cp's OEIS Frontend

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002

For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006

(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]

n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003

For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003

a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004

Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005

Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005

Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005

Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005

One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006

Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006

Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006

If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007

If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006

If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007

a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007

For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011

Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009

a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009

Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010

If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010

In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then

for n > 0, b(n) = a(n)*k-a(n-1); e.g.,

for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1, 64 = 35*2 - 6, 373 = 204*2 - 35;

for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1; 99 = 35*3 - 6; 577 = 204*3 - 35;

for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;

for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.

See also A002315, A054488, A038761, A054489, A054490.

- Charlie Marion, Dec 08 2010

See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012

a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012

a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012

From Richard R. Forberg, Aug 30 2013: (Start)

The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:

a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;

a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)

For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015

Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015

The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and 2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016

a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016

Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017

In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021

Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.

Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571).

Members of Diophantine pairs. Solution to a*(a+1) = 2*b*(b+1) in natural numbers including 0; a = a(n), b = b(n) = A053141(n); The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).

The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.

The three sequences x (A001652), y (A046090) and z (A001653) may be obtained by setting u and v equal to the Pell numbers (A000129) in the formulas x = 2uv, y = u^2 - v^2, z = u^2 + v^2 [Joseph Wiener and Donald Skow]. - Antonio Alberto Olivares, Dec 22 2003

All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy, Aug 14 2006

Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n). - Kenneth J Ramsey, Sep 22 2007

In general, if b(n) = A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g., 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870 = 5741*14+84. - Charlie Marion, Nov 19 2007

Limit_{n -> oo} a(n)/a(n-1) = 3+2*sqrt(2) = A156035. - Klaus Brockhaus, Feb 17 2009

If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with pMohamed Bouhamida, Aug 29 2009

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = y^2 with pMohamed Bouhamida, Sep 02 2009

a(n+k) = A001541(k)*a(n) + A001542(k)*A001653(n+1) + A001108(k). - Charlie Marion, Dec 10 2010

The numbers 3*A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in A075841). - Bruno Berselli, Aug 26 2010

Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,

T(a(n) + 2*k*A001653(n+1)) = 2*T(A053141(n-1) + k*A002315(n)) + k^2 and

T(a(n) + (2*k+1)*A001653(n+1)) = (A001109(n+1) + k*A002315(n))^2 + k*(k+1).

Also (a(n) + k*A001653(n))^2 + (a(n) + k*A001653(n) + 1)^2 = (A001653(n+1) + k*A002315(n))^2 + k^2. - Charlie Marion, Dec 09 2010

For n>0, A143608(n) divides a(n). - Kenneth J Ramsey, Jun 28 2012

Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - Carmine Suriano, Dec 17 2013

The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1) = 1 and c(1) = 2, then b(n) = c(n-1) and c(n) = 2*c(n-1) + b(n-1). Alternatively, b(n) = c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - J. M. Bergot, Oct 09 2014

Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015

Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016

Numbers m such that m^2+(m+1)^2 is a square. - César Aguilera, Aug 14 2017

For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - Charlie Marion, Dec 08 2018

The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - Bernard Schott, Mar 10 2019

Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. - Steven Blasberg, Mar 04 2021

Limit_{n -> oo} b(n+1)/b(n) = 3+2*sqrt(2) for b = A155464, A155465, A155466.

Limit_{n -> oo} b(n)/b(n-1) = 3+2*sqrt(2) for b = A001652, A001653, A002315, A156156, A156157, A156158. - Klaus Brockhaus, Sep 23 2009

From Richard R. Forberg, Aug 14 2013: (Start)

Ratios b(n+1)/b(n) for all sequences of the form b(n) = 6*b(n-1) - b(n-2), for any initial values of b(0) and b(1), converge to this ratio.

Ratios b(n+1)/b(n) for all sequences of the form b(n) = 5*b(n-1) + 5*b(n-2) + b(n-3), for all b(0), b(1) and b(2) also converge to 3 + 2*sqrt(2). For example see A084158 (Pell Triangles).

