,28}] (* James C. McMahon, Apr 16 2025 *)