cp's OEIS Frontend

Number of ways of writing n as a product of primes.

Number of ways of writing n as a sum of distinct powers of 2.

Continued fraction for golden ratio A001622.

Partial sums of A000007 (characteristic function of 0). - Jeremy Gardiner, Sep 08 2002

An example of an infinite sequence of positive integers whose distinct pairwise concatenations are all primes! - Don Reble, Apr 17 2005

Binomial transform of A000007; inverse binomial transform of A000079. - Philippe Deléham, Jul 07 2005

A063524(a(n)) = 1. - Reinhard Zumkeller, Oct 11 2008

For n >= 0, let M(n) be the matrix with first row = (n n+1) and 2nd row = (n+1 n+2). Then a(n) = absolute value of det(M(n)). - K.V.Iyer, Apr 11 2009

The partial sums give the natural numbers (A000027). - Daniel Forgues, May 08 2009

From Enrique Pérez Herrero, Sep 04 2009: (Start)

a(n) is also tau_1(n) where tau_2(n) is A000005.

a(n) is a completely multiplicative arithmetical function.

a(n) is both squarefree and a perfect square. See A005117 and A000290. (End)

Also smallest divisor of n. - Juri-Stepan Gerasimov, Sep 07 2009

Also decimal expansion of 1/9. - Enrique Pérez Herrero, Sep 18 2009; corrected by Klaus Brockhaus, Apr 02 2010

a(n) is also the number of complete graphs on n nodes. - Pablo Chavez (pchavez(AT)cmu.edu), Sep 15 2009

Totally multiplicative sequence with a(p) = 1 for prime p. Totally multiplicative sequence with a(p) = a(p-1) for prime p. - Jaroslav Krizek, Oct 18 2009

n-th prime minus phi(prime(n)); number of divisors of n-th prime minus number of perfect partitions of n-th prime; the number of perfect partitions of n-th prime number; the number of perfect partitions of n-th noncomposite number. - Juri-Stepan Gerasimov, Oct 26 2009

For all n>0, the sequence of limit values for a(n) = n!*Sum_{k>=n} k/(k+1)!. Also, a(n) = n^0. - Harlan J. Brothers, Nov 01 2009

a(n) is also the number of 0-regular graphs on n vertices. - Jason Kimberley, Nov 07 2009

Differences between consecutive n. - Juri-Stepan Gerasimov, Dec 05 2009

From Matthew Vandermast, Oct 31 2010: (Start)

1) When sequence is read as a regular triangular array, T(n,k) is the coefficient of the k-th power in the expansion of (x^(n+1)-1)/(x-1).

2) Sequence can also be read as a uninomial array with rows of length 1, analogous to arrays of binomial, trinomial, etc., coefficients. In a q-nomial array, T(n,k) is the coefficient of the k-th power in the expansion of ((x^q -1)/(x-1))^n, and row n has a sum of q^n and a length of (q-1)*n + 1. (End)

The number of maximal self-avoiding walks from the NW to SW corners of a 2 X n grid.

When considered as a rectangular array, A000012 is a member of the chain of accumulation arrays that includes the multiplication table A003991 of the positive integers. The chain is ... < A185906 < A000007 < A000012 < A003991 < A098358 < A185904 < A185905 < ... (See A144112 for the definition of accumulation array.) - Clark Kimberling, Feb 06 2011

a(n) = A007310(n+1) (Modd 3) := A193680(A007310(n+1)), n>=0. For general Modd n (not to be confused with mod n) see a comment on A203571. The nonnegative members of the three residue classes Modd 3, called [0], [1], and [2], are shown in the array A088520, if there the third row is taken as class [0] after inclusion of 0. - Wolfdieter Lang, Feb 09 2012

Let M = Pascal's triangle without 1's (A014410) and V = a variant of the Bernoulli numbers A027641 but starting [1/2, 1/6, 0, -1/30, ...]. Then M*V = [1, 1, 1, 1, ...]. - Gary W. Adamson, Mar 05 2012

As a lower triangular array, T is an example of the fundamental generalized factorial matrices of A133314. Multiplying each n-th diagonal by t^n gives M(t) = I/(I-t*S) = I + t*S + (t*S)^2 + ... where S is the shift operator A129184, and T = M(1). The inverse of M(t) is obtained by multiplying the first subdiagonal of T by -t and the other subdiagonals by zero, so A167374 is the inverse of T. Multiplying by t^n/n! gives exp(t*S) with inverse exp(-t*S). - Tom Copeland, Nov 10 2012