Ratios of alternating values, b(n+2)/b(n), for all sequences of the form b(n) = 2*b(n-1) + b(n-2), also converge to 3 + 2*sqrt(2). These include A000129 (Pell Numbers). Also see A014176. (End)

Let ABCD be a square inscribed in a circle. When P is the midpoint of the arc AB, then the ratio (PC*PD)/(PA*PB) is equal to 3+2*sqrt(2). See the Mathematical Reflections link. - Michel Marcus, Jan 10 2017

Limit of ratios of successive terms of A001652 when n-> infinity. - Harvey P. Dale, Jun 16 2017; improved by Bernard Schott, Feb 28 2022

A quadratic integer with minimal polynomial x^2 - 6x + 1. - Charles R Greathouse IV, Jul 11 2020

Ratio between radii of the large circumscribed circle R and the small internal circle r drawn on the Sangaku tablet at Isaniwa Jinjya shrine in Ehime Prefecture (pictures in links). - Bernard Schott, Feb 25 2022

Named after the Newman-Shanks-Williams reference.

Also numbers k such that A125650(3*k^2) is an odd perfect square. Such numbers 3*k^2 form a bisection of A125651. - Alexander Adamchuk, Nov 30 2006

For positive n, a(n) corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs). - Lekraj Beedassy, Feb 06 2007

Also numbers m such that m^2 is a centered 16-gonal number; or a number of the form 8k(k+1)+1, where k = A053141(m) = {0, 2, 14, 84, 492, 2870, ...}. - Alexander Adamchuk, Apr 21 2007

The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators=A002315 and denominators=A001653. - Clark Kimberling, Aug 27 2008

The upper intermediate convergents to 2^(1/2) beginning with 10/7, 58/41, 338/239, 1970/1393 form a strictly decreasing sequence; essentially, numerators=A075870, denominators=A002315. - Clark Kimberling, Aug 27 2008

General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->oo} a(n) = x*(k*x+1)^n, k = (a(1)-3), x = (1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008

Numbers k such that (ceiling(sqrt(k*k/2)))^2 = (1+k*k)/2. - Ctibor O. Zizka, Nov 09 2009

A001109(n)/a(n) converges to cos^2(Pi/8) = 1/2 + 2^(1/2)/4. - Gary Detlefs, Nov 25 2009

The values 2(a(n)^2+1) are all perfect squares, whose square root is given by A075870. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010

a(n) represents all positive integers K for which 2(K^2+1) is a perfect square. - Neelesh Bodas (neelesh.bodas(AT)gmail.com), Aug 13 2010

For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(8)'s along the main diagonal, and i's along the superdiagonal and subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

Integers k such that A000217(k-2) + A000217(k-1) + A000217(k) + A000217(k+1) is a square (cf. A202391). - Max Alekseyev, Dec 19 2011

Integer square roots of floor(k^2/2 - 1) or A047838. - Richard R. Forberg, Aug 01 2013

Remark: x^2 - 2*y^2 = +2*k^2, with positive k, and X^2 - 2*Y^2 = +2 reduce to the present Pell equation a^2 - 2*b^2 = -1 with x = k*X = 2*k*b and y = k*Y = k*a. (After a proposed solution for k = 3 by Alexander Samokrutov.) - Wolfdieter Lang, Aug 21 2015

If p is an odd prime, a((p-1)/2) == 1 (mod p). - Altug Alkan, Mar 17 2016

a(n)^2 + 1 = 2*b(n)^2, with b(n) = A001653(n), is the necessary and sufficient condition for a(n) to be a number k for which the diagonal of a 1 X k rectangle is an integer multiple of the diagonal of a 1 X 1 square. If squares are laid out thus along one diagonal of a horizontal 1 X a(n) rectangle, from the lower left corner to the upper right, the number of squares is b(n), and there will always be a square whose top corner lies exactly within the top edge of the rectangle. Numbering the squares 1 to b(n) from left to right, the number of the one square that has a corner in the top edge of the rectangle is c(n) = (2*b(n) - a(n) + 1)/2, which is A055997(n). The horizontal component of the corner of the square in the edge of the rectangle is also an integer, namely d(n) = a(n) - b(n), which is A001542(n). - David Pasino, Jun 30 2016

