

A200513


Least m>0 such that n = y^2  3^x (mod m) has no solution, or 0 if no such m exists.


1



0, 0, 8, 0, 8, 9, 0, 0, 0, 0, 8, 9, 8, 0, 9, 0, 0, 12, 8, 0, 8, 28, 0, 9, 0, 20, 8, 0, 8, 9, 20, 80, 9, 0, 8, 0, 8, 0, 9, 63, 0, 9, 8, 80, 8, 20, 0, 9, 0, 28, 8, 63, 8, 12, 0, 0, 9, 36, 8, 9, 8, 0, 12, 0, 532, 9, 8, 80, 8, 108, 20, 15, 0, 0, 8, 63, 8, 9, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

To prove that an integer n is in A051205, it is sufficient to find integers x,y such that y^2  3^x = n. In that case, a(n)=0. To prove that n is *not* in A051205, it is sufficient to find a modulus m for which the (finite) set of all possible values of 3^x and y^2 (mod m) allows us to deduce that y^2  3^x can never equal n. The present sequence lists the smallest such m>0, if it exists.


LINKS

Table of n, a(n) for n=0..78.


EXAMPLE

See A200512.


PROG

(PARI) A200513(n, b=3, 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: A021557 A242943 A319215 * A200523 A118540 A165110
Adjacent sequences: A200510 A200511 A200512 * A200514 A200515 A200516


KEYWORD

nonn


AUTHOR

M. F. Hasler, Nov 18 2011


STATUS

approved



