A077233 a(n) is smallest natural number satisfying Pell equation b^2- d(n)*a^2= +1 or = -1, with d(n)=A000037(n) (nonsquare). Corresponding smallest b(n)=A077232(n).

1, 1, 1, 2, 3, 1, 1, 3, 2, 5, 4, 1, 1, 4, 39, 2, 12, 42, 5, 1, 1, 5, 24, 13, 2, 273, 3, 4, 6, 1, 1, 6, 4, 3, 5, 2, 531, 30, 24, 3588, 7, 1, 1, 7, 90, 25, 66, 12, 2, 20, 13, 69, 4, 3805, 8, 1, 1, 8, 5967, 4, 936, 30, 413, 2, 125, 5, 3, 6630, 40, 6, 9, 1, 1, 9, 6, 41, 1122, 3, 21, 53, 2, 165, 120, 1260, 221064, 4, 5, 569, 10, 1, 1, 10, 22419

Offset: 1

Author

Wolfdieter Lang, Nov 08 2002

Keywords

• nonn
• nice

Comments

If d(n)=A000037(n) is from A003654 (that is if the regular continued fraction for sqrt(d(n)) has odd (primitive) period length) then the -1 option applies. For such d(n) the minimal b(n) and a(n) numbers for the +1 option are 2*b(n)^2 + 1 and 2*b(n)*a(n), respectively (see Perron I, pp. 94,p5).

For general integer solutions see A077232 comments.

If the trivial solution x=1, y=0 is included, the sequence becomes A006703. - T. D. Noe, May 17 2007

Examples

			d=10=A000037(7)=A003654(3), therefore a(7)=1 and b(7)=A077232(7)=3 give 3^2=10*1^2 -1 and 2*b(7)^2+1=19 and 2*b(7)*a(7)=2*3*1=6 satisfy 19^2 - 10*6^2 = +1.
d=11=A000037(8) is not in A003654, therefore there is no (nontrivial) solution of the b^2 - d*a^2 = -1 Pell equation and a(8)=3 and b(8)=A077232(8)=10 satisfy 10^2 - 11*3^2 = +1. See A077232 for further examples.
		

References

• T. Nagell, "Introduction to Number Theory", Chelsea Pub., New York, 1964, table p. 301.
• O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 26, p. 91 with explanation on pp. 94,95).

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• A. M. Legendre, Fractions les plus simples m/n qui satisfont à l'équation m^2 - an^2 =+-1 pour tout nombre non quarré a depuis 2 jusqu'à 1003, Essai sur la Théorie des Nombres An VI, Table XII. [_Paul Curtz_, Apr 10 2019]

Crossrefs

Cf. A000037, A003654, A003814, A033317, A077232.

Programs

Mathematica

nmax = 500;
nconv = 200; (* The number of convergents 'nconv' should be increased if the linear recurrence is not found for some terms. *)
nonSquare[n_] := n + Round[Sqrt[n]];
b[n_] := b[n] = Module[{lr}, lr = FindLinearRecurrence[ Numerator[ Convergents[ Sqrt[nonSquare[n]], nconv]]]; (1/2) SelectFirst[lr, #>1&]];
a[n_] := If[n == 1, 1, SelectFirst[{Sqrt[(b[n]^2 - 1)/nonSquare[n]], Sqrt[(b[n]^2 + 1)/nonSquare[n]]}, IntegerQ]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, nmax}] (* Jean-François Alcover, Mar 10 2021 *)

Formula

a(n)=sqrt((A077232(n)^2 - (-1)^(c(n)))/A000037(n)) with c(n)=1 if A000037(n)=A003654(k) for some k>=1 else c(n)=0.