cp's OEIS Frontend

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

Equals A073133 with an additional column A(.,0).

If the first column and top row are deleted, antidiagonal reading yields A118243.

Adding a top row of 1's and antidiagonal reading downwards yields A157103.

Antidiagonal sums are 0, 1, 2, 5, 12, 32, 93, 297, 1035, 3911, 15917, 69350, ....

From Jianing Song, Jul 14 2018: (Start)

All rows have strong divisibility, that is, gcd(A(n,k_1), A(n,k_2)) = A(n,gcd(k_1,k_2)) holds for all k_1, k_2 >= 0.

Let E(n,m) be the smallest number l such that m divides A(n,l), we have: for odd primes p that are not divisible by n^2 + 4, E(n,p) divides p - ((n^2+4)/p) if p == 3 (mod 4) and (p - ((n^2+4)/p))/2 if p == 1 (mod 4). E(n,p) = p for odd primes p that are divisible by n^2 + 4. E(n,2) = 2 for even n and 3 for odd n. Here ((n^2+4)/p) is the Legendre symbol. A prime p such that p^2 divides T(n,E(n,p)) is called an n-Wall-Sun-Sun prime.

E(n,p^e) <= p^(e-1)*E(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then E(n,p^e) = p^(e-1)*E(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n,p), so E(n,p^e) = p^e as described above. E(n,m_1*m_2) = lcm(E(n,m_1),E(n,m_2)) if gcd(m_1,m_2) = 1.

Let pi(n,m) be the Pisano period of A(n, k) modulo m, i.e, the smallest number l such that A(n, k+1) == T(n,k) (mod m) holds for all k >= 0, we have: for odd primes p that are not divisible by n^2 - 4, pi(n,p) divides p - 1 if ((n^2+4)/p) = 1 and 2(p+1) if ((n^2+4)/p) = -1. pi(n,p) = 4p for odd primes p that are divisible by n^2 + 4. pi(n,2) = 2 even n and 3 for odd n.

pi(n,p^e) <= p^(e-1)*pi(n,p) for all primes p. If p^2 does not divide A(n, E(n,p)), then pi(n,p^e) = p^(e-1)*pi(n,p) for all exponent e. Specially, for primes p >= 5 that are divisible by n^2 + 4, p^2 is never divisible by A(n, p), so pi(n,p^e) = 4p^e as described above. pi(n,m_1*m_2) = lcm(pi(n,m_1), pi(n,m_2)) if gcd(m_1,m_2) = 1.

If n != 2, the largest possible value of pi(n,m)/m is 4 for even n and 6 for odd n. For even n, pi(n,p^e) = 4p^e; for odd n, pi(n,2p^e) = 12p^e, where p is any odd prime factor of n^2 + 4. For n = 2 it is 8/3, obtained by m = 3^e.

Let z(n,m) be the number of zeros in a period of A(n, k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: z(n,p) = 4 for odd primes p that are divisible by n^2 + 4. For other odd primes p, z(n,p) = 4 if E(n,p) is odd; 1 if E(n,p) is even but not divisible by 4; 2 if E(n,p) is divisible by 4; see the table below. z(n,2) = z(n,4) = 1.

Among all values of z(n,p) when p runs through all odd primes that are not divisible by n^2 + 4, we have:

((n^2+4)/p)...p mod 8....proportion of 1.....proportion of 2.....proportion of 4

......1..........1......1/6 (conjectured)...2/3 (conjectured)...1/6 (conjectured)*

......1..........5......1/2 (conjectured)...........0...........1/2 (conjectured)*

......1.........3,7.............1...................0...................0

.....-1.........1,5.............0...................0...................1

.....-1.........3,7.............0...................1...................0

* The result is that among all odd primes that are not divisible by n^2 + 4, 7/24 of them are with z(n,p) = 1, 5/12 are with z(n,p) = 2 and 7/24 are with z(n,p) = 4 if n^2 + 4 is a twice a square; 1/3 of them are with z(n,p) = 1, 1/3 are with z(n,p) = 2 and 1/3 are with z(n,p) = 4 otherwise. [Corrected by Jianing Song, Jul 06 2019]

z(n,p^e) = z(n,p) for all odd primes p; z(n,2^e) = 1 for even n and 2 for odd n, e >= 3.

(End)

From Michael A. Allen, Mar 06 2023: (Start)

Removing the first (n=0) row of A352361 gives this sequence.

Row n is the n-metallonacci sequence.

A(n,k) is (for k>0) the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)

Period of the sequence defined by reading A000129 modulo n.

Period of the sequence defined by reading A006190 modulo n.

Period of the sequence defined by reading A005668 modulo n.

Period of the sequence defined by reading A052918 modulo n.

a(79)=6240. [John W. Layman, Aug 10 2010]