The original definition of the meter was one ten-millionth of the distance from the Earth's equator to the North Pole. According to that historical definition, the length of one degree of latitude, that is, 60 nautical miles, would be exactly 111111.111... meters. - Jean-François Alcover, Jun 02 2013

Deficiency of 2^n. - Omar E. Pol, Jan 30 2014

Consider n >= 1 nonintersecting spheres each with surface area S. Define point p on sphere S_i to be a "public point" if and only if there exists a point q on sphere S_j, j != i, such that line segment pq INTERSECT S_i = {p} and pq INTERSECT S_j = {q}; otherwise, p is a "private point". The total surface area composed of exactly all private points on all n spheres is a(n)*S = S. ("The Private Planets Problem" in Zeitz.) - Rick L. Shepherd, May 29 2014

For n>0, digital roots of centered 9-gonal numbers (A060544). - Colin Barker, Jan 30 2015

Product of nonzero digits in base-2 representation of n. - Franklin T. Adams-Watters, May 16 2016

Alternating row sums of triangle A104684. - Wolfdieter Lang, Sep 11 2016

A fixed point of the run length transform. - Chai Wah Wu, Oct 21 2016

Length of period of continued fraction for sqrt(A002522) or sqrt(A002496). - A.H.M. Smeets, Oct 10 2017

a(n) is also the determinant of the (n+1) X (n+1) matrix M defined by M(i,j) = binomial(i,j) for 0 <= i,j <= n, since M is a lower triangular matrix with main diagonal all 1's. - Jianing Song, Jul 17 2018

a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j) for 1 <= i,j <= n (see Xavier Merlin reference). - Bernard Schott, Dec 05 2018

a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = tau(gcd(i,j)) for 1 <= i,j <= n (see De Koninck & Mercier reference). - Bernard Schott, Dec 08 2020

Sometimes also called lambda numbers.

Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.

Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002

a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1.

Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2*A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003

Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - Amarnath Murthy, Mar 22 2003

This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003

Number of 132-avoiding two-stack sortable permutations.

From Herbert Kociemba, Jun 02 2004: (Start)

For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.

Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)

Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004

Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. Wilson

Sums of antidiagonals of A038207 [Pascal's triangle squared]. - Ross La Haye, Oct 28 2004

The Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004

a(n) = sum of n-th row of triangle in A008288 = A094706(n) + A000079(n). - Reinhard Zumkeller, Dec 03 2004

Pell trapezoids (cf. A084158); for n > 0, A001109(n) = (a(n-1) + a(n+1))*a(n)/2; e.g., 1189 = (12+70)*29/2. - Charlie Marion, Apr 01 2006

(0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007

Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007

A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008

From Clark Kimberling, Aug 27 2008: (Start)

Related convergents (numerator/denominator):

lower principal convergents: A002315/A001653

upper principal convergents: A001541/A001542

intermediate convergents: A052542/A001333

lower intermediate convergents: A005319/A001541

upper intermediate convergents: A075870/A002315

principal and intermediate convergents: A143607/A002965

lower principal and intermediate convergents: A143608/A079496

upper principal and intermediate convergents: A143609/A084068. (End)

Equals row sums of triangle A143808 starting with offset 1. - Gary W. Adamson, Sep 01 2008

Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008

a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008

Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008

Starting with offset 1 = row sums of triangle A153345. - Gary W. Adamson, Dec 24 2008

From Charlie Marion, Jan 07 2009: (Start)

In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:

let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)

and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);

let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)

and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).

For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.

In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then

k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);

for example, if k=1 and n=3, then a(1,n) = a(n+1) and

1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;

1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;

b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;

b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.

Cf. A001333, A142238, A142239, A153313, A153314, A153315, A153316, A153317, A153318.

(End)

Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009

It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009

Equals eigensequence of triangle A154325. - Gary W. Adamson, Feb 12 2009

Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009

a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009

The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009

As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010

Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010

Numbers k such that 2*k^2 +- 1 is a square. - Vincenzo Librandi, Jul 18 2010

Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010

[u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010

An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010

Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011

Define a t-circle to be a first-quadrant circle 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 next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + (A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012

A001333 and A000129 give the diagonal numbers described by Theon from Smyrna. - Sture Sjöstedt, Oct 20 2012

Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013

a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013

Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./., .\., where . is an empty square. - Jean M. Morales, Sep 18 2013

Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013

An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013

Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014

a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014

For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015

A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017

a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017

Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017

Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017

Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018

(1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019

Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020

Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023

Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023

a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].

Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011

Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012

Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman

a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.

In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017

Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002

a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.

a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).

Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.

Binomial transform of A077957. - Paul Barry, Feb 25 2003

For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004

For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004

Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004

Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006

Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]

Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008

The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008

Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008

Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008

From Charlie Marion, Jan 07 2009: (Start)

In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:

let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);

let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).

For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.

In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then

k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);

for example, if k=1 and n=3, then b(1,n)=a(n+1) and

1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;

1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;

b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;

b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.

Cf. A000129, A142238-A142239, A153313-A153318.

(End)

This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010

Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010

a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010

Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010

From L. Edson Jeffery, Apr 04 2011: (Start)

Let U be the unit-primitive matrix (see [Jeffery])

U = U_(8,2) = (0 0 1 0)

(0 1 0 1)

(1 0 2 0)

(0 2 0 1).

Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)

(End)

For n >= 1, row sums of triangle

m/k.|..0.....1.....2.....3.....4.....5.....6.....7

==================================================

.0..|..1

.1..|..1.....2

.2..|..1.....2.....4

.3..|..1.....4.....4.....8

.4..|..1.....4....12.....8....16

.5..|..1.....6....12....32....16....32

.6..|..1.....6....24....32....80....32....64

.7..|..1.....8....24....80....80...192....64...128

which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012

a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012

The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012

Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012

This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012

a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016

a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018

For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020

For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021

a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022

From Greg Dresden, May 08 2023: (Start)

For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.

_

|| _

|||_|||

|_|

(End)

12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

Twice the odd numbers, also called singly even numbers.

Numbers having equal numbers of odd and even divisors: A001227(a(n)) = A000005(2*a(n)). - Reinhard Zumkeller, Dec 28 2003

Continued fraction for coth(1/2) = (e+1)/(e-1). The continued fraction for tanh(1/2) = (e-1)/(e+1) would be a(0) = 0, a(n) = A016825(n-1), n >= 1.

No solutions to a(n) = b^2 - c^2. - Henry Bottomley, Jan 13 2001

Sequence gives m such that 8 is the largest power of 2 dividing A003629(k)^m-1 for any k. - Benoit Cloitre, Apr 05 2002

k such that Sum_{d|k} (-1)^d = A048272(k) = 0. - Benoit Cloitre, Apr 15 2002

Also k such that Sum_{d|k} phi(d)*mu(k/d) = A007431(k) = 0. - Benoit Cloitre, Apr 15 2002

Also k such that Sum_{d|k} (d/A000005(d))*mu(k/d) = 0, k such that Sum_{d|k}(A000005(d)/d)*mu(k/d) = 0. - Benoit Cloitre, Apr 19 2002

Solutions to phi(x) = phi(x/2); primorial numbers are here. - Labos Elemer, Dec 16 2002

Together with 1, numbers that are not the leg of a primitive Pythagorean triangle. - Lekraj Beedassy, Nov 25 2003

For n > 0: complement of A107750 and A023416(a(n)-1) = A023416(a(n)) <> A023416(a(n)+1). - Reinhard Zumkeller, May 23 2005

Also the minimal value of Sum_{i=1..n+2} (p(i) - p(i+1))^2, where p(n+3) = p(1), as p ranges over all permutations of {1,2,...,n+2} (see the Mihai reference). Example: a(2)=10 because the values of the sum for the permutations of {1,2,3,4} are 10 (8 times), 12 (8 times) and 18 (8 times). - Emeric Deutsch, Jul 30 2005

Except for a(n)=2, numbers having 4 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005

A139391(a(n)) = A006370(a(n)) = A005408(n). - Reinhard Zumkeller, Apr 17 2008

Also a(n) = (n-1) + n + (n+1) + (n+2), so a(n) and -a(n) are all the integers that are sums of four consecutive integers. - Rick L. Shepherd, Mar 21 2009

The denominator in Pi/8 = 1/2 - 1/6 + 1/10 - 1/14 + 1/18 - 1/22 + .... - Mohammad K. Azarian, Oct 13 2011

