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%I #13 Jan 30 2013 11:00:37
%S 0,0,3,0,8,3,0,0,3,7,8,3,8,0,3,16,7,3,8,0,3,16,28,3,0,16,3,16,8,3,7,
%T 16,3,0,8,3,8,7,3,28,0,3,8,16,3,0,19,3,0,0,3,7,8,3,0,20,3,16,7,3,8,
%U 100,3,16,35,3,8,28,3,16,20,3,7,16,3,16,8,3,28
%N Least m>0 such that n = 7^x-y^2 (mod m) has no solution, or 0 if no such m exists.
%C If such an m>0 exists, this proves that n is not in A051218, i.e., not of the form 7^x-y^2. On the other hand, if there are integers x, y such that n=7^x-y^2, then we know that a(n)=0.
%C a(432) is at least of order 10^5.
%H M. F. Hasler, <a href="/A200507/b200507.txt">Table of n, a(n) for n = 0..431</a>
%e a(2)=3 since quadratic residues mod 3 (i.e. possible values for y^2 mod 3) are {0,1}, and 7^x is always congruent to 1 (mod 3), therefore there cannot be any (x,y) such that 7^x-y^2 = 2. The modulus m=3 is the least number for which this equation has no solution in Z/mZ: For m=1 the equation is always true, and for m=2 one always has the solution x=0 and y=0 (for even n) or y=1 (for odd n).
%o (PARI) A200507(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,i,i^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))}
%Y Cf. A051204-A051221, A200505-A200524.
%K nonn
%O 0,3
%A _M. F. Hasler_, Nov 18 2011
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