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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

The identity (16*n + 15)^2 - (16*(n+1)^2 - 2*(n+1))*4^2 = 1 can be written as a(n)^2 - A158058(n+1)*4^2 = 1. - Vincenzo Librandi, Feb 01 2012

a(n-3), n >= 3, appears in the third column of triangle A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014

The companion array and triangle for the odd end numbers N(n, k) is given in A239127.

The two operations on natural numbers m used in the Collatz 3x+1 conjecture are here (following the M. Trümper paper given in the link) denoted by u for 'up' and d for 'down': u m = 3*m+1, if m is odd, and d m = m/2 if m is even. The present array gives all start numbers M(n, k) for the Collatz word (ud)^n = s^n (s = ud is useful because, except for the one letter word u, at least one d follows a letter u), with n >= 1, and k >= 1. Such Collatz sequences have the maximal number of u's (grow fastest).

This rectangular array is M of Example 2.2. with x=y = n, n >= 1, of the M. Trümper reference, pp. 7-8, written as a triangle by taking NE-SW diagonals. The Collatz sequence starting with M(n, k) has length 2*n+1 for each k and it ends in the odd number N(n, k) given in A239127.

The first row sequences of the array M (columns of triangle TM) are A004767, A004771, A125169, A239128, ...

Numbers that are congruent to {0, 1, 2, 3, 4, 5, 6} mod 8.

Numbers n such that binary expansion does not end 111.

Complement of A004771. - Michel Marcus, Sep 11 2015

Numbers such that the n-th triangular number is divisible by 4. - Charles R Greathouse IV, Apr 07 2011

Except for 0, numbers whose binary reflected Gray code (A014550) ends with 00. - Amiram Eldar, May 17 2021

A188172(a(n)) = n and A188172(m) <> n for m < a(n).

Only the numbers 2^m - 1 occur more than once.

a(A005843(n)) = 1; a(A016813(n)) = 3; a(A004771(n)) = 7; a(A008598(n) + 35) = 15; a(A008598(n) + 155) = 31. - Reinhard Zumkeller, Aug 23 2009

A181497(n) = smallest m such that A056753(m) = 2*n + 1. - Reinhard Zumkeller, Oct 24 2010