This sequence gives the positive zeros of i^x + 1 = 0, x real, where i^x = exp(i*x*Pi/2). - Ilya Gutkovskiy, Aug 08 2015

Numbers k such that Sum_{j=1..k} j^3 is not a multiple of k. - Chai Wah Wu, Aug 23 2017

Numbers k such that Lucas(k) is a multiple of 3. - Bruno Berselli, Oct 17 2017

Also numbers k such that t^k == -1 (mod 5), where t is a term of A047221. - Bruno Berselli, Dec 28 2017

The even numbers form a ring, and these are the primes in that ring. Note that unique factorization into primes does not hold, since 60 = 2*30 = 6*10. - N. J. A. Sloane, Nov 11 2019

Also numbers ending with 10 in base 2. - John Keith, May 09 2022

Continued fraction expansion of sqrt(2) is 1 + 1/(2 + 1/(2 + 1/(2 + ...))).

Inverse binomial transform of Mersenne numbers A000225(n+1) = 2^(n+1) - 1. - Paul Barry, Feb 28 2003

A Chebyshev transform of 2^n: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)). - Paul Barry, Oct 31 2004

An inverse Catalan transform of A068875 under the mapping g(x)->g(x(1-x)). A068875 can be retrieved using the mapping g(x)->g(xc(x)), where c(x) is the g.f. of A000108. A040000 and A068875 may be described as a Catalan pair. - Paul Barry, Nov 14 2004

Sequence of electron arrangement in the 1s 2s and 3s atomic subshells. Cf. A001105, A016825. - Jeremy Gardiner, Dec 19 2004

Binomial transform of A165326. - Philippe Deléham, Sep 16 2009

Let m=2. We observe that a(n) = Sum_{k=0..floor(n/2)} binomial(m,n-2*k). Then there is a link with A113311 and A115291: it is the same formula with respectively m=3 and m=4. We can generalize this result with the sequence whose g.f. is given by (1+z)^(m-1)/(1-z). - Richard Choulet, Dec 08 2009

With offset 1: number of permutations where |p(i) - p(i+1)| <= 1 for n=1,2,...,n-1. This is the identical permutation and (for n>1) its reversal.

Equals INVERT transform of bar(1, 1, -1, -1, ...).

Eventual period is (2). - Zak Seidov, Mar 05 2011

Also decimal expansion of 11/90. - Vincenzo Librandi, Sep 24 2011

a(n) = 3 - A054977(n); right edge of the triangle in A182579. - Reinhard Zumkeller, May 07 2012

With offset 1: minimum cardinality of the range of a periodic sequence with (least) period n. Of course the range's maximum cardinality for a purely periodic sequence with (least) period n is n. - Rick L. Shepherd, Dec 08 2014

With offset 1: n*a(1) + (n-1)*a(2) + ... + 2*a(n-1) + a(n) = n^2. - Warren Breslow, Dec 12 2014

With offset 1: decimal expansion of gamma(4) = 11/9 where gamma(n) = Cp(n)/Cv(n) is the n-th Poisson's constant. For the definition of Cp and Cv see A272002. - Natan Arie Consigli, Sep 11 2016

a(n) equals the number of binary sequences of length n where no two consecutive terms differ. Also equals the number of binary sequences of length n where no two consecutive terms are the same. - David Nacin, May 31 2017

a(n) is the period of the continued fractions for sqrt((n+2)/(n+1)) and sqrt((n+1)/(n+2)). - A.H.M. Smeets, Dec 05 2017

Also, number of self-avoiding walks and coordination sequence for the one-dimensional lattice Z. - Sean A. Irvine, Jul 27 2020

a(2*n+1) with b(2*n+1) := A001077(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = -1, a(2*n) with b(2*n) := A001077(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = +1 (cf. Emerson reference).

Bisection: a(2*n+1) = T(2*n+1, sqrt(5))/sqrt(5) = A007805(n), n >= 0 and a(2*n) = 4*S(n-1,18), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. S(n,18)=A049660(n+1). - Wolfdieter Lang, Jan 10 2003

Apart from initial terms, this is the Pisot sequence E(4,17), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).

