cp's OEIS Frontend

If p is a prime, then A006497(p) == 3 (mod p).

This sequence contains the odd composite integers for which the congruence holds.

The generalized Pell-Lucas sequence of integer parameters (a,b) defined by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.

For a=3, b=-1, V(n) recovers the sequence A006497(n) (bronze Fibonacci numbers).

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity

V(p-J(p,D)) == 2*J(p,D) (mod p) when p is prime, b=-1 and D=a^2+4.

This sequence has the odd composite integers with V(m-J(m,D)) == 2*J(m,D) (mod m).

For a=3 and b=-1, we have D=13 and V(m) recovers A006497(m).

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy V(k*p-J(p,D)) == V(k-1)*J(p,D) (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.

The composite integers m with the property V(k*m-J(m,D)) == V(k-1)*J(m,D) (mod m) are called generalized Pell-Lucas pseudoprimes of level k- and parameter a.

Here b=-1, a=3, D=13 and k=2, while V(m) recovers A006497(m).

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy V(k*p-J(p,D)) == V(k-1)*J(p,D) (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.

The composite integers m with the property V(k*m-J(m,D)) == V(k-1)*J(m,D) (mod m) are called generalized Pell-Lucas pseudoprimes of level k- and parameter a.

Here b=-1, a=3, D=13 and k=3, while V(m) recovers A006497(m), with V(2)=11.

Table 1 of Andrica 2021 paper (p. 24), refers to A006497 as "bronze Lucas" numbers.

Also decimal expansion of the positive root of (x+1)^n - x^(2n). (x+1)^n - x^(2n) = 0 has only two real roots x1 = -(sqrt(5)-1)/2 and x2 = (sqrt(5)+1)/2 for all n > 0. - Cino Hilliard, May 27 2004

The golden ratio phi is the most irrational among irrational numbers; its successive continued fraction convergents F(n+1)/F(n) are the slowest to approximate to its actual value (I. Stewart, in "Nature's Numbers", Basic Books, 1997). - Lekraj Beedassy, Jan 21 2005

Let t=golden ratio. The lesser sqrt(5)-contraction rectangle has shape t-1, and the greater sqrt(5)-contraction rectangle has shape t. For definitions of shape and contraction rectangles, see A188739. - Clark Kimberling, Apr 16 2011

The golden ratio (often denoted by phi or tau) is the shape (i.e., length/width) of the golden rectangle, which has the special property that removal of a square from one end leaves a rectangle of the same shape as the original rectangle. Analogously, removals of certain isosceles triangles characterize side-golden and angle-golden triangles. Repeated removals in these configurations result in infinite partitions of golden rectangles and triangles into squares or isosceles triangles so as to match the continued fraction, [1,1,1,1,1,...] of tau. For the special shape of rectangle which partitions into golden rectangles so as to match the continued fraction [tau, tau, tau, ...], see A188635. For other rectangular shapes which depend on tau, see A189970, A190177, A190179, A180182. For triangular shapes which depend on tau, see A152149 and A188594; for tetrahedral, see A178988. - Clark Kimberling, May 06 2011

Given a pentagon ABCDE, 1/(phi)^2 <= (A*C^2 + C*E^2 + E*B^2 + B*D^2 + D*A^2) / (A*B^2 + B*C^2 + C*D^2 + D*E^2 + E*A^2) <= (phi)^2. - Seiichi Kirikami, Aug 18 2011

If a triangle has sides whose lengths form a geometric progression in the ratio of 1:r:r^2 then the triangle inequality condition requires that r be in the range 1/phi < r < phi. - Frank M Jackson, Oct 12 2011

The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - Clark Kimberling, Oct 19 2011

Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - Michel Lagneau, Dec 02 2011

Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - Michel Lagneau, Apr 17 2012

This is the case n=5 of (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)): (1+sqrt(5))/2 = (Gamma(1/5)/Gamma(3/5))*(Gamma(4/5)/Gamma(2/5)). - Bruno Berselli, Dec 14 2012