(a(n)^2)-th triangular number is a square; a(n)^2 = A008843(n) is a subsequence of A001108. - Jaroslav Krizek, Aug 05 2016

a(n-1)/A001653(n) is the closest rational approximation of sqrt(2) with a numerator not larger than a(n-1). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. a(n-1)/A001653(n) < sqrt(2). - A.H.M. Smeets, May 28 2017

Consider the quadrant of a circle with center (0,0) bounded by the positive x and y axes. Now consider, as the start of a series, the circle contained within this quadrant which kisses both axes and the outer bounding circle. Consider further a succession of circles, each kissing the x-axis, the outer bounding circle, and the previous circle in the series. See Holmes link. The center of the n-th circle in this series is ((A001653(n)*sqrt(2)-1)/a(n-1), (A001653(n)*sqrt(2)-1)/a(n-1)^2), the y-coordinate also being its radius. It follows that a(n-1) is the cotangent of the angle subtended at point (0,0) by the center of the n-th circle in the series with respect to the x-axis. - Graham Holmes, Aug 31 2019

There is a link between the two sequences present at the numerator and at the denominator of the fractions that give the coordinates of the center of the kissing circles. A001653 is the sequence of numbers k such that 2*k^2 - 1 is a square, and here, we have 2*A001653(n)^2 - 1 = a(n-1)^2. - Bernard Schott, Sep 02 2019

Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i+4*n+2) - G(i))/(2*G(i+2*n+1)) as long as G(i+2*n+1) != 0. - Klaus Purath, Mar 25 2021

All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

3*a(n-1) is the n-th almost Lucas-cobalancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

In Moret-Blanc (1881) on page 259 some solution of m^2 - 2n^2 = -1 are listed. The values of m give this sequence, and the values of n give A001653. - Michael Somos, Oct 25 2023

From Klaus Purath, May 11 2024: (Start)

For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 6xy + y^2 = 8 = A028884(1). In general, the following applies to all sequences (t) satisfying t(i) = 6t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 6xy + y^2 = A028884(t(1)-6). This includes and interprets the Feb 04 2014 comment on A001541 by Colin Barker as well as the Mar 17 2021 comment on A054489 by John O. Oladokun and the Sep 28 2008 formula on A038723 by Michael Somos. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028884(t(1)-6) always applies.

If (t) is a sequence satisfying t(k) = 7t(k-1) - 7t(k-2) + t(k-3) or t(k) = 6t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) - t(k))/(t(k+n+1) - t(k+n)) always applies, as long as t(k+n+1) - t(k+n) != 0 for integer k and n >= 0. (End)

Chebyshev's T(n,x) polynomials evaluated at x=2.

x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).

Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002

For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002

This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]

a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003

a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006

The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008

The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008

a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009

Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012

Pisano period lengths: 1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012

Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014

Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014

A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.

M =

1, 1, 0, 0, 0, 0, ...

2, 0, 3, 0, 0, 0, ...

2, 0, 0, 3, 0, 0, ...

2, 0, 0, 0, 3, 0, ...

2, 0, 0, 0, 0, 3, ...

...

The triangle generated from M is:

1,

1, 1,

3, 1, 3,

11, 3, 3, 9,

41, 11, 9, 9, 27,

...

The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). - Gary W. Adamson, Jul 25 2016

Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016

For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016

From Timothy L. Tiffin, Oct 21 2016: (Start)

If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.

a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.

a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)

a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018

(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018

Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018

Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020

a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. - Benoit Cloitre, Aug 15 2002

Binomial transformation yields A081908, with A081908(0)=1 dropped. - R. J. Mathar, Oct 05 2008

1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). - Wolfdieter Lang, Mar 01 2013

a(n) is the number of election results for an election with n+2 candidates, say C1, C2, ..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. - Dennis P. Walsh, May 08 2013

This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) = A005248; at n=2, b(k) = 2*A001541; at n=3, b(k)= A057076; at n=4, b(k) = 2*A023039. This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) = A005248. The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also see A000032 for a related family of sequences. - Richard R. Forberg, Nov 22 2014

