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A078986(n+1)^2 - 10*(6*a(n))^2 = +1, n>=0 (Pell equation +1, see A033313 and A033317).

a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,37}. - Milan Janjic, Jan 26 2015

Equally, value of D for incrementally largest values of minimal y satisfying Pell equation x^2-Dy^2=1.

Values of n where A002349 (or A002350) sets a new record.

D = D(n) = A000037(n). - Wolfdieter Lang, Oct 04 2015

a(n+1)^2 - 10*(6*A078987(n))^2 = 1, n >= 0 (Pell equation +1, see A033313 and A033317).

Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(10). - Benoit Cloitre, Feb 14 2004

Numbers n such that 10*(n^2 - 1) is a square. - Vincenzo Librandi, Aug 08 2010

For the fundamental positive solution x(n)^2 - a(n)*y(n)^2 = 2 see (x(n) = A261247(n), y(n) = A261248(n)), for n >= 1.

Conjecture: The sequence consists of all numbers D not a square and even D = 2*d has odd d with prime factors of the form 1 or 7 (mod 8). Odd D has prime factors of the form 1 or 7 (mod 8) but there is an odd number of primes of the form 7 (mod 8). The following will prove that these conditions for D are necessary in order to have solutions.

This conjecture is false. For the odd D case see the counterexamples in A263010, and for the even D in A264352. - Wolfdieter Lang, Nov 12 2015

If there is a solution for D, D not a square, then only one class of solution exists due to Nagell's Theorem 110, p. 208, because then 2 divides 2*D. All solutions will be proper because 2 is a prime.

For the even prime D = p = 2 the positive fundamental solution is [x(1) = 2, y(1) = 1].

For odd primes D = p there can be solutions only for p == +7 (mod 8), that is p from A007522. Then x and y are both odd. Proof: Consider a solution of x^2 - p*y^2 = 2. The parities of x and y have to be either even and even or odd and odd. For odd x one has x^2 == +1 (mod 8) (because x^2 = 8*T(X) + 1 with x = 2*X+1 and the triangular numbers T = A000217); similarly for y^2 if y is odd. In the even-even case x^2 and y^2 are both congruent to 4 (mod 8). The even-even case leads to 4 - 4*p = 2 (mod 8), excluding all odd p, namely p == 1, 3, 5, 7 (mod 8). The odd-odd case is 1 - p*1 = 2 (mod 8), and p == 1, 3, 5 (mod 8) are excluded. Therefore, only p == 7 (mod 8) qualifies for a solution, and then x and y will be both odd.

For D = p == 7 (mod 8) from A007522 one can test if there exists a fundamental positive solution (at most one class can exist, therefore there is either no solution or just one) [2*U(p)+1, 2*V(p)+1] by checking the two inequalities (see Nagell, eq. (4) and (5), p. 206) 0 <= V(p) < floor((Y(p)/sqrt(X(p) + 1) - 1)/2) and 0 <= U(p) <= floor((sqrt(X(p) + 1) - 1)/2), with the positive fundamental solution [X(p), Y(p)] of X^2 - p*Y^2 = +1. These solutions can be found in (A033313(k), A033317(k)) if A000037(k) is the prime p == 7 (mod 8) one is testing.

For composite even D there are solutions only if D/2 is odd. Proof: If D is even then x has to be even, hence x^2 == 0 (mod 4) and then D*y^2 == -2 (mod 4), hence D cannot be 0 (mod 4). Thus an even D can only be of the form D = 2*d with d odd. The modulo 3 and modulo 5 argument used in the next case will show that d can have only prime factors of the form +1 or -1 (mod 8).

For composite odd D one finds like above that the even-even x and y case is excluded, and the odd-odd case needs D == -1 (mod 8) == 7 (mod 8). Hence a candidate for D is from A004771 - A007522. D cannot have any prime factor p of the form 3 or 5 (mod 8) because otherwise x^2 == 2 (mod p), but the Legendre symbol (2/p) = -1 for such p's (see, e.g., Nagell, Theorem 81, p. 136). For example, D = 15 = 3*5 cannot have a solution. Thus the only candidates for D have prime factors p of the form +1 or +7 (mod 8), with the number of the latter ones being odd. E.g., D = 7*17 = 119 qualifies as a candidate and it has indeed solutions, namely the ones obtainable from the fundamental one [11, 1].

