

A200520


Least m>0 such that n = y^2  10^x (mod m) has no solution, or 0 if no such m exists.


13



0, 3, 5, 0, 3, 9, 0, 3, 0, 11, 3, 9, 5, 3, 9, 0, 3, 5, 11, 3, 9, 0, 3, 9, 0, 3, 0, 5, 3, 9, 16, 3, 5, 20, 3, 0, 1001, 3, 9, 0, 3, 9, 5, 3, 0, 56, 3, 5, 0, 3, 9, 11, 3, 11, 0, 3, 9, 5, 3, 9, 112, 3, 5, 0, 3, 9, 16, 3, 9, 0, 3, 0, 5, 3, 9, 11, 3, 5, 16, 3, 0, 3367
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

To prove that an integer n is in A051212, it is sufficient to find integers x,y such that y^2  10^x = n. In that case, a(n)=0. To prove that n is *not* in A051212, it is sufficient to find a modulus m for which the (finite) set of all possible values of 10^x and y^2 allows us to deduce that y^2  10^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) A200520(n, b=10, 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, A200505A200524.
Sequence in context: A153099 A102575 A309091 * A224933 A307209 A243967
Adjacent sequences: A200517 A200518 A200519 * A200521 A200522 A200523


KEYWORD

nonn


AUTHOR

M. F. Hasler, Nov 18 2011


STATUS

approved



