cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A122367 Dimension of 3-variable non-commutative harmonics (twisted derivative) of order n. The dimension of the space of non-commutative polynomials of degree n in 3 variables which are killed by all symmetric differential operators (where for a monomial w, d_{xi} ( xi w ) = w and d_{xi} ( xj w ) = 0 for i != j).

Original entry on oeis.org

1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170, 2971215073, 7778742049, 20365011074, 53316291173, 139583862445, 365435296162, 956722026041
Offset: 0

Views

Author

Mike Zabrocki, Aug 30 2006

Keywords

Comments

Essentially identical to A001519.
From Matthew Lehman, Jun 14 2008: (Start)
Number of monotonic rhythms using n time intervals of equal duration (starting with n=0).
Representationally, let 0 be an interval which is "off" (rest),
1 an interval which is "on" (beep),
1 1 two consecutive "on" intervals (beep, beep),
1 0 1 (beep, rest, beep) and
1-1 two connected consecutive "on" intervals (beeeep).
For f(3)=13:
0 0 0, 0 0 1, 0 1 0, 0 1 1, 0 1-1, 1 0 0, 1 0 1,
1 1 0, 1-1 0, 1 1 1, 1 1-1, 1-1 1, 1-1-1.
(End)
Equivalent to the number of one-dimensional graphs of n nodes, subject to the condition that a node is either 'on' or 'off' and that any two neighboring 'on' nodes can be connected. - Matthew Lehman, Nov 22 2008
Sum_{n>=0} arctan(1/a(n)) = Pi/2. - Jaume Oliver Lafont, Feb 27 2009
This is the limit sequence for certain generalized Pell numbers. - Gregory L. Simay, Oct 21 2024

Examples

			a(1) = 2 because x1-x2, x1-x3 are both of degree 1 and are killed by the differential operator d_x1 + d_x2 + d_x3.
a(2) = 5 because x1*x2 - x3*x2, x1*x3 - x2*x3, x2*x1 - x3*x1, x1*x1 - x2*x1 - x2*x2 + x1*x2, x1*x1 - x3*x1 - x3*x3 + x3*x1 are all linearly independent and are killed by d_x1 + d_x2 + d_x3, d_x1 d_x1 + d_x2 d_x2 + d_x3 d_x3 and Sum_{j = 1..3} (d_xi d_xj, i).

Links

Crossrefs

Cf. A001519, A048575, A055105, A055107, A087903, A074664, A008277, A106729, A112340, A122368, A122369, A122370, A122371, A122372.
Cf. similar sequences listed in A238379.

Programs

  • Magma
    [Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015
  • Maple
    a:=n->if n=0 then 1; elif n=1 then 2 else 3*a(n-1)-a(n-2); fi;
A122367List := proc(m) local A, P, n; A := [1,2]; P := [2];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-1]]);
A := [op(A), P[-1]] od; A end: A122367List(30); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[Fibonacci[2 n + 1], {n, 0, 30}] (* Vincenzo Librandi, Jul 04 2015 *)
  • PARI
    Vec((1-x)/(1-3*x+x^2) + O(x^50)) \\ Michel Marcus, Jul 04 2015

Formula

G.f.: (1-q)/(1-3*q+q^2). More generally, (Sum_{d=0..n} (n!/(n-d)!*q^d)/Product_{r=1..d} (1 - r*q)) / (Sum_{d=0..n} q^d/Product_{r=1..d} (1 - r*q)) where n=3.
a(n) = 3*a(n-1) - a(n-2) with a(0) = 1, a(1) = 2.
a(n) = Fibonacci(2n+1) = A000045(2n+1). - Philippe Deléham, Feb 11 2009
a(n) = (2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5)) + (1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5). - Colin Barker, Oct 14 2015
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(k+i-1, k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = A048575(n-1) for n >= 1. - Georg Fischer, Nov 02 2018
a(n) = Fibonacci(n)^2 + Fibonacci(n+1)^2. - Michel Marcus, Mar 18 2019
a(n) = Product_{k=1..n} (1 + 4*cos(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
From J. M. Bergot, May 27 2022: (Start)
a(n) = F(n)*L(n+1) + (-1)^n where L(n)=A000032(n) and F(n)=A000045(n).
a(n) = (L(n)^2 + L(n)*L(n+2))/5 - (-1)^n.
a(n) = 2*(area of a triangle with vertices at (L(n-1), L(n)), (F(n+1), F(n)), (L(n+1), L(n+2))) - 5*(-1)^n for n > 1. (End)
G.f.: (1-x)/(1-3x+x^2) = 1/(1-2x-x^2-x^3-x^4-...) - Gregory L. Simay, Oct 21 2024
E.g.f.: exp(3*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Nov 07 2024
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/5) * sqrt( 1 - T(2*n+1, - 3/2) ), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(2*n+1/2) = sqrt(5)*a(n)^2 - 2/sqrt(5).
a(3*n+1) = 5*a(n)^3 - 3*a(n); hence a(n) divides a(3*n+1).
a(4*n+3/2) = 5^(3/2)*a(n)^4 - 4*sqrt(5)*a(n)^2 + 2/sqrt(5).
a(5*n+2) = (5^2)*a(n)^5 - 5*5*a(n)^3 + 5*a(n); hence a(n) divides a(5*n+2).
See A034807 for the unsigned coefficients [1, 2; 1, 3; 1, 4, 2; 1, 5, 5; ...].
In general, for k >= 0, a(k*n + (k-1)/2) = a(-1/2) * T(k, a(n)/a(-1/2)), where a(n) = (2^(-1-n)*((3 - sqrt(5))^n *(-1 + sqrt(5)) + (1 + sqrt(5))*(3 + sqrt(5))^n)) / sqrt(5), as given above, and a(-1/2) = 2/sqrt(5).
The aerated sequence [b(n)]n>=1 = [1, 0, 2, 0, 5, 0, 13, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001906(n) - 1/A001906(n+1).) (End)

A004146 Alternate Lucas numbers - 2.

