cp's OEIS Frontend

Columns of the array are generated from Fibonacci polynomials f(x). They are: (1), (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 +1), ... If column headings start 0, 1, 2, ... then the terms in the n-th column are generated from the n-th degree Fibonacci polynomial. For example, column 5 (8, 70, 360, ...) is generated from f(x), x = 1,2,3,...; fifth-degree polynomial x^5 + 4x^3 + 3x; e.g., f(2) = 70 = 2^5 + 4*8 + 3*2. - Gary W. Adamson, Apr 02 2006

The ratio of two consecutive entries of the sequence in the n-th row approaches (n + sqrt(n^2 + 4))/2. Example: The sequence beginning (1, 3, 10, 33, ...) tends to 3.302775... = (3 + sqrt(13))/2. - Gary W. Adamson, Aug 12 2013

As to the array sequences, (n+1)-th sequence is the INVERT transform of the n-th sequence. - Gary W. Adamson, Aug 20 2013

The array can be extended infinitely above the Fibonacci row by taking successive INVERTi transforms, resulting in:

...

1, -2, 5, -12, 29, -70, ...

1, -1, 2, -3, 5, -8, ...

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

1, 1, 2, 3, 5, 8, ...

1, 2, 5, 12, 29, 70, ...

...

This results in an infinite array in which sequences above the (1, 0, 1, 0, ...) are reflections of the sequences below, except for the alternate signs. Any sequence in the (+ sign) row starting (1, n, ...) is the (2*n-th) INVERT transform of the same sequence but with alternate signs. Example: (1, 2, 5, 12, ...) is the (2*2) = fourth INVERT transform of (1, -2, 5, -12, ...) by inspection. Conjecture: This "reflection" principle will result from taking successive INVERT transforms of any aerated sequence starting 1, ... and with positive signs. Likewise, the rows above the aerated sequence are successive INVERTi transforms of the aerated sequence. - Gary W. Adamson, Jul 14 2019

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

Row n is the n-metallonacci sequence.

T(n,k) is 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)

T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), (1,0), (2,0). - Joerg Arndt, Jun 30 2011

T(n,k) is the number of lattice paths of length n, starting from the origin and ending at (n,k), using horizontal steps H=(1,0), up steps U=(1,1) and down steps D=(1,-1), never containing UUU, DD, HD. For instance, for n=4 and k=2, we have the paths; HHUU, HUHU, HUUH, UHHU, UHUH, UUHH, UUDU, UDUU, UUUD. - Emanuele Munarini, Mar 15 2011

Row sums form Pell numbers A000129, T(n,0) forms Fibonacci numbers A000045, T(n,1) forms A001629. T(n+k,n-k) is polynomial sequence of degree k.

T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, etc.).

As a Riordan array, this is (1/(1-x-x^2),x/(1-x-x^2)). An interesting factorization is (1/(1-x^2),x/(1-x^2))*(1/(1-x),x/(1-x)) [abs(A049310) times A007318]. Diagonal sums are the Jacobsthal numbers A001045(n+1). - Paul Barry, Jul 28 2005

T(n,k) = T'(n+1,k+1), T' given by [0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 19 2005

Equals A049310 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007

This triangle may also be obtained from the coefficients of the Morgan-Voyce polynomials defined by: Mv(x, n) = (x + 1)*Mv(x, n - 1) + Mv(x, n - 2). - Roger L. Bagula, Apr 09 2008

Row sums are A000129. - Roger L. Bagula, Apr 09 2008

Absolute value of coefficients of the characteristic polynomial of tridiagonal matrices with 1's along the main diagonal, and i's along the superdiagonal and the subdiagonal (where i=sqrt(-1), see Mathematica program). - John M. Campbell, Aug 23 2011

A037027 is jointly generated with A122075 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+1)*v(n-1)x and v(n,x)=u(n-1,x)+x*v(n-1,x). See the Mathematica section at A122075. - Clark Kimberling, Mar 05 2012

For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Row n, for n>=0, shows the coefficients of the polynomial u(n) = c(0) + c(1)*x + ... + c(n)*x^n which is the denominator of the n-th convergent of the continued fraction [x+1, x+1, x+1, ...]; see A230000. - Clark Kimberling, Nov 13 2013

T(n,k) is the number of ternary words of length n having k letters 2 and avoiding a runs of odd length for the letter 0. - Milan Janjic, Jan 14 2017

Let T(m, n, k) be an m-bonacci Pascal's triangle, where T(m, n, 0) gives the values of F(m, n), the n-th m-bonacci number, and T(m, n, k) gives the values for the k-th convolution of F(m, n). Then the classic Pascal triangle is T(1, n, k) and this sequence is T(2, n, k). T(m, n, k) is the number of compositions of n using only the positive integers 1, 1' and 2 through m, with the part 1' used exactly k times. G.f. for k-th column of T(m, n, k): x/(1 - x - x^2 - ... - x^m)^k. The row sum for T(m, n, k) is the number of compositions of n using only the positive integers 1, 1' and 2 through m. G.f. for row sum of T(m, n, k): 1/(1 - 2x - x^2 - ... - x^m). - Gregory L. Simay, Jul 24 2021

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)

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

In the following guide to p-INVERT sequences using s = (1,0,1,0,1,...) = A000035, in some cases t(1,0,1,0,1,...) is a shifted version of the indicated sequence.

p(S) t(1,0,1,0,1,...)

