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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A194005 Triangle of the coefficients of an (n+1)-th order differential equation associated with A103631.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 3, 3, 1, 1, 5, 4, 6, 3, 1, 1, 6, 5, 10, 6, 4, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
Offset: 0

Views

Author

Johannes W. Meijer and A. Hirschberg (a.hirschberg(AT)tue.nl), Aug 11 2011

Keywords

Comments

This triangle is a companion to Parks' triangle A103631.
The coefficients of triangle A103631(n,k) appear in appendix 2 of Park’s remarkable article “A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov” if we assume that the b(n) coefficients are all equal to 1, see the second Maple program.
The a(n,k) coefficients of the triangle given above are related to the coefficients of a linear (n+1)-th order differential equation for the case b(n)=1, see the examples.
a(n,k) is also the number of symmetric binary strings of odd length n with Hamming weight k>0 and no consecutive 1's. - Christian Barrientos and Sarah Minion, Feb 27 2018

Examples

			For the 5th-order linear differential equation the coefficients a(k) are: a(0) = 1, a(1) = a(4,0) = 1, a(2) = a(4,1) = 4, a(3) = a(4,2) = 3, a(4) = a(4,3) = 3 and a(5) = a(4,4) = 1.
The corresponding Hurwitz matrices A(k) are, see Parks: A(5) = Matrix([[a(1),a(0),0,0,0], [a(3),a(2),a(1),a(0),0], [a(5),a(4),a(3),a(2),a(1)], [0,0,a(5),a(4),a(3)], [0,0,0,0,a(5)]]), A(4) = Matrix([[a(1),a(0),0,0], [a(3),a(2),a(1),a(0)], [a(5),a(4),a(3),a(2)], [0,0,a(5),a(4)]]), A(3) = Matrix([[a(1),a(0),0], [a(3),a(2),a(1)], [a(5),a(4),a(3)]]), A(2) = Matrix([[a(1),a(0)], [a(3),a(2)]]) and A(1) = Matrix([[a(1)]]).
The values of b(k) are, see Parks: b(1) = d(1), b(2) = d(2)/d(1), b(3) = d(3)/(d(1)*d(2)), b(4) = d(1)*d(4)/(d(2)*d(3)) and b(5) = d(2)*d(5)/(d(3)*d(4)).
These a(k) values lead to d(k) = 1 and subsequently to b(k) = 1 and this confirms our initial assumption, see the comments.
'
Triangle starts:
  [0] 1;
  [1] 1, 1;
  [2] 1, 2, 1;
  [3] 1, 3, 2,  1;
  [4] 1, 4, 3,  3,  1;
  [5] 1, 5, 4,  6,  3,  1;
  [6] 1, 6, 5, 10,  6,  4,  1;
  [7] 1, 7, 6, 15, 10, 10,  4,  1;
  [8] 1, 8, 7, 21, 15, 20, 10,  5, 1;
  [9] 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;
		

Crossrefs

Cf. A065941 and A103631.
Triangle sums (see A180662): A000071 (row sums; alt row sums), A075427 (Kn22), A000079 (Kn3), A109222(n+1)-1 (Kn4), A000045 (Fi1), A034943 (Ca3), A001519 (Gi3), A000930 (Ze3)
Interesting diagonals: T(n,n-4) = A189976(n+5) and T(n,n-5) = A189980(n+6)
Cf. A052509.

Programs

  • Haskell
    a194005 n k = a194005_tabl !! n !! k
    a194005_row n = a194005_tabl !! n
    a194005_tabl = [1] : [1,1] : f [1] [1,1] where
       f row' row = rs : f row rs where
         rs = zipWith (+) ([0,1] ++ row') (row ++ [0])
    -- Reinhard Zumkeller, Nov 22 2012
  • Maple
    A194005 := proc(n, k): binomial(floor((2*n+1-k)/2), n-k) end:
    for n from 0 to 11 do seq(A194005(n, k), k=0..n) od;
    seq(seq(A194005(n,k), k=0..n), n=0..11);
    nmax:=11: for n from 0 to nmax+1 do b(n):=1 od:
    A103631 := proc(n,k) option remember: local j: if k=0 and n=0 then b(1)
    elif k=0 and n>=1 then 0 elif k=1 then b(n+1) elif k=2 then b(1)*b(n+1)
    elif k>=3 then expand(b(n+1)*add(procname(j,k-2), j=k-2..n-2)) fi: end:
    for n from 0 to nmax do for k from 0 to n do
    A194005(n,k):= add(A103631(n1,k), n1=k..n) od: od:
    seq(seq(A194005(n,k),k=0..n), n=0..nmax);
  • Mathematica
    Flatten[Table[Binomial[Floor[(2n+1-k)/2],n-k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Apr 15 2012 *)