Also the number of maximum matchings in the n-gear graph. - Eric W. Weisstein, Dec 31 2017

Also the Wiener index of the n-dipyramidal graph. - Eric W. Weisstein, Jun 14 2018

Numbers of the form n^2+2 have no factors that are congruent to 7 (mod 8). - Gordon E. Michaels, Sep 12 2019

For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {n, 2n}]. - Magus K. Chu, Sep 10 2022

Consider the equation core(x) = core(2x+1) where core(x) is the smallest number such that x*core(x) is a square: solutions are given by a(n)^2, n > 0. - Benoit Cloitre, Apr 06 2002

Terms > 0 give numbers k which are solutions to the inequality |round(sqrt(2)*k)/k - sqrt(2)| < 1/(2*sqrt(2)*k^2). - Benoit Cloitre, Feb 06 2006

Also numbers m such that A125650(6*m^2) is an even perfect square, where A124650(m) is a numerator of m*(m+3)/(4*(m+1)*(m+2)) = Sum_{k=1..m} 1/(k*(k+1)*(k+2)). Sequence A033581 is a bisection of A125651. - Alexander Adamchuk, Nov 30 2006

The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators = A001541 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008

Even Pell numbers. - Omar E. Pol, Dec 10 2008

Numbers k such that 2*k^2+1 is a square. - Vladimir Joseph Stephan Orlovsky, Feb 19 2010

These are the integer square roots of the Half-Squares, A007590(k), which occur at values of k given by A001541. Also the numbers produced by adding m + sqrt(floor(m^2/2) + 1) when m is in A002315. See array in A227972. - Richard R. Forberg, Aug 31 2013

A001541(n)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n), and 2*a(n)/A001541(n) is the closest rational approximation of sqrt(2) with a numerator not larger than 2*a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations of sqrt(2) with restricted numerator as well as denominator. - A.H.M. Smeets, May 28 2017

Conjecture: Numbers k such that c/m < k for all natural a^2 + b^2 = c^2 (Pythagorean triples), a < b < c and a+b+c = m. Numbers which correspondingly minimize c/m are A002939. - Lorraine Lee, Jan 31 2020

All of the positive integer solutions of a*b + 1 = x^2, a*c + 1 = y^2, b*c + 1 = z^2, x + z = 2*y, 0 < a < b < c are given by a=a(n), b=A005319(n), c=a(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

b(0)=0, c(0)=1, b(i+1)=b(i)+c(i), c(i+1)=b(i+1)+b(i); then a(i) (the number in the sequence) is 2b(i)^2 if i is even, c(i)^2 if i is odd and b(n)=A000129(n) and c(n)=A001333(n). - Darin Stephenson (stephenson(AT)cs.hope.edu) and Alan Koch

For n > 1 gives solutions to A007913(2x) = A007913(x+1). - Benoit Cloitre, Apr 07 2002

If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X are in the sequence.

For n >= 2, a(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is A001109. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006

This is the r=8 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Also, 1^3 + 2^3 + 3^3 + ... + a(n)^3 = k(n)^4 where k(n) is A001109. - Anton Vrba (antonvrba(AT)yahoo.com), Nov 18 2006

If T_x = y^2 is a triangular number which is also a square, the least number which is both triangular and square and greater than T_x is T_(3*x + 4*y + 1) = (2*x + 3*y + 1)^2 (W. Sierpiński 1961). - Richard Choulet, Apr 28 2009

If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or (a+1) is a perfect square. If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or a/8 is a perfect square. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c, then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution, too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c < e, then (8*d^2, d*(f-b)) is a solution, too. - Mohamed Bouhamida, Aug 29 2009

If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with p < r, then r = 3p + 4q + 1 and s = 2p + 3q + 1. - Mohamed Bouhamida, Sep 02 2009

Also numbers k such that (ceiling(sqrt(k*(k+1)/2)))^2 - k*(k+1)/2 = 0. - Ctibor O. Zizka, Nov 10 2009

From Lekraj Beedassy, Mar 04 2011: (Start)