Also decimal expansion of the only number x>1 such that (x^x)^(x^x) = (x^(x^x))^x = x^((x^x)^x). - Jaroslav Krizek, Feb 01 2014

For n >= 1, round(phi^prime(n)) == 1 (mod prime(n)) and, for n >= 3, round(phi^prime(n)) == 1 (mod 2*prime(n)). - Vladimir Shevelev, Mar 21 2014

The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - Giovanni Zedda, Jun 22 2019

Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - Diego Rattaggi, Apr 17 2021

Given any complex p such that real(p) > -1, phi is the only real solution of the equation z^p+z^(p+1)=z^(p+2), and the only attractor of the complex mapping z->M(z,p), where M(z,p)=(z^p+z^(p+1))^(1/(p+2)), convergent from any complex plane point. - Stanislav Sykora, Oct 14 2021

The only positive number such that its decimal part, its integral part and the number itself (x-[x], [x] and x) form a geometric progression is phi, with respectively (phi -1, 1, phi) and a ratio = phi. This is the answer to the 4th problem of the 7th Canadian Mathematical Olympiad in 1975 (see IMO link and Doob reference). - Bernard Schott, Dec 08 2021

The golden ratio is the unique number x such that f(n*x)*c(n/x) - f(n/x)*c(n*x) = n for all n >= 1, where f = floor and c = ceiling. - Clark Kimberling, Jan 04 2022

In The Second Scientific American Book Of Mathematical Puzzles and Diversions, Martin Gardner wrote that, by 1910, Mark Barr (1871-1950) gave phi as a symbol for the golden ratio. - Bernard Schott, May 01 2022

Phi is the length of the equal legs of an isosceles triangle with side c = phi^2, and internal angles (A,B) = 36 degrees, C = 108 degrees. - Gary W. Adamson, Jun 20 2022

The positive solution to x^2 - x - 1 = 0. - Michal Paulovic, Jan 16 2023

The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025

This is the largest real zero x of (x^4 + x^2 + 1)^2 = 2*(x^8 + x^4 + 1). - Thomas Ordowski, May 14 2025

Cf. A000204 for Lucas numbers beginning with 1.

Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014

Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017

Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017

Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017

This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003

For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.

a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).

Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007

From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)

The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.

The first six multiplication formulas are:

L(2n) = L(n)^2 - 2*(-1)^n;

L(3n) = L(n)^3 - 3*(-1)^n*L(n);

L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;

L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);

L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.

Generally, L(n) | L(mn) if and only if m is odd.

In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)

From John Blythe Dobson, Nov 15 2007: (Start)

The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:

For a >= b and odd b, L(a + b) + L(a - b) = 5*F(a)*F(b).

For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).

For a >= b and odd b, L(a + b) - L(a - b) = L(a)*L(b).

For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).

A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)

From John Blythe Dobson, Nov 15 2007: (Start)

The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:

L(n) - L(n - 3) = 2*L(n - 2);

L(n) - L(n - 4) = 5*F(n - 2);

L(n) - L(n - 6) = 4*L(n - 3);

L(n) - L(n - 12) = 40*F(n - 6);

L(n) - L(n - 60) = 4160200*F(n - 30).

These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).

The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)

Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009

A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011

The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014

Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014

If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015

A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015

A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016

Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017

Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017

For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019

From Klaus Purath, Apr 19 2019: (Start)

While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:

First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.

Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.

Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)

L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019

If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019

From Kai Wang, Dec 18 2019: (Start)

L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).

Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)

From Peter Bala, Dec 23 2021: (Start)

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.

For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)

For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024

For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

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

Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003

a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003

Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004

a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006

a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006

Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007

Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011

These numbers could also be called "bronze Fibonacci numbers". Indeed, 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; analogously, for Pell numbers (A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013

Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013

The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014

a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014

For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

With offset 1 is the INVERTi transform of A001076. - Gary W. Adamson, Jul 24 2015

Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017

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)

Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018

Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020

From Michael A. Allen, Jan 25 2023: (Start)

Also called the 3-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 3 kinds of squares available. (End)

a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023