A303604 Numbers n such that both n-1 and n are nonsquares and the least positive solutions to the Pell equations x1^2 - n*y1^2 =1 and x0^2-(n-1)*y0^2 = 1 have a record for rho(n)=log(x1)/log(x0).
3, 6, 7, 13, 61, 157, 241, 409, 421, 1321, 1621, 3541, 4129, 5209, 5701, 8269, 9241, 9769, 11701, 12601, 13729, 18181, 27061, 32341, 39901, 78121, 78541, 118681, 129361, 153469, 189661, 207481, 314161, 431869, 451669, 455701, 507301, 655561, 842521, 979969
Offset: 1
Keywords
Examples
n = 61 is in the sequence since the least positive solution to x^2-60*y^2 = 1 has x = 31, and the least positive solution to x^2-61*y^2 = 1 has x = 1766319049, so rho(61) = log(1766319049)/log(31) = 6.200... larger than for any smaller n.
Links
- Michael J. Jacobson and Hugh C. Williams, The size of the fundamental solutions of consecutive Pell equations, Experimental Mathematics, Vol. 9, No. 4 (2000), pp. 631-640. alternative link.
Programs
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Mathematica
$MaxExtraPrecision= 1000; a[n_]:=If[IntegerQ[Sqrt[n]],0,For[y=1, !IntegerQ[ Sqrt[n*y^2+1]], y++, Null]; y];PellSolve[(m_Integer)?Positive] := Module[ {cf, n, s}, cof = ContinuedFraction[Sqrt[m]]; n = Length[ Last[cof]]; If[ OddQ[n], n = 2*n]; s = FromContinuedFraction[ ContinuedFraction[ Sqrt[m], n]]; {Numerator[s], Denominator[s]}]; f[n_] := If[ !IntegerQ[ Sqrt[n]], PellSolve[n][[1]], 0]; rho[x0_,x1_]:=If[x0==0||x1==0,0,Log[x1]/Log[x0]]; x0=2; n=3; rhom=0; seq={};Do[x1=f[n]; rho1 = rho[x0,x1]; If[rho1 > rhom, AppendTo[seq, n];rhom=rho1];x0=x1;n++,{k,1,1000}]; seq
Comments