1,2

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 a(n) and b(n) numbers for the +1 option are 2*a(n)^2+1 and 2*a(n)*b(n), respectively (see Perron I, pp. 94,95).

If d(n)=A000037(n)= k^2+1, k=1,2,.., then the a^2 - d(n)*b^2 = -1 Pell equation has the minimal solution a(n)=k and b(n)=1. If d(n)=A000037(n)= k^2-1, k=2,3,..., then the a^2 - d(n)*b^2 = +1 Pell equation has the minimal solution a=k and b=1.

The general integer solutions (up to signs) of Pell equation a^2 - d(n)*b^2 = +1 with d(n)=A000037(n), but not from A003654, are a(n,p)= T(p+1,a(n)) and b(n,p)= b(n)*S(p,2*a(n)), p=0,1,... If d(n)=A000037(n) is also from A003654 then these solutions are a(n,p)= T(p+1,2*a(n)^2+1) and b(n,p)= 2*a(n)*b(n)*S(p,2*(2*a(n)^2+1)), p=0,1,... Here T(n,x), resp. S(n,x) := U(n,x/2), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310.

The general integer solutions (up to signs) of the Pell equation a^2 - d(n)*b^2 = -1 with d(n)=A000037(n)= A003654(k), for some k>=1, are a(n,p) = a(n)*(S(n,2*(2*a(n)^2)+1) + S(n-1,2*(2*a(n)^2)+1)) and b(n,p) = b(n)*(S(n,2*(2*a(n)^2)+1) - S(n-1,2*(2*a(n)^2)+1)) with the S(n,x) := U(n,x/2) Chebyshev polynomials. S(-1,x) := 0.

If the trivial solution x=1, y=0 is included, the sequence becomes A006702. - T. D. Noe, May 17 2007

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).

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]

a(n) = sqrt(A000037(n)*A077233(n)^2 + (-1)^(c(n))) with c(n)=1 if A000037(n)=A003654(k) for some k>=1 else c(n)=0.

d=10=A000037(7)=A003654(3), therefore a(7)^2=10*b(7)^2 -1, i.e. 3^2=10*1^2 -1 and 2*a(7)^2+1=19 and 2*a(7)*b(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 a^2 - d*b^2 = -1 Pell equation and a(8)=10 and b(8)=A077233(8)=3 satisfy 10^2 - 11*3^2 = +1.

10=d(7)=A000037(7)=A003654(3)=3^2+1 hence a(7)=3 and b(7)=1 are the smallest numbers satisfying a^2-10*b^2=-1.

8=d(6)=A000037(6)=3^2-1 (not in A003654) hence a(6)=3 and b(6)=1 are the smallest numbers satisfying a^2-8*b^2=+1.

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]];

a[n_] := a[n] = Module[{lr}, lr = FindLinearRecurrence[ Numerator[ Convergents[ Sqrt[nonSquare[n]], nconv]]]; (1/2) SelectFirst[lr, #>1&]];

Table[Print[n, " ", a[n]]; a[n], {n, 1, nmax}] (* Jean-François Alcover, Mar 10 2021 *)

(Python)

from math import isqrt

from sympy.solvers.diophantine.diophantine import diop_DN

def A077232(n):

d = n+(k:=isqrt(n))+(n>k*(k+1))

return min(a[0] for a in diop_DN(d, -1)+diop_DN(d, 1)) # Chai Wah Wu, Dec 21 2025

Cf. A000037, A003654, A003814, A006702, A033313, A077233.

Sequence in context: A265819 A254746 A011021 * A193891 A193906 A224791

Adjacent sequences: A077229 A077230 A077231 * A077233 A077234 A077235

nonn,nice

Wolfdieter Lang, Nov 08 2002

approved