This is also the Horadam sequence (0,1,1,4), having the recurrence relation a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 1. a(n) / a(n-1) converges to 5^1/2 + 2 as n approaches infinity. 5^(1/2) + 2 can also be written as (2 * Phi) + 1 and Phi^2 + Phi. - Ross La Haye, Aug 18 2003

Numerators of continued fraction [4, 4, 4, ...], where the convergents to [4, 4, 4, ...] = (4/1, 17/4, 72/17, ...). Let X = the 2 X 2 matrix [0, 1; 1, 4]; then X^n = [a(n-1), a(n); a(n), a(n+1)]; e.g., X^3 = [4, 17; 17, 72]. Let C = the limit of a(n)/a(n-1) = 2 + sqrt(5) = 4.236067977...; then C^n = a(n+1) + (1/C)*a(n), where (1/C) = 0.236067977... . Example: C^3 = 76.01315556..., = 72 + 17*(0.2360679...). - Gary W. Adamson, Dec 15 2007, corrected by Greg Dresden, Sep 16 2019, corrected by Alex Mark, Jul 21 2020

Sqrt(5) = 4/2 + 4/17 + 4/(17*305) + 4/(305*5473) + 4/(5473*98209) + ... . - Gary W. Adamson, Dec 15 2007

a(p) == 20^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009

a(n) = A167808(3*n). - Reinhard Zumkeller, Nov 12 2009

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 4's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) (see also A033887). This fact can be proved similarly like the proof of Paul Barry's remark in A033887 by using the following scaling identity for delta-Fibonacci numbers: y^n b(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) b(k;x) and the fact that b(n;2) = (1-(-1)^n) 5^floor(n/2). - Roman Witula, Jul 12 2012

Binomial transform of 0, 1, 2, 8, 24, 80, 256, ... (A063727 with offset 1). - R. J. Mathar, Feb 05 2014

For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,4} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

With offset 1 is the INVERT transform of A006190: (1, 3, 10, 33, 109, 360, ...). - Gary W. Adamson, Jul 24 2015

From Rogério Serôdio, Mar 30 2018: (Start)

This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).

gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)

The initial 0 of this sequence is in contradiction with the fact that 0 is no valid denominator and according to all standard references, the first convergent of a continued fraction is p(0)/q(0) = b(0)/1 where b(0) is the first term of the continued fraction, given by the integer part of the number. One may artificially define q(-1) = 0 to have a recurrent relation q(n) = b(n)*q(n-1) + q(n-2), n >= 1, but then its index should be -1. - M. F. Hasler, Nov 01 2019

Number of 4-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020

From Michael A. Allen, Feb 15 2023: (Start)

Also called the 4-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 4 kinds of squares available. (End)

a(n) is the smallest nonnegative integer that is the sum of n, but no fewer, Fibonacci numbers including negative-index Fibonacci numbers (A039834), with that sum being a(n) = Sum_{i=0..n-1} A000045(3*i+1). a(n) is also the smallest nonnegative integer that is the sum of n, but no fewer, terms each of which is either a Fibonacci number or the negative of a Fibonacci number. (See A027941 for negatives disallowed.) - Mike Speciner, Oct 08 2023

From Enrique Navarrete, Dec 16 2023: (Start)

a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n > = 1, where P(k) = A006190(k) is the k-th 3-metallonacci number (see example below).

In general, the number of compositions with k-metallonacci number of parts is counted by the (k+1)-st metallonacci sequence (note k=1 and k=2 are the Fibonacci and the Pell numbers, respectively). (End).

a(n) is the number of tilings of a 2 X n rectangle missing the top right 1 X 1 cell, using 1 X 1 squares, dominoes and right trominoes. Compare to A110679 which is the same problem but without the missing top right cell. - Greg Dresden and Yilin Zhu, Jul 10 2025

Continued fraction for sqrt(3).

Also coefficient of the highest power of q in the expansion of the polynomial nu(n) defined by: nu(0)=1, nu(1)=b and for n>=2, nu(n)=b*nu(n-1)+lambda*(n-1)_q*nu(n-2) with (b,lambda)=(1,1), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

nu(0)=1 nu(1)=1; nu(2)=2; nu(3)=3+q; nu(4)=5+3q+2q^2; nu(5)=8+7q+6q^2+4q^3+q^4; nu(6)=13+15q+16q^2+14q^3+11q^4+5q^5+2q^6.