The general proper positive solutions for D(n) = a(n) are obtained from the fundamental ones [x(n), y(n)] given in A261247 and A261248 with the help of powers of the matrix M(n) = [[u(n), D(n)*v(n)], [v(n), u(n)]], where u(n) and v(n) are the positive fundamental solutions of U(n) - D(n)*V(n) = 1, by (x(n; k), y(n; k))^T = M(n)^k (x(n), y(n))^T (T for transposed), for k >= 0. [u(n), v(n)] = [A033313(j(n)), A033317(j(n))] if A000037(j(n)) = D(n) = a(n).

Observation: All degrees (7, 47, 79, 103, 119, 127) of the modular equations derived for solving Ramanujan's question 699 by Galkin & Kozirev (see reference and A318732) are terms of this sequence. - Hugo Pfoertner, Sep 24 2023

These are the nonsquare odd numbers D that admit proper solutions (x, y) to the generalized Pell equation x^2 - D*y^2 = +8 with both x and y odd. They are given by D == 1 (mod 8), not a square, no prime factors 3 or 5 (mod 8) in the composite case (see A263011), and they are not exceptional values which are given in A264348. Up to the number 2000 these exceptional values are 257, 401, 577, 697, 761, 1009, 1129, 1297, 1393, 1489, 1601, 1897. [sequence reference corrected by Peter Munn, Jun 19 2020]

The corresponding positive proper fundamental solutions (x1(D), y1(D)) for the first class are given in A264349 and A264350. There always seem to be two conjugacy classes. The positive proper fundamental solution of the second class (x2, y2) is, for given D, obtained by applying the matrix M(D) = matrix[[x0(D), D*y0(D)],[y0(D), x0(D)]] on (x1(D), -y1(D))^T (T for transposed). Here (x0(D), y0(D)) is the positive fundamental solution of the Pell equation x^2 - D*y^2 = +1 (which is always proper). See the appropriate entries of A033313 and A033317 for these solutions. There would be only one class (the ambiguous case) if this application of M(D) would lead to (x1(D), y1(D))^T. This does not seem to happen. The positive proper fundamental solutions (x2(D), y2(D)) of the second class are given in A264351 and A264353.

The case of odd D with both y and x even leads to improper solutions obtained from the +2 Pell equation (see A261246), e.g., D = 7 has the fundamental positive improper solution (6, 2) = 2*(3, 1) obtained from the proper solution (3, 1) of x^2 - 7*y^2 = +2 (see A261247(2) and A261248(2)). There is only one class of solutions (ambiguous case).

The case of even D with y odd and x even needs D == 0 (mod 4). See 4*A261246 = A264354 for the even D values that admit proper solutions. There appear one or two classes of solutions in this case.

The improper solutions with even D and both x and y even, come from X^2 - D*Y2 = +2 which needs D/2 odd without prime factors 3 or 5 (mod 8) in the composite case. Such D values that do not admit a solution are called exceptional and are given by A264352.

This is a proper subsequence of A263011.

The i-th solution pair V(i) = [x(i), y(i)] to the Pellian x^2 - D*y^2 = 1 for a given least solution x(1) = n may be generated through the recurrence V(i+2) = 2*n*V(i+1) - V(i) taking V(0) = [1, 0] and V(1) = [n, sqrt((n^2-1)/a(n))]. V(i) stands for the numerator and denominator of the 2i-th convergent of the continued fraction expansion of sqrt(D).

Thus setting n = 3, for instance, we have D = a(3) = 2 and V(1) = [3, 2] so that along with V(0) = [1, 0] recurrence V(i+2) = 6*V(i+1) - V(i) generates [A001333(2k), A000129(2k)]. Similarly, setting n = 9 generates [A023039, A060645], respectively the numerator and denominator of the 2i-th convergent of sqrt(a(9)), i.e., sqrt(5). - Lekraj Beedassy, Feb 26 2002

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

2*a(n) = y0(n) is the positive fundamental solution satisfying the Pell equation x0(n)^2 + D(n)*y0(n)^2 = +1 with D(n) coinciding apparently with Conway's rectangular numbers r(n) = A007969(n). The corresponding x0 values are given in A262024.

For a proof of this coincidence see the W. Lang link under A007969. - Wolfdieter Lang, Oct 04 2015