Original entry on oeis.org

0, 1, 5, 16, 45, 121, 320, 841, 2205, 5776, 15125, 39601, 103680, 271441, 710645, 1860496, 4870845, 12752041, 33385280, 87403801, 228826125, 599074576, 1568397605, 4106118241, 10749957120, 28143753121, 73681302245, 192900153616, 505019158605, 1322157322201
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

This is the r=5 member in the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Number of spanning trees of the wheel W_n on n+1 vertices. - Emeric Deutsch, Mar 27 2005
Also number of spanning trees of the n-helm graph. - Eric W. Weisstein, Jul 16 2011
a(n) is the smallest number requiring n terms when expressed as a sum of Lucas numbers (A000204). - David W. Wilson, Jan 10 2006
This sequence has a primitive prime divisor for all terms beyond the twelfth. - Anthony Flatters (Anthony.Flatters(AT)uea.ac.uk), Aug 17 2007
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series of gamma matrix with determinant 1:
a(n) = Determinant(A + A^2 + A^3 + A^4 + A^5 + ... + A^n)
where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant
A = [[1,1,1,1,1,1,...],[1,2,1,2,1,2,...],[1,2,3,1,2,3,...],[1,2,3,4,1,2,...],
[1,2,3,4,5,1,...],[1,2,3,4,5,6,...],...]. Note: Determinant A(1..n,1..n)= (n-1)!.
See A158039, A158040, A158041, A158042, A158043, A158044, for sequences of matrix 2!,3!,... (End)
The previous comment could be rephrased as: a(n) = -det(A^n - I) where I is the 2 X 2 identity matrix and A = [1, 1; 1, 2]. - Peter Bala, Mar 20 2015
a(n) is also the number of points of Arnold's "cat map" that are on orbits of period n-1. This is a map of the two-torus T^2 into itself. If we regard T^2 as R^2 / Z^2, the action of this map on a two vector in R^2 is multiplication by the unit-determinant matrix A = [2, 1;1, 1], with the vector components taken modulo one. As such, an explicit formula for the n-th entry of this sequence is -det(I-A^n). - Bruce Boghosian, Apr 26 2009
7*a(n) gives the total number of vertices in a heptagonal hyperbolic lattice {7,3} with n total levels, in which an open heptagon is centered at the origin. - Robert M. Ziff, Apr 10 2011
The sequence is the case P1 = 5, P2 = 6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
Determinants of the spiral knots S(3,k,(1,-1)). a(k) = det(S(3,k,(1,-1))). These knots are also the weaving knots W(k,3) and the Turk's Head Links THK(3,k). - Ryan Stees, Dec 14 2014
Even-indexed Fibonacci numbers (1, 3, 8, 21, ...) convolved with (1, 2, 2, 2, 2, ...). - Gary W. Adamson, Aug 09 2016
a(n) is the number of ways to tile a bracelet of length n with 1-color square, 2-color dominos, 3-color trominos, etc. - Yu Xiao, May 23 2020
a(n) is the number of face-labeled unfoldings of a pyramid whose base is a simple n-gon. Cf. A103536. - Rick Mabry, Apr 17 2023

Examples

			For k=3, b(3) = sqrt(5)*b(2) - b(1) = 5 - 1 = 4, so det(S(3,3,(1,-1))) = 4^2 = 16.
G.f. = x + 5*x^2 + 16*x^3 + 45*x^4 + 121*x^5 + 320*x^4 + 841*x^5 + ... - _Michael Somos_, Feb 10 2023

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (p. 193, Problem 3.3.40 (a)).
  • N. Hartsfield and G. Ringel, Pearls in Graph Theory, p. 102. Academic Press: 1990.
  • B. Hasselblatt and A. Katok, "Introduction to the Modern Theory of Dynamical Systems," Cambridge University Press, 1997.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

This is the r=5 member of the family S_r(n) defined in A092184.
Cf. A005248. Partial sums of A002878. Pairwise sums of A027941. Bisection of A074392.
Sequence A032170, the Möbius transform of this sequence, is then the number of prime periodic orbits of Arnold's cat map. - Bruce Boghosian, Apr 26 2009
Cf. A000032, A100047, A001350.
Cf. also A158039, A158040, A158041, A158042, A158043, A158044.
Cf. A103536 for the number of geometrically distinct edge-unfoldings of the regular pyramid. - Rick Mabry, Apr 17 2023

Programs

  • Magma
    [Lucas(n)-2: n in [0..60 by 2]]; // Vincenzo Librandi, Mar 20 2015
  • Mathematica
    Table[LucasL[2*n] - 2, {n, 0, 20}]
(* Second program: *)
LinearRecurrence[{4, -4, 1}, {0, 1, 5}, 30] (* Jean-François Alcover, Jan 08 2019 *)
  • PARI
    a(n) = { we = quadgen(5);((1+we)^n) + ((2-we)^n) - 2;} /* Michel Marcus, Aug 18 2012 */

