A247221 Numbers n such that 2*n^2 + 1 divides 2^n + 1.

0, 1, 67653, 2124804

Offset

1,3

Comments

Numbers n such that (2^n + 1)/(2*n^2 + 1) is an integer.

a(5) > 2*10^10. - Chai Wah Wu, Dec 07 2014

Links

Table of n, a(n) for n=1..4.

Example

0 is in this sequence because 2^0 + 1 = 2 divides 2*0^2 + 1 = 1,
1 is in this sequence because 2^1 + 1 = 3 divides 2*1^2 + 1 = 3.

Mathematica

a247221[n_Integer] := Select[Range[n], Divisible[2^# + 1, 2*#^2 + 1] &]; a247221[2500000] (* Michael De Vlieger, Nov 30 2014 *)

Prog

(Magma) [n: n in [1..300000] | Denominator((2^n+1)/(2*n^2+1)) eq 1];
(PARI) for(n=0, 10^9, if(Mod(2, 2*n^2+1)^n==-1, print1(n, ", "))); \\ Joerg Arndt, Nov 30 2014
(Python)
A247221_list = [n for n in range(10**6) if pow(2, n, 2*n*n+1) == 2*n*n]
# Chai Wah Wu, Dec 07 2014

Crossrefs

Cf. A247205, A247220.

Sequence in context: A249208 A234120 A101697 * A224655 A234655 A104948

Adjacent sequences: A247218 A247219 A247220 * A247222 A247223 A247224

Keyword

nonn,more

Author

Juri-Stepan Gerasimov, Nov 30 2014

Status

approved