cp's OEIS Frontend

Molien series of 2-dimensional representation of cyclic group of order 3 over GF(2).

One step back, two steps forward.

The crossing number of the graph C(n, {1,3}), n >= 8, is [n/3] + n mod 3, which gives this sequence starting at the first 4. [Yang Yuansheng et al.]

A Chebyshev transform of A078008. The g.f. is the image of (1-x)/(1-x-2*x^2) (g.f. of A078008) under the Chebyshev transform A(x)-> (1/(1+x^2))*A(x/(1+x^2)). - Paul Barry, Oct 15 2004

A047878 is an essentially identical sequence. - Anton Chupin, Oct 24 2009

Rhyme scheme of Dante Alighieri's "Divine Comedy." - David Gaita, Feb 11 2011

A194960 results from deleting the first four terms of A008611. Note that deleting the first term or first four terms of A008611 leaves a concatenation of segments (n, n+1, n+2); for related concatenations, see

A008619, (n,n+1) after deletion of first term;

A053737, (n,n+1,n+2,n+3) beginning with n=0;

A053824, (n to n+4) beginning with n=0. - Clark Kimberling, Sep 07 2011

It appears that a(n) is the number of roots of x^(n+1) + x + 1 inside the unit circle. - Michel Lagneau, Nov 02 2012

Also apparently for n >= 2: a(n) is the largest remainder r that results from dividing n+2 by 1..n+2 more than once, i.e., a(n) = max(i, A072528(n+2,i)>1). - Ralf Stephan, Oct 21 2013

Number of n-element subsets of [n+1] whose sum is a multiple of 3. a(4) = 1: {1,2,4,5}. - Alois P. Heinz, Feb 06 2017

It appears that a(n) is the number of roots of the Fibonacci polynomial F(n+2,x) strictly inside the unit circle of the complex plane. - Michel Lagneau, Apr 07 2017

For the proof of the preceding conjecture see my comments under A008615 and A049310. Chebyshev S(n,x) = i^n*F(n+1,-i*x), with i = sqrt(-1). - Wolfdieter Lang, May 06 2017

The sequence is the interleaving of three sequences: the positive integers (A000027), the nonnegative integers (A001477), and the positive integers, in that order. - Guenther Schrack, Nov 07 2020

a(n) is the number of multiples of 3 between n and 2n. - Christian Barrientos, Dec 20 2021

a(n) is the least number of football games a team has to play to be able to get n-1 points, where a win is 3 points, a draw is 1 point, and a loss is 0 points. - Sigurd Kittilsen, Dec 01 2022

a(n) = 3^n*b(n;2/3) = -b(n;-2), but we have 3^n*a(n;2/3) = F(3n+1) = A033887 and a(n;-2) = F(3n-1) = A015448, where a(n;d) and b(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (the argument "d" of a(n;d) and b(n;d) is abbreviation of the symbol "delta") defined by the following equivalent relations: (1 + d*((sqrt(5) - 1)/2))^n = a(n;d) + b(n;d)*((sqrt(5) - 1)/2) equiv. a(0;d)=1, b(0;d)=0, a(n+1;d) = a(n;d) + d*b(n;d), b(n+1;d) = d*a(n;d) + (1-d)b(n;d) equiv. a(0;d)=a(1;d)=1, b(0;1)=0, b(1;d)=d, and x(n+2;d) + (d-2)*x(n+1;d) + (1-d-d^2)*x(n;d) = 0 for every n=0,1,...,d, and x=a,b equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k-1)*(-d)^k, and b(n;d) = Sum_{k=0..n} C(n,k)*(-1)^(k-1)*F(k)*d^k equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k+1)*(1-d)^(n-k)*d^k, and b(n;d) = Sum_{k=1..n} C(n;k)*F(k)*(1-d)^(n-k)*d^k. The sequences a(n;d) and b(n;d) for special values d are connected with many known sequences: A000045, A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A163073 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012

For any odd k, Fibonacci(k*n) = sqrt(Fibonacci((k-1)*n) * Fibonacci((k+1)*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012

The ratio of consecutive terms approaches the continued fraction 4 + 1/(4 + 1/(4 +...)) = A098317. - Hal M. Switkay, Jul 05 2020

In general, sequences with recurrence a(n) = k*a(n-1) + a(n-2) and a(0)=1 (and a(-1)=0) have the generating function 1/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n) = b(5n), a(3n+1) = b(5n+3), a(3n+2) = 2*b(5n+4) where b(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n) is the sequence of denominators of continued fraction convergents to sqrt(k^2/4+1).]

a(p-1) == 53^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]

From Johannes W. Meijer, Jun 12 2010: (Start)

For the sequence given above k=7 which implies that it is associated with A041091.