Formula

a(n) = A005248(n) - 2 = A000032(2*n) - 2.
a(n+1) = 3*a(n) - a(n-1) + 2.
G.f.: x*(1+x)/(1-4*x+4*x^2-x^3) = x*(1+x)/((1-x)*(1-3*x+x^2)).
a(n) = 2*(T(n, 3/2)-1) with Chebyshev's polynomials T(n, x) of the first kind. See their coefficient triangle A053120.
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=5.
a(n) = 2*T(n, 3/2) - 2, with twice the Chebyshev polynomials of the first kind, 2*T(n, x=3/2) = A005248(n).
a(n) = b(n) + b(n-1), n>=1, with b(n):=A027941(n-1), n>=1, b(-1):=0, the partial sums of S(n, 3) = U(n, 3/2) = A001906(n+1), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind.
a(2n) = A000204(2n)^2 - 4 = 5*A000045(2n)^2; a(2n+1) = A000204(2n+1)^2. - David W. Wilson, Jan 10 2006
a(n) = ((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jun 09 2001
a(n) = b(n-1) + b(n-2), n>=1, with b(n):=A027941(n), b(-1):=0, partial sums of S(n, 3) = U(n, 3/2) = A001906(n+1), Chebyshev's polynomials of the second kind.
a(n) = n*Sum_{k=1..n} binomial(n+k-1,2*k-1)/k, n > 0. - Vladimir Kruchinin, Sep 03 2010
a(n) = floor(tau^(2*n)*(tau^(2*n) - floor(tau^(2*n)))), where tau = (1+sqrt(5))/2. - L. Edson Jeffery, Aug 26 2013
From Peter Bala, Apr 03 2014: (Start)
a(n) = U(n-1,sqrt(5)/2)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -3/2; 1, 5/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(k) = det(S(3,k,(1,-1))) = b(k)^2, where b(1)=1, b(2)=sqrt(5), b(k)=sqrt(5)*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + Sum_{n >= 1} Fibonacci(2*n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(phi^2*x) + exp(x/phi^2) - 2*exp(x), where phi = (1 + sqrt(5))/2. - G. C. Greubel, Aug 24 2015
a(n) = a(-n) for all n in Z. - Michael Somos, Aug 27 2015
From Peter Bala, Jun 03 2016: (Start)
a(n) = Lucas(2*n) - Lucas(0*n);
a(n)^2 = Lucas(4*n) - 3*Lucas(2*n) + 3*Lucas(0*n) - Lucas(-2*n);
a(n)^3 = Lucas(6*n) - 5*Lucas(4*n) + 10*Lucas(2*n) - 10*Lucas(0*n) + 5*Lucas(-2*n) - Lucas(-4*n) and so on (follows from Binet's formula for Lucas(2*n) and the algebraic identity (x + 1/x - 2)^m = f(x) + f(1/x) where f(x) = (x - 1)^(2*m - 1)/x^(m-1) ). (End)
Limit_{n->infinity} a(n+1)/a(n) = (3 + sqrt(5))/2 = A104457. - Ilya Gutkovskiy, Jun 03 2016
a(n) = (phi^n - phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020
a(n) = 4*sinh(n*A002390)^2, where A002390 = arcsinh(1/2). - Gleb Koloskov, Sep 18 2021
a(n) = 5*F(n)^2 = L(n)^2 - 4 if n even and a(n) = L(n)^2 = 5*F(n)^2 - 4 if n odd. - Michael Somos, Feb 10 2023
a(n) = n^2 + Sum_{i=1..n-1} i*a(n-i). - Fern Gossow, Dec 03 2024

Extensions

Correction to formula from Nephi Noble (nephi(AT)math.byu.edu), Apr 09 2002
Chebyshev comments from Wolfdieter Lang, Sep 10 2004

A027465 Cube of lower triangular normalized binomial matrix.

Original entry on oeis.org

1, 3, 1, 9, 6, 1, 27, 27, 9, 1, 81, 108, 54, 12, 1, 243, 405, 270, 90, 15, 1, 729, 1458, 1215, 540, 135, 18, 1, 2187, 5103, 5103, 2835, 945, 189, 21, 1, 6561, 17496, 20412, 13608, 5670, 1512, 252, 24, 1, 19683, 59049, 78732, 61236, 30618, 10206, 2268
Offset: 0

Views

Author

Olivier Gérard, N. J. A. Sloane

Keywords

Comments

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

Examples

			Example: n = 3 offers 2^3 = 8 different binary vectors (0,0,0), (0,0,1), ..., (1,1,0), (1,1,1). a(3,2) = 9 of the 2^4 = 64 pairs have overlap k = 2: (0,1,1)*(0,1,1) = (1,0,1)*(1,0,1) = (1,1,0)*(1,1,0) = (1,1,1)*(1,1,0) = (1,1,1)*(1,0,1) = (1,1,1)*(0,1,1) = (0,1,1)*(1,1,1) = (1,0,1)*(1,1,1) = (1,1,0)*(1,1,1) = 2.
For example, T(2,1)=6 since there are 6 subsets of {1,2,3,4} that have exactly 1 symmetric pair, namely, {1,4}, {2,3}, {1,2,3}, {1,2,4}, {1,3,4}, and {2,3,4}.
The present sequence formatted as a triangular array:
     1
     3     1
     9     6     1
    27    27     9     1
    81   108    54    12    1
   243   405   270    90   15    1
   729  1458  1215   540  135   18   1
  2187  5103  5103  2835  945  189  21  1
  6561 17496 20412 13608 5670 1512 252 24 1
  ...
A013610 formatted as a triangular array:
  1
  1  3
  1  6   9
  1  9  27   27
  1 12  54  108   81
  1 15  90  270  405   243
  1 18 135  540 1215  1458   729
  1 21 189  945 2835  5103  5103  2187
  1 24 252 1512 5670 13608 20412 17496 6561
   ...
A099097 formatted as a square array:
      1     0     0    0   0 0 0 0 0 0 0 ...
      3     1     0    0   0 0 0 0 0 0 ...
      9     6     1    0   0 0 0 0 0 ...
     27    27     9    1   0 0 0 0 ...
     81   108    54   12   1 0 0 ...
    243   405   270   90  15 1 ...
    729  1458  1215  540 135 ...
   2187  5103  5103 2835 ...
   6561 17496 20412 ...
  19683 59049 ...
  59049 ...

Links

Crossrefs

Cf. A000244, A007318, A013610, A013610, A099097, A027471, A027472, A036216, A036217, A036219, A036220, A036221, A036222, A036223.

