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A200517 Least m>0 such that n = y^2 - 7^x (mod m) has no solution, or 0 if no such m exists. 1
0, 3, 0, 0, 3, 7, 8, 3, 0, 0, 3, 16, 7, 3, 8, 0, 3, 16, 0, 3, 8, 16, 3, 16, 0, 3, 7, 16, 3, 0, 8, 3, 0, 7, 3, 0, 8, 3, 8, 16, 3, 28, 0, 3, 8, 19, 3, 7, 0, 3, 20, 0, 3, 16, 7, 3, 100, 0, 3, 16, 8, 3, 8, 0, 3, 16, 28, 3, 7, 16, 3, 16, 0, 3, 0, 7, 3, 19, 8, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

To prove that an integer n is in A051209, it is sufficient to find integers x,y such that y^2 - 7^x = n. In that case, a(n)=0. To prove that n is *not* in A051209, it is sufficient to find a modulus m for which the (finite) set of all possible values of 7^x and y^2 allows us to deduce that y^2-7^x can never equal n. The present sequence lists the smallest such m>0, if it exists.

LINKS

M. F. Hasler, Table of n, a(n) for n = 0..1000

EXAMPLE

See A200512 for motivation and detailed examples.

PROG

(PARI) A200517(n, b=7, 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, bx=bx*b%m), for(i=1, #qr, qr[i]<bx & next; qr[i]>bx & break; next(3))); return(m))}

CROSSREFS

Cf. A051204-A051221, A200505-A200520.

Sequence in context: A005082 A050452 A267875 * A151665 A171793 A079201

Adjacent sequences:  A200514 A200515 A200516 * A200518 A200519 A200520

KEYWORD

nonn

AUTHOR

M. F. Hasler, Nov 18 2011

STATUS

approved

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Last modified November 15 15:59 EST 2018. Contains 317239 sequences. (Running on oeis4.)