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 A200515 Least m>0 such that n = y^2 - 5^x (mod m) has no solution, or 0 if no such m exists. 1
 0, 4, 4, 0, 0, 4, 4, 5, 0, 4, 4, 0, 5, 4, 4, 0, 24, 4, 4, 0, 0, 4, 4, 35, 0, 4, 4, 5, 15, 4, 4, 0, 5, 4, 4, 0, 24, 4, 4, 0, 24, 4, 4, 15, 0, 4, 4, 5, 0, 4, 4, 0, 5, 4, 4, 39, 0, 4, 4, 0, 24, 4, 4, 0, 24, 4, 4, 5, 35, 4, 4, 0, 5, 4, 4, 0, 0, 4, 4, 63, 0, 4, 4 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS To prove that an integer n is in A051207, it is sufficient to find integers x,y such that y^2 - 5^x = n. In that case, a(n)=0. To prove that n is *not* in A051207, it is sufficient to find a modulus m for which the (finite) set of all possible values of 5^x and y^2 (mod m) allows us to deduce that y^2 - 5^x can never equal n. The present sequence lists the smallest such m>0, if it exists. LINKS FORMULA A200515(A051207(k))=0 for all k in N. A200515(1+4k)=A200515(2+4k)=4 for all k>=0. EXAMPLE See A200512 for motivation and detailed examples. PROG (PARI) A200515(n, b=5, 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 & break; next(3))); return(m))} CROSSREFS Cf. A051204-A051221, A200505-A200520. Sequence in context: A282866 A098445 A290696 * A200505 A285242 A143266 Adjacent sequences:  A200512 A200513 A200514 * A200516 A200517 A200518 KEYWORD nonn AUTHOR M. F. Hasler, Nov 18 2011 STATUS approved

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Last modified November 14 03:52 EST 2018. Contains 317159 sequences. (Running on oeis4.)