Formula

T(n,k) = binomial(floor((2*n+1-k)/2), n-k).
T(n,k) = sum(A103631(n1,k), n1=k..n), 0<=k<=n and n>=0.
T(n,k) = sum(binomial(floor((2*n1-k-1)/2), n1-k), n1=k..n).
T(n,0) = T(n,n) = 1, T(n,k) = T(n-2,k-2) + T(n-1,k), 0 < k < n. - Reinhard Zumkeller, Nov 23 2012

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
Offset: 0

Views

Author

Len Smiley, Nov 29 2001

Keywords

Comments

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]
Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011
Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011
From Gary W. Adamson, Aug 25 2019: (Start)
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:
3...............1
5...............1, 2
7...............1, 2, 3
9...............1, 2, 4
...............3 (singleton terms reduce to "1") (9 has two rows)
11...............1, 2, 4, 3, 5
13...............1, 2, 4, 5, 3, 6
15...............1, 2, 4, 7
................3, 6 (dividing through by the gcd gives (1, 2))
................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)
From Gary W. Adamson, Sep 30 2019: (Start)
The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)
From Gary W. Adamson, Aug 21 2019: (Start)
Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:
....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,
....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)
Received from Herb Conn, Jan 2004: (Start)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x - 1
sin(7A)/sin A = x^3 + x - 2x - 1
sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1
... (signed ++--++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012
The matrix inverse is given by
1;
1, 1;
0, -1, 1;
0, 1, -2, 1;
0, 0, 1, -2, 1;
0, 0, -1, 3, -3, 1;
0, 0, 0, -1, 3, -3, 1;
0, 0, 0, 1, -4, 6, -4, 1;
0, 0, 0, 0, 1, -4, 6, -4, 1;
... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

Examples

			Triangle T(n, k) begins:
n\k 0  1  2  3   4   5  6   7  8  9 ...
---------------------------------------
[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2,  1,
[5] 1, 1, 4, 3,  3,  1,
[6] 1, 1, 5, 4,  6,  3,  1,
[7] 1, 1, 6, 5, 10,  6,  4,  1,
[8] 1, 1, 7, 6, 15, 10, 10,  4,  1,
[9] 1, 1, 8, 7, 21, 15, 20, 10,  5, 1,
---------------------------------------
From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
  N =  5:  x^2 - 5;
  N =  7:  x^3 - 7x + 7;
  N = 11:  x^5 - 11x^3 + 11x^2 + 11x - 11;
  N = 13:  x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
  N = 17:  x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
  N = 19:  x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)
		

Crossrefs

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).

Programs

  • Haskell
    a065941 n k = a065941_tabl !! n !! k
    a065941_row n = a065941_tabl !! n
    a065941_tabl = iterate (\row ->
       zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
    -- Reinhard Zumkeller, May 07 2012
    
  • Magma
    [Binomial(n - Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    A065941 := proc(n,k): binomial(n-floor((k+1)/2),floor(k/2)) end: seq(seq(A065941(n,k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
    A065941 := proc(n,k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j,k-2), j=k-2..n-2) fi: end: seq(seq(A065941(n,k), k=0..n), n=0..15);  # Johannes W. Meijer, Aug 11 2011
    # The function qStirling2 is defined in A333143.
    seq(print(seq(qStirling2(n, k, -1), k=0..n)), n=0..9);
    # Peter Luschny, Mar 09 2020
  • Mathematica
    Flatten[Table[Binomial[n-Floor[(k+1)/2],Floor[k/2]],{n,0,15},{k,0,n}]] (* Harvey P. Dale, Dec 11 2011 *)
  • PARI
    T065941(n, k) = binomial(n-(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014
    
  • Sage
    [[binomial(n - floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019