Programs

  • Haskell
    a027465 n k = a027465_tabl !! n !! k
a027465_row n = a027465_tabl !! n
a027465_tabl = iterate (\row ->
   zipWith (+) (map (* 3) (row ++ [0])) (map (* 1) ([0] ++ row))) [1]
-- Reinhard Zumkeller, May 26 2013
  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*3^(i-j), j = 0 .. i) od; # Zerinvary Lajos, Nov 25 2007
# Uses function PMatrix from A357368. Adds column 1, 0, 0, ... to the left.
PMatrix(10, n -> 3^(n-1)); # Peter Luschny, Oct 09 2022
  • Mathematica
    t[n_, k_] := Binomial[n, k]*3^(n-k); Table[t[n, n-k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 19 2012 *)
  • PARI
    {T(n, k) = polcoeff( (3 + x)^n, k)}; /* Michael Somos, Feb 14 2002 */

Formula

Numerators of lower triangle of (b^2)[ i, j ] where b[ i, j ] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
Triangle whose (i, j)-th entry is binomial(i, j)*3^(i-j).
a(n, m) = 4^(n-1)*Sum_{j=m..n} b(n, j)*b(j, m) = 3^(n-m)*binomial(n-1, m-1), n >= m >= 1; a(n, m) := 0, n < m. G.f. for m-th column: (x/(1-3*x))^m (m-fold convolution of A000244, powers of 3). - Wolfdieter Lang, Feb 2006
G.f.: 1 / (1 - x(3+y)).
a(n,k) = 3*a(n-1,k) + a(n-1,k-1) - R. J. Mathar, Mar 17 2006
From the formalism of A133314, the e.g.f. for the row polynomials of A027465 is exp(x*t)*exp(3x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(-3x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*3x). The results generalize for 3 replaced by any number. - Tom Copeland, Aug 18 2008
T(n,k) = A164942(n,k)*(-1)^k. - Philippe Deléham, Oct 09 2011
Let P and P^T be the Pascal matrix and its transpose and H = P^3 = A027465. Then from the formalism of A132440 and A218272,
exp[x*z/(1-3z)]/(1-3z) = exp(3z D_z z) e^(x*z)= exp(3D_x x D_x) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T = Sum_{n>=0} (3z)^n L_n(-x/3), where D is the derivative operator and L_n(x) are the regular (not normalized) Laguerre polynomials. - Tom Copeland, Oct 26 2012
E.g.f. for column k: x^k/k! * exp(3x). - Geoffrey Critzer, Sep 04 2013

A053122 Triangle of coefficients of Chebyshev's S(n,x-2) = U(n,x/2-1) polynomials (exponents of x in increasing order).

Original entry on oeis.org

1, -2, 1, 3, -4, 1, -4, 10, -6, 1, 5, -20, 21, -8, 1, -6, 35, -56, 36, -10, 1, 7, -56, 126, -120, 55, -12, 1, -8, 84, -252, 330, -220, 78, -14, 1, 9, -120, 462, -792, 715, -364, 105, -16, 1, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 1, 11, -220, 1287, -3432, 5005, -4368, 2380, -816, 171, -20
Offset: 0

Views

Author

Wolfdieter Lang

Keywords

Comments

Apart from signs, identical to A078812.
Another version with row-leading 0's and differing signs is given by A285072.
G.f. for row polynomials S(n,x-2) (signed triangle): 1/(1+(2-x)*z+z^2). Unsigned triangle |a(n,m)| has g.f. 1/(1-(2+x)*z+z^2) for row polynomials.
Row sums (signed triangle) A049347(n) (periodic(1,-1,0)). Row sums (unsigned triangle) A001906(n+1)=F(2*(n+1)) (even-indexed Fibonacci).
In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The (unsigned) column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for m=0..5, resp. For m=6..23 they are A010966..(+2)..A011000 and for m=24..49 they are A017713..(+2)..A017763.
Riordan array (1/(1+x)^2,x/(1+x)^2). Inverse array is A039598. Diagonal sums have g.f. 1/(1+x^2). - Paul Barry, Mar 17 2005. Corrected by Wolfdieter Lang, Nov 13 2012.
Unsigned version is in A078812. - Philippe Deléham, Nov 05 2006
Also row n gives (except for an overall sign) coefficients of characteristic polynomial of the Cartan matrix for the root system A_n. - Roger L. Bagula, May 23 2007
From Wolfdieter Lang, Nov 13 2012: (Start)
The A-sequence for this Riordan triangle is A115141, and the Z-sequence is A115141(n+1), n>=0. For A- and Z-sequences for Riordan matrices see the W. Lang link under A006232 with details and references.
S(n,x^2-2) = sum(r(j,x^2),j=0..n) with Chebyshev's S-polynomials and r(j,x^2) := R(2*j+1,x)/x, where R(n,x) are the monic integer Chebyshv T-polynomials with coefficients given in A127672. Proof from comparing the o.g.f. of the partial sum of the r(j,x^2) polynomials (see a comment on the signed Riordan triangle A111125) with the present Riordan type o.g.f. for the row polynomials with x -> x^2. (End)
S(n,x^2-2) = S(2*n+1,x)/x, n >= 0, from the odd part of the bisection of the o.g.f. - Wolfdieter Lang, Dec 17 2012
For a relation to a generator for the Narayana numbers A001263, see A119900, whose columns are unsigned shifted rows (or antidiagonals) of this array, referring to the tables in the example sections. - Tom Copeland, Oct 29 2014
The unsigned rows of this array are alternating rows of a mirrored A011973 and alternating shifted rows of A030528 for the Fibonacci polynomials. - Tom Copeland, Nov 04 2014
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): a(n, m) = (2*(m + 1)/(n - m))*Sum_{k = m..n-1} (-1)^(n-k)*a(k, m), with input a(n, n) = 1, and a(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020
Row n gives the characteristic polynomial of the (n X n)-matrix M where M[i,j] = 2 if i = j, -1 if |i-j| = 1 and 0 otherwise. The matrix M is positive definite and has 2-condition number (cot(Pi/(2*n+2)))^2. - Jianing Song, Jun 21 2022
Also the convolution triangle of (-1)^(n+1)*n. - Peter Luschny, Oct 07 2022

