

A200516


Least m>0 such that n = y^2  6^x (mod m) has no solution, or 0 if no such m exists.


1



0, 5, 5, 0, 7, 9, 5, 5, 0, 0, 0, 5, 5, 0, 9, 0, 5, 5, 7, 0, 9, 5, 5, 9, 0, 7, 5, 5, 0, 9, 0, 5, 5, 36, 16, 0, 5, 5, 9, 7, 0, 5, 5, 0, 21, 0, 5, 5, 0, 43, 9, 5, 5, 7, 16, 16, 5, 5, 0, 9, 7, 5, 5, 0, 0, 9, 5, 5, 9, 36, 16, 5, 5, 0, 7, 0, 5, 5, 32, 24, 0, 5, 5
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OFFSET

0,2


COMMENTS

To prove that an integer n is in A051208, it is sufficient to find integers x,y such that y^2  6^x = n. In that case, a(n)=0. To prove that n is *not* in A051208, it is sufficient to find a modulus m for which the (finite) set of all possible values of 6^x and y^2 (mod m) allows us to deduce that y^2  6^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) A200516(n, b=6, 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

Cf. A051204A051221, A200505A200520.
Sequence in context: A021649 A200257 A236023 * A233383 A144184 A119380
Adjacent sequences: A200513 A200514 A200515 * A200517 A200518 A200519


KEYWORD

nonn


AUTHOR

M. F. Hasler, Nov 18 2011


STATUS

approved