Formula

T(n, k) = binomial(n-floor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k). - Paul Barry, Mar 11 2003
Triangle is a reflection of that in A066170 (absolute values). - Gary W. Adamson, Feb 16 2004
Recurrences: T(k, 0) = 1, T(k, n) = T(k-1, n) + T(k-2, n-2), or T(k, n) = T(k-1, n) + T(k-1, n-1) if n even, T(k-1, n-1) if n odd. - Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1-y-x^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1-y)^[k/2]. - Ralf Stephan, May 17 2004
T(n, k) = A108299(n, k)*A087960(k) = abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
T(n,k) = A046854(n, n-k) = abs(A066170(n, n-k)).
T(n+k, n-k) = A109223(n,k).
T(n, k) = sum(T(j, k-2), j=k-2..n-2), 2 <= k <= n, n>=2;
T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)
For n > 1: T(n, k) = T(n-2, k) + T(n-1, k), 1 < k < n. - Reinhard Zumkeller, Apr 24 2013

A133607 Triangle read by rows: T(n, k) = qStirling2(n, k, q) for q = -1, with 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, -1, 0, 1, -1, -1, 0, 1, -1, -2, 1, 0, 1, -1, -3, 2, 1, 0, 1, -1, -4, 3, 3, -1, 0, 1, -1, -5, 4, 6, -3, -1, 0, 1, -1, -6, 5, 10, -6, -4, 1, 0, 1, -1, -7, 6, 15, -10, -10, 4, 1, 0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1
Offset: 0

Views

Author

Philippe Deléham, Dec 27 2007

Keywords

Comments

Previous name: Triangle T(n,k), 0<=k<=n, read by rows given by [0, 1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, -2, 1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Examples

			Triangle begins:
  1;
  0, 1;
  0, 1, -1;
  0, 1, -1, -1;
  0, 1, -1, -2, 1;
  0, 1, -1, -3, 2, 1;
  0, 1, -1, -4, 3, 3, -1;
  0, 1, -1, -5, 4, 6, -3, -1;
  0, 1, -1, -6, 5, 10, -6, -4, 1;
  0, 1, -1, -7, 6, 15, -10, -10, 4, 1;
  0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  0, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...
Triangle A103631 begins:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 1, 2, 1;
  0, 1, 1, 3, 2, 1;
  0, 1, 1, 4, 3, 3, 1;
  0, 1, 1, 5, 4, 6, 3, 1;
  0, 1, 1, 6, 5, 10, 6, 4, 1;
  0, 1, 1, 7, 6, 15, 10, 10, 4, 1;
  0, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1;
  0, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;
  0, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1;
  ...
Triangle A108299 begins:
  1;
  1, -1;
  1, -1, -1;
  1, -1, -2, 1;
  1, -1, -3, 2, 1;
  1, -1, -4, 3, 3, -1;
  1, -1, -5, 4, 6, -3, -1;
  1, -1, -6, 5, 10, -6, -4, 1;
  1, -1, -7, 6, 15, -10, -10, 4, 1;
  1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...
		

Crossrefs

Another version is A108299.
Unsigned version is A103631 (T(n,k) = A103631(n,k)*A057077(k)).

Programs

  • Mathematica
    m = 13
    (* DELTA is defined in A084938 *)
    DELTA[Join[{0, 1}, Table[0, {m}]], Join[{1, -2, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)
    qStirling2[n_, k_, q_] /; 1 <= k <= n := q^(k-1) qStirling2[n-1, k-1, q] + Sum[q^j, {j, 0, k-1}] qStirling2[n-1, k, q];
    qStirling2[n_, 0, _] := KroneckerDelta[n, 0];
    qStirling2[0, k_, _] := KroneckerDelta[0, k];
    qStirling2[, , _] = 0;
    Table[qStirling2[n, k, -1], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2020 *)
  • Sage
    from sage.combinat.q_analogues import q_stirling_number2
    for n in (0..9):
        print([q_stirling_number2(n,k).substitute(q=-1) for k in [0..n]])
    # Peter Luschny, Mar 09 2020

Formula

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k)= A057077(n), A010892(n), A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n-1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 respectively .
Sum_{k, 0<=k<=n}T(n,k)*x^k = A000007(n), A010892(n), A133631(n), A133665(n), A133666(n), A133667(n), A133668(n), A133669(n), A133671(n), A133672(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively .
G.f.: (1-x+y*x)/(1-x+y^2*x^2). - Philippe Deléham, Mar 14 2012
T(n,k) = T(n-1,k) - T(n-2,k-2), T(0,0) = T(1,1) = T(2,1) = 1, T(1,0) = T(2,0) = 0, T(2,2) = -1 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 14 2012

