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A200513
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Least m>0 such that n = y^2 - 3^x (mod m) has no solution, or 0 if no such m exists.
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1
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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
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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PROG
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(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^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))}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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