For a similar statement about sequences with recurrence a(n) = k*a(n-1) + a(n-2) but with a(0) = 2, and a(-1) = 0, see A086902; a sequence that is associated with A041090.

For more information follow the Khovanova link and see A087130, A140455 and A178765.

(End)

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 7's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

a(n) equals the number of words of length n on alphabet {0,1,...,7} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015

From Michael A. Allen, Feb 21 2023: (Start)

Also called the 7-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.

a(n) 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 7 kinds of squares available. (End)

With different sign pattern, see A000748.

Conjecture: Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n) - A057682(n)*M + z(n)*M^2, where z(0) = z(1) = 0 and, apparently, z(n+2) = a(n). - Stanislav Sykora, Jun 10 2012

Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.

Apart from signs, identical to A053122.

Coefficient array for Morgan-Voyce polynomial B(n,x); see A085478 for references. - Philippe Deléham, Feb 16 2004

T(n,k) is the number of compositions of n having k parts when there are q kinds of part q (q=1,2,...). Example: T(4,2) = 10 because we have (1,3),(1,3'),(1,3"), (3,1),(3',1),(3",1),(2,2),(2,2'),(2',2) and (2',2'). - Emeric Deutsch, Apr 09 2005

T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008

This sequence is jointly generated with A085478 as a triangular array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = u(n-1,x) + x*v(n-1)x and v(n,x) = u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012

Concerning Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016

Subtriangle of the triangle T(n,k), 0 <= k <= n, read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 27 2012

From Wolfdieter Lang, Aug 30 2012: (Start)

With offset [0,0] the triangle with entries R(n,k) = T(n+1,k+1):= binomial(n+k+1, 2*k+1), n >= k >= 0, and zero otherwise, becomes the Riordan lower triangular convolution matrix R = (G(x)/x, G(x)) with G(x):=x/(1-x)^2 (o.g.f. of A000027). This means that the o.g.f. of column number k of R is (G(x)^(k+1))/x. This matrix R is the inverse of the signed Riordan lower triangular matrix A039598, called in a comment there S.

The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598.

For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)

From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014

For 1 <= k <= n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01. - Milan Janjic, Dec 20 2016

The infinite sum (Sum_{i >= 0} (T(s+i,1+i) / 2^(s+2*i)) * zeta(s+1+2*i)) = 1 allows any zeta(s+1) to be expressed as a sum of rational multiples of zeta(s+1+2*i) having higher arguments. For example, zeta(3) can be expressed as a sum involving zeta(5), zeta(7), etc. The summation for each s >= 1 uses the s-th diagonal of the triangle. - Robert B Fowler, Feb 23 2022

The convolution triangle of the nonnegative integers. - Peter Luschny, Oct 07 2022

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, x*(1+x)). - Paul Barry, Oct 17 2005

Central coefficients of rows with odd numbers of term are A052227.

From Wolfdieter Lang, Jun 26 2011: (Start)

T(k,s) appears as T_s(k) in the Knuth reference, p. 285.

This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s) - T(k-1,s) = A156308(k,s), k>=s>=1 (identity on p. 286). T(k,s) = A156308(k+1,s+1) - A156308(k,s+1), k>=s>=0 (identity on p. 286).

(End)

A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)= v(1,x)= 1; for n>1, u(n,x)= u(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,x) +1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,-1,-2,-1). - Clark Kimberling, Feb 28 2012

This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of F-numbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = Sum_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), k >= 0, n >= 0. - Wolfdieter Lang, Aug 24 2012

From Wolfdieter Lang, Oct 18 2012: (Start)

This triangle T(k,s) appears in the formula x^(2*k+1) - x^(-(2*k+1)) = Sum_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).

Alternating row sums are A057079.

The Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)

The signed triangle ((-1)^(k-s))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x. - Wolfdieter Lang, Oct 23 2012

From Wolfdieter Lang, Oct 04 2013: (Start)

The signed triangle S(k,s) := ((-1)^(k-s))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))-gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) = 2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:

s(4*(k+1)) = Sum_{s=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).

This is to be computed modulo C(4*(k+1), rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular n-gon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)-gon. (End)

From Wolfdieter Lang, Aug 14 2014: (Start)

The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = Sum_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.

This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (-1)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (-1)^k*(2*k+1). Note that the recurrence derived from the Riordan A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), k >= s >=1.

(End)

From Tom Copeland, Nov 07 2015: (Start)

Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.

For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 20 and 21) and Damianou and Evripidou (p. 7).