Examples

			The triangle a(n,m) begins:
n\m   0    1    2     3     4     5     6    7    8  9
0:    1
1:   -2    1
2:    3   -4    1
3:   -4   10   -6     1
4:    5  -20   21    -8     1
5:   -6   35  -56    36   -10     1
6:    7  -56  126  -120    55   -12     1
7:   -8   84 -252   330  -220    78   -14    1
8:    9 -120  462  -792   715  -364   105  -16    1
9:  -10  165 -792  1716 -2002  1365  -560  136  -18  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012
E.g., fourth row (n=3) {-4,10,-6,1} corresponds to the polynomial S(3,x-2) = -4+10*x-6*x^2+x^3.
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: a(5,1) = 35 = 1*5 + (-2)*(-20) -1*(10).
Recurrence from Z-sequence [-2,-1,-2,-5,...]: a(5,0) = -6 = (-2)*5 + (-1)*(-20) + (-2)*21 + (-5)*(-8) + (-14)*1.
Recurrence from A-sequence [1,-2,-1,-2,-5,...]: a(5,1) = 35 = 1*5  + (-2)*(-20) + (-1)*21 + (-2)*(-8) + (-5)*1.
(End)
E.g., the fourth row (n=3) {-4,10,-6,1} corresponds also to the polynomial S(7,x)/x = -4 + 10*x^2 - 6*x^4 + x^6. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -56 = a(5, 2) = 2*(-1*1 + 1*(-6) - 1*21) = -2*28 = -56. - _Wolfdieter Lang_, Jun 03 2020

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
  • R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 62.
  • Sigurdur Helgasson, Differential Geometry, Lie Groups and Symmetric Spaces, Graduate Studies in Mathematics, volume 34. A. M. S.: ISBN 0-8218-2848-7, 1978, p. 463.

Links

Crossrefs

Cf. A005248, A127677, A053123, A049310.
Cf. A011973, A030528, A034867, A119900.
Cf. A285072 (version with row-leading 0's and differing signs). - Eric W. Weisstein, Apr 09 2017

Programs

  • Maple
    seq(seq((-1)^(n+m)*binomial(n+m+1,2*m+1),m=0..n),n=0..10); # Robert Israel, Oct 15 2014
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> -(-1)^n*n); # Peter Luschny, Oct 07 2022
  • Mathematica
    T[n_, m_, d_] := If[ n == m, 2, If[n == m - 1 || n == m + 1, -1, 0]]; M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}]; a = Join[M[1], Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]]; Flatten[a] (* Roger L. Bagula, May 23 2007 *)
(* Alternative code for the matrices from MathWorld: *)
sln[n_] := 2IdentityMatrix[n] - PadLeft[PadRight[IdentityMatrix[n - 1], {n, n - 1}], {n, n}] - PadLeft[PadRight[IdentityMatrix[n - 1], {n - 1, n}], {n, n}] (* Roger L. Bagula, May 23 2007 *)
  • Sage
    @CachedFunction
def A053122(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A053122(n-1,k-1)-A053122(n-2,k)-2*A053122(n-1,k)
for n in (0..9): [A053122(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Formula

a(n, m) := 0 if n
a(n, m) = -2*a(n-1, m) + a(n-1, m-1) - a(n-2, m), a(n, -1) := 0 =: a(-1, m), a(0, 0)=1, a(n, m) := 0 if n
O.g.f. for m-th column (signed triangle): ((x/(1+x)^2)^m)/(1+x)^2.
From Jianing Song, Jun 21 2022: (Start)
T(n,k) = [x^k]f_n(x), where f_{-1}(x) = 0, f_0(x) = 1, f_n(x) = (x-2)*f_{n-1}(x) - f_{n-2}(x) for n >= 2.
f_n(x) = (((x-2+sqrt(x^2-4*x))/2)^(n+1) - ((x-2-sqrt(x^2-4*x))/2)^(n+1))/sqrt(x^2-4x).
The roots of f_n(x) are 2 + 2*cos(k*Pi/(n+1)) = 4*(cos(k*Pi/(2*n+2)))^2 for 1 <= k <= n. (End)

A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264
Offset: 0

Views

Author

Vladimir Joseph Stephan Orlovsky, May 24 2011

Keywords

Comments

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Links

Crossrefs

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
1..A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
2..A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
3..A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
4..A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
5..A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
6..A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
7..A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
8..A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
9..A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
10.A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011
  • Mathematica
    LinearRecurrence[{2,-10}, {0,1}, 50]
  • PARI
    a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016
  • SageMath
    [lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

Formula

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A059304 a(n) = 2^n * (2*n)! / (n!)^2.

Original entry on oeis.org

1, 4, 24, 160, 1120, 8064, 59136, 439296, 3294720, 24893440, 189190144, 1444724736, 11076222976, 85201715200, 657270374400, 5082890895360, 39392404439040, 305870434467840, 2378992268083200, 18531097667174400
Offset: 0

Views

Author

Henry Bottomley, Jan 25 2001

Keywords

Comments

Number of lattice paths from (0,0) to (n,n) using steps (0,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011
The convolution square root of this sequence is A004981. - T. D. Noe, Jun 11 2002
Also main diagonal of array: T(i,1)=2^(i-1), T(1,j)=1, T(i,j) = T(i,j-1) + 2*T(i-1,j). - Benoit Cloitre, Feb 26 2003
The Hankel transform (see A001906 for definition) of this sequence with interpolated zeros(1, 0, 4, 0, 24, 0, 160, 0, 1120, ...) = is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
The Hankel transform of this sequence gives A103488. - Philippe Deléham, Dec 02 2007
Equals the central column of the triangle A038207. - Zerinvary Lajos, Dec 08 2007
Equals number of permutations whose reverse shares the same recording tableau in the Robinson-Schensted correspondence with n=(k-1)/2 for k odd. - Dang-Son Nguyen, Jul 02 2024
Number of ternary strings of length 2*n that have the same number of 0's as the combined number of 1's and 2's. For example, a(2)=24 since the strings of length 4 are the 6 permutations of 0011, the 12 permutations of 0012, and the 6 permutations of 0022. - Enrique Navarrete, Jul 30 2025

Links

Crossrefs

Diagonal of A013609.
Column k=0 of A067001.
Cf. A000079, A000984, A001906, A036442, A038207, A103488.