1 - S A000045 (Fibonacci numbers)

1 - S^2 A147600

1 - S^3 A291217

1 - S^5 A291218

1 - S - S^2 A289846

1 - S - S^3 A291219

1 - S - S^4 A291220

1 - S^3- S^6 A291221

1 - S^2- S^3 A291222

1 - S^3- S^4 A291223

1 - 2S A052542

1 - 3S A006190

(1 - S)^2 A239342

(1 - S)^3 A276129

(1 - S)^4 A291224

(1 - S)^5 A291225

(1 - S)^6 A291226

1 - S - 2 S^2 A291227

1 - 2 S - 2 S^2 A291228

1 - 3 S - 2 S^2 A060801

(1 - S)(1 - 2 S) A291229

(1 - S)(1 - 2 S)(1 - 3 S) A291230

(1 - S)(1 - 2 S)(1 - 3 S)( 1 - 4 S) A291231

(1 - 2 S)^2 A291264

(1 - 3 S)^2 A291232

1 - S - S^2 - S^3 A291233

1 - S - S^2 - S^3 - S^4 A291234

1 - S - S^2 - S^3 - S^4 - S^5 A291235

(1 - S)(1 - 3 S) A291236

(1 - S)(1 - 2S)( 1 - 4S) A291237

(1 - S)^2 (1 - 2S) A291238

(1 - S^2) (1 - 2S) A291239

(1 - S^3)^2 A291240

1 - S - S^2 + S^3 A291241

1 - 2 S - S^2 + S^3 A291242

1 - 3 S + S^2 A291243

1 - 4 S + S^2 A291244

1 - 5 S + S^2 A291245

1 - 6 S + S^2 A291246

1 - S - S^2 - S^3 + S^4 A291247

1 - S - S^2 - S^3 - S^4 + S^5 A291248

1 - S - S^2 - S^3 + S^4 + S^5 A291249

1 - S - 2 S^2 + 2 S^3 A291250

1 - 3 S^2 + 2 S^3 A291251 (includes negative terms)

(1 - S^3)^3 A291252

(1 - S - S^2)^2 A291253

(1 - 2 S - S^2)^2 A291254

(1 - S - 2 S^2)^2 A291255

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

From Michael A. Allen, Mar 26 2023: (Start)

Row n is the n-metallonacci sequence for n>0.

A(n,k), for n > 0 and k > 0, is 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)

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)

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)

Rows of A013610 reversed. - Michael Somos, Feb 14 2002

Row sums are powers of 4 (A000302), antidiagonal sums are A006190 (a(n) = 3*a(n-1) + a(n-2)). - Gerald McGarvey, May 17 2005

Triangle of coefficients in expansion of (3+x)^n.

Also: Pure Galton board of scheme (3,1). Also: Multiplicity (number) of pairs of n-dimensional binary vectors with dot product (overlap) k. There are 2^n = A000079(n) binary vectors of length n and 2^(2n) = 4^n = A000302(n) different pairs to form dot products k = Sum_{i=1..n} v[i]*u[i] between these, 0 <= k <= n. (Since dot products are symmetric, there are only 2^n*(2^n-1)/2 different non-ordered pairs, actually.) - R. J. Mathar, Mar 17 2006

Mirror image of A013610. - Zerinvary Lajos, Nov 25 2007

T(i,j) is the number of i-permutations of 4 objects a,b,c,d, with repetition allowed, containing j a's. - Zerinvary Lajos, Dec 21 2007

The antidiagonals of the sequence formatted as a square array (see Examples section) and summed with alternating signs gives a bisection of Fibonacci sequence, A001906. Example: 81-(27-1)=55. Similar rule applied to rows gives A000079. - Mark Dols, Sep 01 2009

Triangle T(n,k), read by rows, given by (3,0,0,0,0,0,0,0,...)DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 09 2011

T(n,k) = binomial(n,k)*3^(n-k), the number of subsets of [2n] with exactly k symmetric pairs, where elements i and j of [2n] form a symmetric pair if i+j=2n+1. Equivalently, if n couples attend a (ticketed) event that offers door prizes, then the number of possible prize distributions that have exactly k couples as dual winners is T(n,k). - Dennis P. Walsh, Feb 02 2012

T(n,k) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that the intersection of A and B contains exactly k elements. For example, T(2,1) = 6 because we have ({1},{1}); ({1},{1,2}); ({2},{2}); ({2},{1,2}); ({1,2},{1}); ({1,2},{2}). Sum_{k=0..n} T(n,k)*k = A002697(n) (see comment there by Ross La Haye). - Geoffrey Critzer, Sep 04 2013

Also the convolution triangle of A000244. - Peter Luschny, Oct 09 2022

From Michael A. Allen, Mar 30 2023: (Start)

Column k is the k-metallonacci sequence for k > 0.

T(n,k) is, for n > 0 and k > 0, 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 k kinds of squares available. (End)