Extensions

New name from Peter Luschny, Mar 09 2020

A374439 Triangle read by rows: the coefficients of the Lucas-Fibonacci polynomials. T(n, k) = T(n - 1, k) + T(n - 2, k - 2) with initial values T(n, k) = k + 1 for k < 2.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 3, 4, 1, 1, 2, 4, 6, 3, 2, 1, 2, 5, 8, 6, 6, 1, 1, 2, 6, 10, 10, 12, 4, 2, 1, 2, 7, 12, 15, 20, 10, 8, 1, 1, 2, 8, 14, 21, 30, 20, 20, 5, 2, 1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1, 1, 2, 10, 18, 36, 56, 56, 70, 35, 30, 6, 2
Offset: 0

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Author

Peter Luschny, Jul 22 2024

Keywords

Comments

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.
A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.
The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).
Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.
The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

Examples

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 1]
  [ 3] [1, 2, 2,  2]
  [ 4] [1, 2, 3,  4,  1]
  [ 5] [1, 2, 4,  6,  3,  2]
  [ 6] [1, 2, 5,  8,  6,  6,  1]
  [ 7] [1, 2, 6, 10, 10, 12,  4,  2]
  [ 8] [1, 2, 7, 12, 15, 20, 10,  8,  1]
  [ 9] [1, 2, 8, 14, 21, 30, 20, 20,  5,  2]
  [10] [1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1]
.
Table of interpolated sequences:
  |  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |
  |  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|
  |    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |
  |  0 |        -1         |     1   |       1     |       1    |
  |  1 |         1         |     3   |       0     |       4    |
  |  2 |         0         |     4   |       1     |       9    |
  |  3 |         1         |     7   |       2     |      22    |
  |  4 |         1         |    11   |       5     |      53    |
  |  5 |         2         |    18   |      12     |     128    |
  |  6 |         3         |    29   |      29     |     309    |
  |  7 |         5         |    47   |      70     |     746    |
  |  8 |         8         |    76   |     169     |    1801    |
  |  9 |        13         |   123   |     408     |    4348    |
		

Crossrefs

Triangles related to Lucas polynomials: A034807, A114525, A122075, A061896, A352362.
Triangles related to Fibonacci polynomials: A162515, A053119, A168561, A049310, A374441.
Sums include: A000204 (Lucas numbers, row), A000045 & A212804 (even sums, Fibonacci numbers), A006355 (odd sums), A039834 (alternating sign row).
Type m^n*P(n, 1/m): A000129 & A048654 (Pell, m=2), A108300 & A003688 (m=3), A001077 & A048875 (m=4).
Adding and subtracting the values in a row of the table (plus halving the values obtained in this way): A022087, A055389, A118658, A052542, A163271, A371596, A324969, A212804, A077985, A069306, A215928.
Columns include: A040000 (k=1), A000027 (k=2), A005843 (k=3), A000217 (k=4), A002378 (k=5).
Diagonals include: A000034 (k=n), A029578 (k=n-1), abs(A131259) (k=n-2).
Cf. A029578 (subdiagonal), A124038 (row reversed triangle, signed).

Programs

  • Magma
    function T(n,k) // T = A374439
      if k lt 0 or k gt n then return 0;
      elif k le 1 then return k+1;
      else return T(n-1,k) + T(n-2,k-2);
      end if;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 23 2025
    
  • Maple
    A374439 := (n, k) -> ifelse(k::odd, 2, 1)*binomial(n - irem(k, 2) - iquo(k, 2), iquo(k, 2)):
    # Alternative, using the function qStirling2 from A333143:
    T := (n, k) -> 2^irem(k, 2)*qStirling2(n, k, -1):
    seq(seq(T(n, k), k = 0..n), n = 0..10);
  • Mathematica
    A374439[n_, k_] := (# + 1)*Binomial[n - (k + #)/2, (k - #)/2] & [Mod[k, 2]];
    Table[A374439[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Paolo Xausa, Jul 24 2024 *)
  • Python
    from functools import cache
    @cache
    def T(n: int, k: int) -> int:
        if k > n: return 0
        if k < 2: return k + 1
        return T(n - 1, k) + T(n - 2, k - 2)
    
  • Python
    from math import comb as binomial
    def T(n: int, k: int) -> int:
        o = k & 1
        return binomial(n - o - (k - o) // 2, (k - o) // 2) << o
    