Programs

  • Magma
    [2^n*Binomial(2*n,n): n in [0..25]]; // Vincenzo Librandi, Oct 08 2015
  • Maple
    seq(binomial(2*n,n)*2^n,n=0..19); # Zerinvary Lajos, Dec 08 2007
  • Mathematica
    Table[2^n Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Dec 16 2014 *)
  • PARI
    {a(n)=if(n<0, 0, 2^n*(2*n)!/n!^2)} /* Michael Somos, Jan 31 2007 */
  • PARI
    { for (n = 0, 200, write("b059304.txt", n, " ", 2^n * (2*n)! / n!^2); ) } \\ Harry J. Smith, Jun 25 2009
  • PARI
    /* as lattice paths: same as in A092566 but use */
steps=[[1, 0], [1, 0], [0, 1]]; /* note the double [1, 0] */
/* Joerg Arndt, Jul 01 2011 */
  • SageMath
    def A059304(n): return pow(2,n)*binomial(2*n,n)
print([A059304(n) for n in range(41)]) # G. C. Greubel, Jan 18 2025

Formula

a(n) = 2^n * C(2*n,n).
D-finite with recurrence a(n) = 4*(2-1/n)*a(n-1).
a(n) = A000079(n)*A000984(n)
G.f.: 1/sqrt(1-8*x) - T. D. Noe, Jun 11 2002
E.g.f.: exp(4*x)*BesselI(0, 4*x). - Vladeta Jovovic, Aug 20 2003
a(n) = A038207(n,n). - Joerg Arndt, Jul 01 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+1)/(4*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 4*x/(4*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: Q(0)/(1+2*sqrt(x)), where Q(k) = 1 + 2*sqrt(x)/(1 - 2*sqrt(x)*(2*k+1)/(2*sqrt(x)*(2*k+1) + (k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 09 2013
O.g.f.: hypergeom([1/2], [], 8*x). - Peter Luschny, Oct 08 2015
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(3*n-2*k,n)* binomial(n+k,n). - Peter Bala, Aug 04 2016
a(n) ~ 8^n/sqrt(Pi*n). - Ilya Gutkovskiy, Aug 04 2016
From Amiram Eldar, Jul 21 2020: (Start)
Sum_{n>=0} 1/a(n) = 8/7 + 8*sqrt(7)*arcsin(1/sqrt(8))/49.
Sum_{n>=0} (-1)^n/a(n) = (8/27)*(3 - arcsinh(1/sqrt(8))). (End)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)*binomial(k,n). In general, for m >= 1, Sum_{k = n..m*n} binomial(m*n,k)*binomial(k,n) = 2^((m-1)*n)*binomial(m*n,n). - Peter Bala, Mar 25 2023
Conjecture: a(n) = Sum_{0 <= j, k <= n} binomial(n, j)*binomial(n, k)* binomial(k+j, n). - Peter Bala, Jul 16 2024

A094440 Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 8, 6, 4, 5, 15, 20, 10, 5, 8, 30, 45, 40, 15, 6, 13, 56, 105, 105, 70, 21, 7, 21, 104, 224, 280, 210, 112, 28, 8, 34, 189, 468, 672, 630, 378, 168, 36, 9, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 89, 605, 1870, 3465, 4290, 3696, 2310, 990, 330, 55, 11
Offset: 1

Views

Author

Clark Kimberling, May 03 2004

Keywords

Comments

Row sums yield the even-subscripted Fibonacci numbers (A001906).
Row n shows the coefficients of the numerator of the n-th derivative of c(n)/(x^2+x-1), where c(n) = ((-1)^(n + 1))/n!; see the Mathematica program. - Clark Kimberling, Oct 22 2019

Examples

			Triangle starts:
   1;
   1,  2;
   2,  3,   3;
   3,  8,   6,   4;
   5, 15,  20,  10,  5;
   8, 30,  45,  40, 15,  6;
  13, 56, 105, 105, 70, 21, 7;
  ...
T(4,3) = F(2)*C(4,2) = 1*6 = 6.

Links

Crossrefs

Cf. A000032, A000045, A001906, A030191, A081574, A094435, A132148.
Cf. A094435, A094436, A094437, A094438, A094439, A094441, A094442, A094443, A094444.
Cf. A367208, A367209, A367210, A367211.

Programs

  • GAP
    Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k-1)* Fibonacci(n-k+1) ))); # G. C. Greubel, Oct 30 2019
  • Magma
    /* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 15 2017
  • Maple
    with(combinat): T:=(n,k)->binomial(n,k-1)*fibonacci(n+1-k): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # Emeric Deutsch
  • Mathematica
    Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 14 2016 *)
(* Next program outputs polynomials having coefficients T(n,k) *)
g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[1/(1 - x - x^2), {x, n}]]]
Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* Clark Kimberling, Oct 22 2019 *)
  • PARI
    T(n,k) = binomial(n,k-1)*fibonacci(n-k+1);
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019
  • Sage
    [[binomial(n,k-1)*fibonacci(n-k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Oct 30 2019