Let x=a(n) be the index of the associated triangular number T_x=1+2+3+...+x and y=A001109(n) be the base of the associated perfect square S_y=y^2. Now using the identity S_y = T_y + T_{y-1}, the defining T_x = S_y may be rewritten as T_y = T_x - T_{y-1}, or 1+2+3+...+y = y+(y+1)+...+x. This solves the Strand Magazine House Number problem mentioned in A001109 in references from Poo-Sung Park and John C. Butcher. In a variant of the problem, solving the equation 1+3+5+...+(2*x+1) = (2*x+1)+(2*x+3)+...+(2*y-1) implies S_(x+1) = S_y - S_x, i.e., with (x,x+1,y) forming a Pythagorean triple, the solutions are given by pairs of x=A001652(n), y=A001653(n). (End)

If P = 8*n +- 1 is a prime, then P divides a((P-1)/2); e.g., 7 divides a(3) and 41 divides a(20). Also, if P = 8*n +- 3 is prime, then 4*P divides (a((P-1)/2) + a((P+1)/2) + 3). - Kenneth J Ramsey, Mar 05 2012

Starting at a(2), a(n) gives all the dimensions of Euclidean k-space in which the ratio of outer to inner Soddy hyperspheres' radii for k+1 identical kissing hyperspheres is rational. The formula for this ratio is (1+3k+2*sqrt(2k*(k+1)))/(k-1) where k is the dimension. So for a(3) = 49, the ratio is 6 in the 49th dimension. See comment for A010502. - Frank M Jackson, Feb 09 2013

Conjecture: For n>1 a(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015

For n=2*k, k>0, a(n) is divisible by 8 (deficient), so since all proper divisors of deficient numbers are deficient, then a(n) is deficient. For n=2*k+1, k>0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. sigma(a(5)) = 1723 < 3362 = 2*a(5). In either case, a(n) is deficient. - Muniru A Asiru, Apr 14 2016

The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016

For n > 0, numbers for which the number of odd divisors of both n and of n + 1 is odd. - Gionata Neri, Apr 30 2018

a(n) will be solutions to some (A000217(k) + A000217(k+1))/2. - Art Baker, Jul 16 2019

For n >= 2, a(n) is the base for which A058331(A001109(n)) is a length-3 repunit. Example: for n=2, A001109(2)=6 and A058331(6)=73 and 73 in base a(2)=8 is 111. See Grantham and Graves. - Michel Marcus, Sep 11 2020

Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n).

The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).

Also the indices of triangular numbers that are half other triangular numbers [a of T(a) such that 2T(a)=T(b)]. The T(a)'s are in A075528, the T(b)'s are in A029549 and the b's are in A001652. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002

Sequences A053141 (this entry), A016278, A077259, A077288 and A077398 are part of an infinite series of sequences. Each depends upon the polynomial p(n) = 4k*n^2 + 4k*n + 1, when 4k is not a perfect square. Equivalently, they each depend on the equation k*t(x)=t(z) where t(n) is the triangular number formula n(n+1)/2. The dependencies are these: they are the sequences of positive integers n such that p(n) is a perfect square and there exists a positive integer m such that k*t(n)=t(m). A053141 is for k=2, A016278 is for k=3, A077259 is for k=5. - Robert Phillips (bobanne(AT)bellsouth.net), Oct 11 2007, Nov 27 2007

Jason Holt observes that a pair drawn from a drawer with A053141(n)+1 red socks and A001652(n) - A053141(n) blue socks will as likely as not be matching reds: (A053141+1)*A053141/((A001652+1)*A001652) = 1/2, n>0. - Bill Gosper, Feb 07 2010

The values x(n)=A001652(n), y(n)=A046090(n) and z(n)=A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)=A011900(n) and c(n)=A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009

Behera & Panda call these the balancers and A001109 are the balancing numbers. - Michel Marcus, Nov 07 2017

Two times Chebyshev polynomials of the first kind evaluated at 3.

Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003

Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003

Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004

Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009

It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - Kenneth J Ramsey, Feb 14 2012 and Mar 05 2012

For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - Kival Ngaokrajang, Sep 04 2013

Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - Colin Barker, Feb 08 2014