  • Python
    def P(n, x):
        if n < 0: return P(n, x)
        return sum(T(n, k)*x**k for k in range(n + 1))
    def sgn(x: int) -> int: return (x > 0) - (x < 0)
    # Table of interpolated sequences
    print("|  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |")
    print("|  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|")
    print("|    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |")
    f = "| {0:2d} | {1:9d}         |  {2:4d}   |   {3:5d}     |    {4:4d}    |"
    for n in range(10): print(f.format(n, -P(n, -1), P(n, 1), int(2**n*P(n, -1/2)), int(2**n*P(n, 1/2))))
    
  • SageMath
    from sage.combinat.q_analogues import q_stirling_number2
    def A374439(n,k): return (-1)^((k+1)//2)*2^(k%2)*q_stirling_number2(n+1, k+1, -1)
    print(flatten([[A374439(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 23 2025

Formula

T(n, k) = 2^k' * binomial(n - k' - (k - k') / 2, (k - k') / 2) where k' = 1 if k is odd and otherwise 0.
T(n, k) = (1 + (k mod 2))*qStirling2(n, k, -1), see A333143.
2^n*P(n, -1/2) = A000129(n - 1), Pell numbers, P(-1) = 1.
2^n*P(n, 1/2) = A048654(n), dual Pell numbers.
T(2*n, n) = (1/2)*(-1)^n*( (1+(-1)^n)*A005809(n/2) - 2*(1-(-1)^n)*A045721((n-1)/2) ). - G. C. Greubel, Jan 23 2025

A103632 Expansion of (1 - x + x^2)/(1 - x - x^4).

Original entry on oeis.org

1, 0, 1, 1, 2, 2, 3, 4, 6, 8, 11, 15, 21, 29, 40, 55, 76, 105, 145, 200, 276, 381, 526, 726, 1002, 1383, 1909, 2635, 3637, 5020, 6929, 9564, 13201, 18221, 25150, 34714, 47915, 66136, 91286, 126000, 173915, 240051, 331337, 457337, 631252, 871303, 1202640
Offset: 0

Views

Author

Paul Barry, Feb 11 2005

Keywords

Comments

Diagonal sums of A103631.
The Kn11 sums, see A180662, of triangle A065941 equal the terms of this sequence without a(0) and a(1). - Johannes W. Meijer, Aug 11 2011
For n >= 2, a(n) is the number of palindromic compositions of n-2 with parts in {2,1,3,5,7,9,...}. The generating function follows easily from Theorem 1,2 of the Hoggatt et al. reference. Example: a(9) = 8 because we have 7, 151, 11311, 232, 313, 12121, 21112, and 1111111. - Emeric Deutsch, Aug 17 2016
Essentially the same as A003411. - Georg Fischer, Oct 07 2018

Crossrefs

Cf. A275446.

Programs

  • GAP
    a:=[1,0,1,1];;  for n in [5..50] do a[n]:=a[n-1]+a[n-4]; od; a; # Muniru A Asiru, Oct 07 2018
    
  • Magma
    I:=[1,0,1,1]; [n le 4 select I[n] else Self(n-1) + Self(n-4): n in [1..50]]; // G. C. Greubel, Mar 10 2019
    
  • Maple
    A103632 := proc(n): add( binomial(floor((2*n-3*k-1)/2), n-2*k), k=0..floor(n/2)) end: seq(A103632(n), n=0..46); # Johannes W. Meijer, Aug 11 2011
  • Mathematica
    LinearRecurrence[{1,0,0,1}, {1,0,1,1}, 50] (* G. C. Greubel, Mar 10 2019 *)
  • PARI
    my(x='x+O('x^50)); Vec((1-x+x^2)/(1-x-x^4)) \\ G. C. Greubel, Mar 10 2019
    
  • Sage
    ((1-x+x^2)/(1-x-x^4)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Mar 10 2019

Formula

G.f.: (1 - x + x^2)/(1 - x - x^4).
a(n) = a(n-1) + a(n-4) with a(0)=1, a(1)=0, a(2)=1 and a(3)=1.
a(n) = Sum_{k=0..floor(n/2)} binomial(floor((2*n-3*k-1)/2), n-2*k).
a(n) = A003269(n+1) - A003269(n-4), n > 4.