From Jaroslav Krizek, May 28 2010: (Start)

a(n) = denominators of arithmetic means of the first n positive integers for n >= 1.

See A026741(n+1) or A145051(n) - denominators of arithmetic means of the first n positive integers. (End)

From R. J. Mathar, Feb 16 2011: (Start)

This is a prototype of multiplicative sequences defined by a(p^e)=1 for odd primes p, and a(2^e)=c with some constant c, here c=2. They have Dirichlet generating functions (1+(c-1)/2^s)*zeta(s).

Examples are A153284, A176040 (c=3), A040005 (c=4), A021070, A176260 (c=5), A040011, A176355 (c=6), A176415 (c=7), A040019, A021059 (c=8), A040029 (c=10), A040041 (c=12). (End)

a(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = A000325(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012

For n > 0: denominators of row sums of the triangular enumeration of rational numbers A226314(n,k) / A054531(n,k), 1 <= k <= n; see A226555 for numerators. - Reinhard Zumkeller, Jun 10 2013

From Jianing Song, Nov 01 2022: (Start)

For n > 0, a(n) is the minimal gap of distinct numbers coprime to n. Proof: denote the minimal gap by b(n). For odd n we have A058026(n) > 0, hence b(n) = 1. For even n, since 1 and -1 are both coprime to n we have b(n) <= 2, and that b(n) >= 2 is obvious.

The maximal gap is given by A048669. (End)

The first 5821569425 terms were computed by Eric W. Weisstein on Sep 18 2011.

The first 10672905501 terms were computed by Eric W. Weisstein on Jul 17 2013.

The first 15000000000 terms were computed by Eric W. Weisstein on Jul 27 2013.

The first 30113021586 terms were computed by Syed Fahad on Apr 27 2021.

The metagolden rectangle is introduced here as a rectangle such that if a golden rectangle is removed from one end, the remaining rectangle is metagolden. That is, using mu to represent the metagolden ratio, mu - phi = 1/mu, so that when the phi X 1 golden rectangle is removed, the remaining rectangle is 1/mu X 1, the same metagolden shape as the original mu X 1 rectangle. - Geoffrey Caveney, Apr 20 2014

(Recall that a rectangle is golden if the removal of a square from one end leaves a rectangle having the same shape as the original.)

To generalize, suppose that a rectangle R of length L and width W is to be partitioned into a set of squares. One way to do this matches the ordinary continued fraction of the shape (defined as L/W). Explicitly, if L/W = [a(0), a(1), a(2), ...] (finite or infinite), then a(0) squares are removed in step 1, then a(1) squares in step 2, then a(2) squares, and so on. An example is the golden rectangle, of shape [1,1,1,...].

To generalize from squares to rectangles all having a common shape s >= 1, in such a manner that a(n) of these rectangles are removed at step k, the original rectangle must have shape given by the continued fraction [s*a(0), s*a(1), s*a(2), ...]. This result follows easily from the Euclidean algorithm for two positive real numbers (as in the paper cited below on golden triangles, generalized to match continued fractions). Here, the terms of the continued fraction need not be integers, and by Seidel's theorem of 1846, the continued fraction converges (e.g., Roberts, pages 97-105). We shall refer to the geometric partitioning of the original rectangle as the continued fraction partition (matching the continued fraction [s*a(0), s*a(1), s*a(2), ...]).

Thus, the constant at A136319 is the shape of a rectangle whose continued fraction partition is an infinite collection of golden rectangles.

a(2*n+1) with b(2*n+1) := A005667(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 10*a^2 = -1, a(2*n) with b(2*n) := A005667(2*n), n>=1, give all (positive integer) solutions to Pell equation b^2 - 10*a^2 = +1 (cf. Emerson reference).

Bisection: a(2*n)= 6*S(n-1,2*19) = 6*A078987(n-1), n >= 0 and a(2*n+1) = A097315(n), n >= 0, with S(n,x) Chebyshev's polynomials of the second kind. S(-1,x)=0. See A049310.

Sqrt(10) = 6/2 + 6/37 + 6/(37*1405) + 6/(1405*53353) + ... - Gary W. Adamson, Dec 21 2007

a(p) == 40^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009

For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 6's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

For n>=1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,6} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

From Michael A. Allen, Feb 15 2023: (Start)

Also called the 6-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 6 kinds of squares available. (End)