Formula

From Peter Bala, Aug 17 2007: (Start)
With an offset of 0, the row polynomials F(n,x) = Sum_{k = 0..n} C(n,k)* Fibonacci(n-k)*x^k satisfy F(n,x)*L(n,x) = F(2*n,x), where L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k.
Other identities and formulas include:
F(n+1,x)^2 - F(n,x)*F(n+2,x) = (x^2 + x - 1)^n;
Sum_{k = 0..n} C(n,k)*F(n-k,x)*L(k,x) = (2^n)*F(n,x);
F(n,2*x) = Sum_{k = 0..n} C(n,k)*F(n-k,x)*x^k;
F(n,3*x) = Sum_{k = 0..n} C(n,k)*F(n-k,2*x)*x^k, etc.
The sequence {F(n,r)}n>=1 gives the r-th binomial transform of the Fibonacci numbers: r = 1 gives A001906, r = 2 gives A030191, r = 3 gives A099453, r = 4 gives A081574, r = 5 gives A081575.
F(n,1/phi) = (-1)^(n-1)*F(n,-phi) = sqrt(5)^(n-1) for n >= 1, where phi = (1 + sqrt(5))/2.
The polynomials F(n,-x) satisfy a Riemann hypothesis: the zeros of F(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
G.f.: t/(1 - (2*x + 1)*t + (x^2 + x - 1)*t^2) = t + (1 + 2*x)*t^2 + (2 + 3*x + 3*x^2)*t^3 + (3 + 8*x + 6*x^2 + 4*x^3)*t^4 + ... . (End)
From Peter Bala, Jun 29 2016: (Start)
Working with an offset of 0, the n-th row polynomial F(n,x) = 1/sqrt(5)*( (x + phi)^n - (x - 1/phi)^n ), where phi = (1 + sqrt(5))/2.
d/dx(F(n,x)) = n*F(n-1,x).
F(-n,x) = -F(n,x)/(x^2 + x - 1)^n.
F(n,x - 1) = (-1)^(n-1)*F(n,-x).
F(n,x) is a divisibility sequence of polynomials, that is, if n divides m then F(n,x) divides F(m,x) in the polynomial ring Z[x]. (End)
From G. C. Greubel, Oct 30 2019: (Start)
Sum_{k = 1..n} T(n,k) = Fibonacci(2*n).
Sum_{k = 1..n} (-1)^k * T(n,k) = (-1)^n * Fibonacci(n). (End)
From Clark Kimberling, Oct 30 2019: (Start)
F(n,x) is a strong divisibility sequence of polynomials in Z[x]; that is,
gcd(F(x,h),F(x,k)) = F(x,gcd(h,k)) for h,k >= 1. Thus, if x is an integer, then F(n,x) is a strong divisibility sequence of integers; e.g., for x=3, we have A099453. (End)
Let p(n) denote the polynomial F(x,n). Then p(n) = k(b^n - c^n), where k = -1/sqrt(5), b = (1/2)(2x + 1 - sqrt(5)), c = (1/2)(2x + 1 + sqrt(5)), and for n >=3, p(n) = u*p(n - 1) + v*p(n - 2), where u = 1 + 2 x, v = 1 - x - x^2. - Clark Kimberling, Nov 11 2023

Extensions

Error in expansion of generating function corrected by Peter Bala, Sep 24 2008

A056570 Third power of Fibonacci numbers (A000045).

Original entry on oeis.org

0, 1, 1, 8, 27, 125, 512, 2197, 9261, 39304, 166375, 704969, 2985984, 12649337, 53582633, 226981000, 961504803, 4073003173, 17253512704, 73087061741, 309601747125, 1311494070536, 5555577996431, 23533806109393, 99690802348032, 422297015640625
Offset: 0

Views

Author

Wolfdieter Lang, Jul 10 2000

Keywords

Comments

Divisibility sequence; that is, if n divides m, then a(n) divides a(m).
In general, cubing the terms of a Horadam sequence with signature (c,d) will result in a fourth-order recurrence with signature (c^3+2*c*d, c^4*d+3*(c*d)^2+2*d^3, -(c*d)^3-2*c*d^4, -d^6). - Gary Detlefs, Nov 12 2021
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/3,2/3)-fences and third-squares (1/3 X 1 pieces, always placed so that the shorter sides are horizontal). A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/6,1/3)-fences and (1/6,5/6)-fences. - Michael A. Allen, Jan 11 2022

Examples

			a(4) = 27 because the fourth Fibonacci number is 3 and 3^3 = 27.
a(5) = 125 because the fifth Fibonacci number is 5 and 5^3 = 125.

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).

Links

Crossrefs

Cf. A000045, A007598, A056588, A055870, A066259, A066258.
Cf. A346513 (first differences), A005968 (partial sums).
Third row of array A103323.

Programs

Formula

a(n) = A000045(n)^3.
G.f.: x*p(3, x)/q(3, x) with p(3, x) = Sum_{m=0..2} A056588(2, m)*x^m = 1 -2*x -x^2 and q(3, x) = Sum_{m=0..4} A055870(4, m)*x^m = 1 -3*x -6*x^2 +3*x^3 +x^4 = (1+x-x^2)*(1-4*x-x^2) (factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): 1*a(n) -3*a(n-1) -6*a(n-2) +3*a(n-3) +1*a(n-4) = 0, n >= 4, a(0)=0, a(1)=a(2)=1, a(3)=2^3. See 5th row of signed Fibonomial triangle for coefficients: A055870(4, m), m=0..4
a(n) = (Fibonacci(3n) - 3(-1)^n*Fibonacci(n))/5. - Ralf Stephan, May 14 2004
a(n) and a(n+1) are found as rightmost and leftmost terms (respectively) in M^n * [1 0 0 0] where M = the 4 X 4 upper triangular Pascal's triangle matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., Ma(4) = 27, a(5) = 125. M^4 * [1 0 0 0] = [125 75 45 27]; where 75 = A066259(4) and 45 = A066258(3). The characteristic polynomial of M = x^4 - 3x^3 - 6x^2 + 3x + 1. a(n)/a(n-1) of the sequence and companions tend to 2+sqrt(5) = 4.2360679..., an eigenvalue of M and a root of the polynomial. - Gary W. Adamson, Oct 31 2004
From R. J. Mathar, Oct 16 2006: (Start)
Sum_{j=0..n} binomial(n,j)*a(j) = (2^n*A001906(n) + 3*A000045(n))/5.
Sum_{j=0..n} (-1)^j*binomial(n,j)*a(j) = ((-2)^n*A000045(n) - 3*A001906(n))/5. (End)
G.f.: x*(1-2*x-x^2)/((1+x-x^2)*(1-4*x-x^2)). - Colin Barker, Feb 28 2012
a(n) = F(n-2)*F(n+1)^2 + F(n-1)*(-1)^n. - J. M. Bergot, Mar 17 2016
a(n) = ((-3*(1/2*(-1-sqrt(5)))^n-(2-sqrt(5))^n+3*(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)) / (5*sqrt(5)). - Colin Barker, Jun 04 2016
a(n) = F(n-1)*F(n)*F(n+1) + F(n)*(-1)^(n-1). - Tony Foster III, Apr 11 2018
5*a(n) = L(2*n-1)*F(n+2) - L(2*n+1)*F(n-2) - 7*(-1)^n*F(n), where L(n) = A000032(n). - Peter Bala, Nov 12 2019
F(n+1)*F(n)*F(n-1) = 2*Sum_{j=1..n-1} P(j)*a(n-j) for n>0, where Pell number P(n) = A000129(n). - Michael A. Allen, Jan 11 2022

A047656 a(n) = 3^((n^2-n)/2).