Extensions

Formula corrected by Johannes W. Meijer, Aug 11 2011

A122994 a(n) = a(n-1)+9*a(n-2) initialized with a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 12, 39, 147, 498, 1821, 6303, 22692, 79419, 283647, 998418, 3551241, 12537003, 44498172, 157331199, 557814747, 1973795538, 6994128261, 24758288103, 87705442452, 310530035379, 1099879017447, 3894649335858, 13793560492881, 48845404515603, 172987448951532
Offset: 0

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Author

Roger L. Bagula, Sep 22 2006

Keywords

Comments

The two roots of the denominator of the g.f. (for Binet's formula) are -0.393486... and 0.2823756...
Pisano period lengths: 1, 3, 1, 6, 6, 3, 6, 12, 1, 6, 10, 6, 84, 6, 6, 24,144, 3, 72, 6,... - R. J. Mathar, Aug 10 2012

Crossrefs

Cf. A026597.

Programs

Formula

G.f.: -(1+2*x)/(-1+x+9*x^2). a(n) = A015445(n)+2*A015445(n-1). [R. J. Mathar, Aug 12 2009]
a(n) = (1/2+5*sqrt(37)/74) *(1/2+sqrt(37)/2)^(n-1) +(1/2-5*sqrt(37)/74) *(1/2-sqrt(37)/2)^(n-1). [Antonio Alberto Olivares, Jun 07 2011]
a(n) = Sum_{k, 0<=k<=n} A103631(n,k)*3^k. - Philippe Deléham, Dec 17 2011
a(n) = A015445(n) + 2*A015445(n-1), n>0. - Ralf Stephan, Jul 21 2013

Extensions

Definition replaced with the Deleham recurrence of Mar 2009 by the Assoc. Editors of the OEIS, Mar 12 2010

A123029 a(2*n-1) = Product_{i=1..n} Fibonacci(2*i-1) and a(2*n) = Product_{i=1..n} Fibonacci(2*i).

Original entry on oeis.org

1, 1, 2, 3, 10, 24, 130, 504, 4420, 27720, 393380, 3991680, 91657540, 1504863360, 55911099400, 1485300136320, 89290025741800, 3838015552250880, 373321597626465800, 25964175210977203200, 4086378207619294646800
Offset: 1

Views

Author

Roger L. Bagula, Sep 25 2006

Keywords

Comments

From Johannes W. Meijer, Aug 21 2011: (Start)
An appropriate name for this sequence is Fibonacci double factorial, cf. A006882.
In Parks' article appendix 2, a number triangle T(n,k) with T(n,n) = a(n+1), n>=0, appears if we assume that b(r) = Fibonacci(r); see A103631 and A194005. (End)
The original name of this sequence was: A000045 inside a second linear differential equation recursion: b(n) = b(n-1) + b(n-2) --> Binet(n) of A000045 a(n) = b(n)*a(n-2)/(n*(n-1)).
Bagula also stated that using the solutions to these second order differential equations Markov/ linear recursions can be encoded as analog functions.
Partial products of the odd-indexed Fibonacci numbers interleaved with the partial products of the even-indexed Fibonacci numbers. - Harvey P. Dale, Mar 14 2012

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 10*x^5 + 24*x^6 + 130*x^7 + 504*x^8 + ...
		

Crossrefs

Programs

  • Magma
    function a(n)
      if n lt 2 then return 1;
      else return Fibonacci(n)*a(n-2);
      end if; return a;
    end function;
    [a(n): n in [1..30]]; // G. C. Greubel, Jul 20 2021
    
  • Maple
    with(combinat): A123029 :=proc(n): if type(n,even) then mul(fibonacci(2*i), i=1..n/2) else mul(fibonacci(2*i-1), i= 1..(n+1)/2) fi: end: seq(A123029(n), n=1..21); # Johannes W. Meijer, Aug 21 2011
  • Mathematica
    a[n_]:= a[n]= If[n<2, 1, Fibonacci[n]*a[n-2]]; Table[a[n], {n, 30}] (* modified by G. C. Greubel, Jul 20 2021 *)
    With[{nn=21},Riffle[FoldList[Times,1,Fibonacci[Range[3,nn,2]]],FoldList[ Times,1, Fibonacci[ Range[4,nn+1,2]]]]] (* Harvey P. Dale, Mar 14 2012 *)
  • PARI
    {a(n) = if( n<0, 0, prod(k=0, (n-1)\2, fibonacci(n - 2*k)))}; /* Michael Somos, Oct 07 2014 */
    
