A200514 Least m>0 such that n = y^2 - 4^x (mod m) has no solution, or 0 if no such m exists.

0, 3, 4, 0, 3, 0, 4, 3, 0, 0, 3, 16, 0, 3, 4, 0, 3, 0, 4, 3, 0, 0, 3, 16, 0, 3, 4, 16, 3, 40, 4, 3, 0, 0, 3, 0, 0, 3, 4, 16, 3, 63, 4, 3, 63, 0, 3, 20, 0, 3, 4, 20, 3, 40, 4, 3, 80, 0, 3, 16, 0, 3, 4, 0, 3, 0, 4, 3, 0, 40, 3, 16, 80, 3, 4, 16, 3, 0, 4, 3, 0

Offset: 0

Author

M. F. Hasler, Nov 18 2011

Keywords

• nonn

Comments

To prove that an integer n is in A051206, it is sufficient to find integers x,y such that y^2 - 4^x=n. In that case, a(n)=0. To prove that n is *not* in A051206, it is sufficient to find a modulus m for which the (finite) set of all possible values of 4^x and y^2 (mod m) allows us to deduce that y^2 - 4^x can never equal n. The present sequence lists the smallest such m>0, if it exists.

Examples

			See A200512.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 0..688

Crossrefs

Cf. A051204-A051221, A200505-A200520.

Programs

PARI

A200514(n,b=4,p=3)={ my( x=0, qr, bx, seen ); for( m=3,9e9, while( x^p < m, issquare(b^x+n) & return(0); x++); qr=vecsort(vector(m,y,y^2-n)%m,,8); seen=0; bx=1; until( bittest(seen+=1<bx & break; next(3))); return(m))}