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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).
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%I #4 Jun 18 2018 18:13:58

%S 3,6,7,13,61,157,241,409,421,1321,1621,3541,4129,5209,5701,8269,9241,

%T 9769,11701,12601,13729,18181,27061,32341,39901,78121,78541,118681,

%U 129361,153469,189661,207481,314161,431869,451669,455701,507301,655561,842521,979969

%N 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).

%C Jacobson & Williams proved that rho(n) can be arbitrarily large, therefore this sequence is infinite.

%C Of the first 40 terms only 6 is composite.

%H Michael J. Jacobson and Hugh C. Williams, <a href="https://doi.org/10.1080/10586458.2000.10504666">The size of the fundamental solutions of consecutive Pell equations</a>, Experimental Mathematics, Vol. 9, No. 4 (2000), pp. 631-640. <a href="http://pages.cpsc.ucalgary.ca/~jacobs/PDF/Pell.pdf">alternative link</a>.

%e 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.

%t $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

%Y Cf. A000037, A002349, A002350, A033313, A033317.

%K nonn

%O 1,1

%A _Amiram Eldar_, Apr 26 2018