  • Sage
    def a(n): return 1 if (n<2) else fibonacci(n)*a(n-2)
    [a(n) for n in (1..30)] # G. C. Greubel, Jul 20 2021

Formula

a(n) = n!*c(n) with c(n) = b(n)*c(n-2)/(n*(n-1)), c(0) = 1, c(1) = 1; b(n) = b(n-1) + b(n-2), b(0) = 0, b(1) = 1 and b(n) = F(n) with F(n) = A000045(n).
From Johannes W. Meijer, Aug 21 2011: (Start)
a(n) = F(n)*a(n-2).
a(2*n) = A194157(n) and a(2*n-1) = A194158(n). (End)
a(0) = 1 by convention since empty products equal 1. - Michael Somos, Oct 07 2014
0 = a(n)*(a(n+2)*a(n+3) - a(n+1)*a(n+4)) + a(n+1)*(+a(n+2)^2) for all n>=0. - Michael Somos, Oct 07 2014

Extensions

Edited by Johannes W. Meijer, Aug 21 2011

A374441 Triangle read by rows: T(n, k) = binomial(n - floor(k/2), ceiling(k/2)) - binomial(n - ceiling(k/2), floor(k/2)).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 3, 0, 1, 0, 0, 4, 0, 3, 0, 0, 0, 5, 0, 6, 0, 1, 0, 0, 6, 0, 10, 0, 4, 0, 0, 0, 7, 0, 15, 0, 10, 0, 1, 0, 0, 8, 0, 21, 0, 20, 0, 5, 0, 0, 0, 9, 0, 28, 0, 35, 0, 15, 0, 1, 0, 0, 10, 0, 36, 0, 56, 0, 35, 0, 6, 0, 0, 0, 11, 0, 45, 0, 84, 0, 70, 0, 21, 0, 1, 0
Offset: 0

Views

Author

Peter Luschny, Jul 19 2024

Keywords

Comments

Member of the family of Fibonacci polynomials (A011973, A162515, ...) and Chebyshev polynomials (A053119).

Examples

			Triangle starts:
  [ 0]  0;
  [ 1]  0, 0;
  [ 2]  0, 1, 0;
  [ 3]  0, 2, 0,  0;
  [ 4]  0, 3, 0,  1, 0;
  [ 5]  0, 4, 0,  3, 0,  0;
  [ 6]  0, 5, 0,  6, 0,  1, 0;
  [ 7]  0, 6, 0, 10, 0,  4, 0,  0;
  [ 8]  0, 7, 0, 15, 0, 10, 0,  1, 0;
  [ 9]  0, 8, 0, 21, 0, 20, 0,  5, 0, 0;
  [10]  0, 9, 0, 28, 0, 35, 0, 15, 0, 1, 0;
		

Crossrefs

Cf. A374440 (odd columns agree).
Cf. A000071 (row sums), A065941, A194005, A103631, A007318.

Programs

  • Maple
    T := (n, k) -> if k::even then 0 else binomial(n - (k + 1)/2, (k + 1)/2) fi:
    # Or as a recurrence:
    T := proc(n, k) option remember; if k::even or k > n then 0 elif k = 1 then n - 1 else T(n - 1, k) + T(n - 2, k - 2) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..12);
  • Mathematica
    A374441[n_, k_] := If[OddQ[k], Binomial[n - (k + 1)/2, (k + 1)/2], 0];
    Table[A374441[n, k], {n, 0, 15}, {k, 0, n}] (* Paolo Xausa, Nov 16 2024 *)
  • Python
    from math import isqrt, comb
    def A374441(n):
        a = (m:=isqrt(k:=n+1<<1))-(k<=m*(m+1))
        b = n-comb(a+1,2)
        return comb(a-(b+1>>1),b+1>>1) if b&1 else 0 # Chai Wah Wu, Nov 14 2024
    
  • Python
    from math import comb as binomial
    def row(n: int) -> list[int]:
        return [binomial(n - (k+1)//2, (k+1)//2) if k%2 else 0 for k in range(n+1)]
    for n in range(11): print(row(n))  # Peter Luschny, Nov 21 2024

Formula

T(n, k) = [x^(n-k)][z^n] (x / (1 - x*z - z^2)).
T(n, k) = binomial(n - (k + 1)/2, (k + 1)/2) if k is odd, and otherwise 0.
Sum_{k=0..n} T(n, k) = Fibonacci(n + 1) - 1.
Columns with odd index agree with the odd indexed columns of A374440.