As suggested by the equations on p. 7 of Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., n*x)/x, where A(n,x) are the polynomials of A127677 and F(n, ...) are the Faber polynomials of A263196. Cf. A127672 and A127677.

(End)

The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by

P(k, x) = R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 16's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011

For n>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,15}. - Milan Janjic, Jan 23 2015

Binomial transform of A063727, and second binomial transform of (1,1,5,5,25,25,...), which is A074872 with offset 0. - Paul Barry, Jul 16 2003

Equals INVERT transform of A104934: (1, 2, 8, 28, 100, 356, ...) and INVERTi transform of A005054: (1, 4, 20, 100, 500, ...). - Gary W. Adamson, Jul 22 2010

a(n) is the number of compositions of n when there are 3 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010

F(3*n+1) = 3^n*a(n;2/3), where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers by Witula et al.). - Roman Witula, Jul 12 2012

We note that the remark above by Paul Barry can be easily obtained from the following scaling identity for delta-Fibonacci numbers y^n a(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) a(k;x) and the fact that a(n;2)=5^floor(n/2). Indeed, for x=y=2 we get 2^n a(n;1) = Sum_{k=0..n} binomial(n,k) a(k;2) and, by A000045: Sum_{k=0..n} binomial(n,k) 2^k a(k;1) = Sum_{k=0..n} binomial(n,k) F(k+1) 2^k = 3^n a(n;2/3) = F(3n+1). - Roman Witula, Jul 12 2012

Except for the first term, this sequence can be generated by Corollary 1 (iv) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015

Number of 1’s in the substitution system {0 -> 110, 1 -> 11100} at step n from initial string "1" (1 -> 11100 -> 111001110011100110110 -> ...). - Ilya Gutkovskiy, Apr 10 2017

The o.g.f. of {F(m*n + 1)}A000045%20and%20L%20=%20A000032.%20-%20_Wolfdieter%20Lang">{n>=0}, for m = 1, 2, ..., is G(m,x) = (1 - F(m-1)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and L = A000032. - _Wolfdieter Lang, Feb 06 2023

Number of domino tilings in K_3 X P_2n (or in S_4 X P_2n).

Number of perfect matchings in graph C_{3} X P_{2n}.

Number of perfect matchings in S_4 X P_2n.

In general, Sum_{k=0..n} binomial(2*n-k, k)*j^(n-k) = (-1)^n * U(2*n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005

a(n) = L(n,5), where L is defined as in A108299; see also A030221 for L(n,-5). - Reinhard Zumkeller, Jun 01 2005

Number of 01-avoiding words of length n on alphabet {0,1,2,3,4} which do not end in 0 (e.g., at n=2, we have 02, 03, 04, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44). - Tanya Khovanova, Jan 10 2007

(sqrt(21)+5)/2 = 4.7912878... = exp(arccosh(5/2)) = 4 + 3/4 + 3/(4*19) + 3/(19*91) + 3/(91*436) + ... - Gary W. Adamson, Dec 18 2007

a(n+1) is the number of compositions of n when there are 4 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010

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

Right-shifted Binomial Transform of the left-shifted A030195. - R. J. Mathar, Oct 15 2012

Values of x (or y) in the solutions to x^2 - 5xy + y^2 + 3 = 0. - Colin Barker, Feb 04 2014

From Wolfdieter Lang, Oct 15 2020: (Start)

All positive solutions of the Diophantine equation x^2 + y^2 - 5*x*y = -3 (see the preceding comment) are given by [x(n) = S(n, 5) - S(n-1, 5), y(n) = x(n-1)], for n =-oo..+oo, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.

This binary indefinite quadratic form has discriminant D = +21. There is only this family representing -3 properly with x and y positive, and there are no improper solutions.

See the formula for a(n) = x(n-1), for n >= 1, in terms of S-polynomials below.

This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

From Wolfdieter Lang, Feb 08 2021: (Start)

All proper and improper solutions of the generalized Pell equation X^2 - 21*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n+1) from the preceding comment by X(n) = x(n) + x(n-1) = S(n-1, 5) - S(n-2, 5) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 5), for all integer numbers n. For positive integers X(n) = A003501(n) and Y(n) = A004254(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.

The two conjugated proper families of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer n. This follows from the mentioned paper by Robert K. Moniot. (End)

Equivalent definition: a(n) = ceiling(a(n-1)^2 / a(n-2)), with a(1)=1, a(2)=4, a(3)=19. The problem for USA Olympiad (see Andreescu and Gelca reference) asks to prove that a(n)-1 is always a multiple of 3. - Bernard Schott, Apr 13 2022

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)