

A200514


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


1



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
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OFFSET

0,2


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.


LINKS



EXAMPLE



PROG

(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^2n)%m, , 8); seen=0; bx=1; until( bittest(seen+=1<<bx, bx=bx*b%m), for(i=1, #qr, qr[i]<bx & next; qr[i]>bx & break; next(3))); return(m))}


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