A103633 Triangle read by rows: triangle of repeated stepped binomial coefficients.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 2, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 0, 0, 0, 0, 0, 0, 1, 6, 15
Offset: 0

Views

Author

Paul Barry, Feb 11 2005

Keywords

Comments

Row sums are Sum_{k=0..n} binomial(floor(n/2),n-k) = (1,1,2,2,4,4,...). Diagonal sums have g.f. (1+x^2)/(1-x^3-x^4) (see A079398). Matrix inverse of the signed triangle (-1)^(n-k)T(n,k) is A103631. Matrix inverse of T(n,k) is the alternating signed version of A103631.
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ....] DELTA [1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 08 2005

Examples

			Triangle begins:
  1;
  0,  1;
  0,  1,  1;
  0,  0,  1,  1;
  0,  0,  1,  2,  1;
  0,  0,  0,  1,  2,  1;
  0,  0,  0,  1,  3,  3,  1;
  0,  0,  0,  0,  1,  3,  3,  1;
  0,  0,  0,  0,  1,  4,  6,  4,  1;
  0,  0,  0,  0,  0,  1,  4,  6,  4,  1;
  0,  0,  0,  0,  0,  1,  5, 10, 10,  5,  1;
  0,  0,  0,  0,  0,  0,  1,  5, 10, 10,  5,  1;
  0,  0,  0,  0,  0,  0,  1,  6, 15, 20, 15,  6,  1; ...
		

Formula

Number triangle T(n, k) = binomial(floor(n/2), n-k).
Sum_{n>=0} T(n, k) = A000045(k+2) = Fibonacci(k+2). - Philippe Deléham, Oct 08 2005
Sum_{k=0..n} T(n,k) = 2^floor(n/2) = A016116(n). - Philippe Deléham, Dec 03 2006
G.f.: (1+x*y)/(1-x^2*y-x^2*y^2). - Philippe Deléham, Nov 10 2013
T(n,k) = T(n-2,k-1) + T(n-2,k-2) for n > 2, T(0,0) = T(,1) = T(2,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0, T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Nov 10 2013

A153764 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,-1,0,0,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 3, 1, 1, 0, 1, 3, 3, 4, 1, 1, 0, 1, 3, 6, 4, 5, 1, 1, 0, 1, 4, 6, 10, 5, 6, 1, 1, 0, 1, 4, 10, 10, 15, 6, 7, 1, 1, 0, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 0, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 0, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 0
Offset: 0

Views

Author

Philippe Deléham, Jan 01 2009

Keywords

Comments

A130595*A153342 as infinite lower triangular matrices. Reflected version of A103631. Another version of A046854. Row sums are Fibonacci numbers (A000045).
A055830*A130595 as infinite lower triangular matrices.

Examples

			Triangle begins:
  1;
  1, 0;
  1, 1, 0;
  1, 1, 1, 0;
  1, 2, 1, 1, 0;
  1, 2, 3, 1, 1, 0;
  1, 3, 3, 4, 1, 1, 0;
  ...
		

Crossrefs

Programs

  • Magma
    /* As triangle */ [[Binomial(Floor((n+k-1)/2),k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 28 2016
  • Mathematica
    Table[Binomial[Floor[(n + k - 1)/2], k], {n, 0, 45}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 27 2016 *)

Formula

T(n,k) = binomial(floor((n+k-1)/2),k).
Sum_{k=0..n} T(n,k)*x^k = A122335(n-1), A039834(n-2), A000012(n), A000045(n+1), A001333(n), A003688(n), A015448(n), A015449(n), A015451(n), A015453(n), A015454(n), A015455(n), A015456(n), A015457(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively. - Philippe Deléham, Dec 17 2011
Sum_{k=0..n} T(n,k)*x^(n-k) = A152163(n), A000007(n), A000045(n+1), A026597(n), A122994(n+1), A158608(n), A122995(n+1), A158797(n), A122996(n+1), A158798(n), A158609(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Dec 17 2011
G.f.: (1+(1-y)*x)/(1-y*x-x^2). - Philippe Deléham, Dec 17 2011
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = T(1,0) = T(2,0) = T(2,1) = 1, T(1,1) = T(2,2) = 0, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 09 2013
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