A215937 Numbers n such that 2^n + 1 can be written in the form a^2 + 5*b^2.

2, 3, 7, 10, 11, 19, 23, 31, 43, 47, 50, 58, 71, 79, 82, 107, 127, 167, 178, 179, 191, 199, 250, 290, 298, 311, 347, 359, 410, 487, 563, 599, 683, 751, 802, 890, 907, 1051

Offset: 1

Author

V. Raman, Aug 27 2012

Keywords

• nonn

Comments

These 2^n + 1 numbers can only have prime factors of the form 1 (mod 20) or 3 (mod 20) or 5 (mod 20) or 7 (mod 20) or 9 (mod 20) raised to an odd power, but their overall product 2^n+1 can only be 1 (mod 20) or 5 (mod 20) or 9 (mod 20). This statement is limited to odd numbers.

In general,

A number n can be written in the form a^2+5*b^2 if and only if n is 0,

or of the form 2^(2i) 5^j Prod_{p==1 or 9 mod 20} p^k Prod_{q==3 or 7 mod 20) q^(2m)

or of the form 2^(2i+1) 5^j Prod_{p==1 or 9 mod 20} p^k Prod_{q==3 or 7 mod 20) q^(2m+1),

for integers i,j,k,m, for primes p,q.

Examples

			3 is in the sequence because 2^3 + 1 = 9 can be written as 2^2 + 5 * 1^2 = 9.
		

Links

• Samuel S. Wagstaff, Jr., The Cunningham Project, Factorizations of 2^n-1, for odd n's < 1200

Crossrefs

Cf. A000051, A000978, A215800, A215801.

Cf. A020669, A033205 (numbers and primes of the form x^2 + 5*y^2).

Programs

PARI

for(i=2, 500, a=factorint(2^i+1)~; has=0; for(j=1, #a, if(((a[1, j]%20>10)||(i%4<2))&&a[2, j]%2==1, has=1; break)); if(has==0, print(i",")))

PARI

for(i=2, 500, a=factorint(2^i+1)~; flag=0; flip=0; for(j=1, #a, if(((a[1, j]%20>10))&&a[2, j]%2==1, flag=1); if(((a[1, j]%20==2)||(a[1, j]%20==3)||(a[1, j]%20==7))&&a[2, j]%2==1, flip=flip+1)); if(flag==0&&flip%2==0, print(i",")))

Extensions

Terms corrected by V. Raman, Sep 20 2012