Original entry on oeis.org

1, 1, 3, 27, 729, 59049, 14348907, 10460353203, 22876792454961, 150094635296999121, 2954312706550833698643, 174449211009120179071170507, 30903154382632612361920641803529, 16423203268260658146231467800709255289, 26183890704263137277674192438430182020124347
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The number of outcomes of a chess tournament with n players.
For n >= 1, a(n) is the size of the Sylow 3-subgroup of the Chevalley group A_n(3) (sequence A053290). - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 30 2001
The number of binary relations on an n-element set that are both reflexive and antisymmetric. - Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005
The sequence a(n+1) = [1,3,27,729,59049,14348907,...] is the Hankel transform (see A001906 for definition) of A047891 = 1, 3, 12, 57, 300, 1586, 9912, ... . - Philippe Deléham, Aug 29 2006
a(n) is the number of binary relations on a set with n elements that are total relations, i.e., for a relation on a set X it holds for all a and b in X that a~b or b~a (or both). E.g., a(2) = 3 because there are three total relations on a set with two elements: {(a,a),(a,b),(b,a),(b,b)}, {(a,a),(a,b),(b,b)}, and {(a,a),(b,a),(b,b)}. - Geoffrey Critzer, May 23 2008
The number of semicomplete digraphs (or weak tournaments) on n labeled nodes. - Rémy-Robert Joseph, Nov 12 2012
The number of n X n binary matrices A that have a(i,j)=0 whenever a(j,i)=1 for i!=j and zeros on the diagonal. We need only consider the (n^2-n)/2 non-diagonal entry pairs . Since each pair is of the form <0,0>, <0,1>, or <1,0>, a(n) = 3^((n^2-n)/2). - Dennis P. Walsh, Apr 03 2014
a(n) is the number of symmetric (-1,0,1)-matrices of dimension (n-1) X (n-1). - Eric W. Weisstein, Jan 03 2021

Examples

			The a(2)=3 binary 2 X 2 matrices are [0 0; 0 0], [0 1; 0 0], and [0 0; 1 0]. - _Dennis P. Walsh_, Apr 03 2014

References

  • P. A. MacMahon, Chess tournaments and the like treated by the calculus of symmetric functions, Coll. Papers I, MIT Press, 344-375.

Links

Crossrefs

Cf. A007747.

Programs

Formula

a(n+1) is the determinant of an n X n matrix M_(i, j) = C(3*i,j). - Benoit Cloitre, Aug 27 2003
Sequence is given by the Hankel transform (see A001906 for definition) of A007564 = {1, 1, 4, 19, 100, 562, 3304, ...}; example: det([1, 1, 4, 19; 1, 4, 19, 100; 4, 19, 100, 562; 19, 100, 562, 3304]) = 3^6 = 729. - Philippe Deléham, Aug 20 2005
The sequence a(n+1) = [1,3,27,729,59049,14348907,...] is the Hankel transform (see A001906 for definition) of A047891 = 1, 3, 12, 57, 300, 1586, 9912, ... . - Philippe Deléham, Aug 29 2006
a(n) = 3^binomial(n,2). - Zerinvary Lajos, Jun 16 2007
G.f. A(x) satisfies: A(x) = 1 + x * A(3*x). - Ilya Gutkovskiy, Jun 04 2020
a(n) = a(n-1)*3^(n-1), a(0) = 1. - Mehdi Naima, Mar 09 2022

A124758 Product of the parts of the compositions in standard order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 4, 2, 3, 2, 2, 1, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 6, 5, 8, 4, 9, 6, 6, 3, 8, 6, 8, 4, 6, 4, 4, 2, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 7, 6, 10, 5, 12, 8, 8, 4, 12, 9, 12, 6, 9, 6, 6, 3, 10, 8, 12, 6, 12, 8, 8, 4, 8, 6, 8, 4, 6, 4, 4, 2, 6, 5, 8, 4, 9, 6
Offset: 0

Views

Author

Franklin T. Adams-Watters, Nov 06 2006

Keywords

Comments

The standard order of compositions is given by A066099.
A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. - Gus Wiseman, Apr 03 2020

Examples

			Composition number 11 is 2,1,1; 2*1*1 = 2, so a(11) = 2.
The table starts:
  1
  1
  2 1
  3 2 2 1
  4 3 4 2 3 2 2 1
  5 4 6 3 6 4 4 2 4 3 4 2 3 2 2 1
The 146-th composition in standard order is (3,3,2), with product 18, so a(146) = 18. - _Gus Wiseman_, Apr 03 2020

Links

Crossrefs

Cf. A066099, A118851, A011782 (row lengths), A001906 (row sums).
The lengths of standard compositions are given by A000120.
The version for prime indices is A003963.
The version for binary indices is A096111.
Taking the sum instead of product gives A070939.
The sum of binary indices is A029931.
The sum of prime indices is A056239.
Taking GCD instead of product gives A326674.
Positions of first appearances are A331579.
Cf. A001519, A011782, A048793, A114994, A124767, A228351, A272919, A329369 (similar recurrence), A333218, A333219.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Times@@stc[n],{n,0,100}] (* Gus Wiseman, Apr 03 2020 *)

Formula

For a composition b(1),...,b(k), a(n) = Product_{i=1}^k b(i).
a(A164894(n)) = a(A246534(n)) = n!. - Gus Wiseman, Apr 03 2020
a(A233249(n)) = a(A333220(n)) = A003963(n). - Gus Wiseman, Apr 03 2020
From Mikhail Kurkov, Jul 11 2021: (Start)
Some conjectures:
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + 1/A001511(n))*a(n) = 2*a(n) + a(n - 2^f(n)) - a(2n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
From the 1st formula for a(2n) we get a(4n+2) = 2*a(n), a(4n) = 2*a(2n) - a(n).
Sum_{k=0..2^n - 1} a(k) = A001519(n+1) for n >= 0.
a((4^n - 1)/3) = A011782(n) for n >= 0.
a(2^m*(2^n - 1)) = m + 1 for n > 0, m >= 0. (End)
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