A036355 -id:A036355 - cp's OEIS frontend

A051159 Triangle read by rows: T(n, k) = binomial(n mod 2, k mod 2) * binomial(n div 2, k div 2), where 'div' denotes integer division.

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

Offset: 0

Michael Somos, Oct 14 1999

Previous name: Triangular array made of three copies of Pascal's triangle.
Computing each term modulo 2 also gives A047999, i.e., a(n) mod 2 = A007318(n) mod 2 for all n. (The triangle is paritywise isomorphic to Pascal's Triangle.) - Antti Karttunen
5th row/column gives entries of A000217 (triangular numbers C(n+1,2)) repeated twice and every other entry in 6th row/column form A000217. 7th row/column gives entries of A000292 (Tetrahedral (or pyramidal) nos: C(n+3,3)) repeated twice and every other entry in 8th row/column form A000292. 9th row/column gives entries of A000332 (binomial coefficients binomial(n,4)) repeated twice and every other entry in 10th row/column form A000332. 11th row/column gives entries of A000389 (binomial coefficients C(n,5)) repeated twice and every other entry in 12th row/column form A000389. - Gerald McGarvey, Aug 21 2004
If Sum_{k=0..n} A(k)*T(n,k) = B(n), the sequence B is the S-D transform of the sequence A. - Philippe Deléham, Aug 02 2006
Number of n-bead black-white reversible strings with k black beads; also binary grids; string is palindromic. - Yosu Yurramendi, Aug 07 2008
Row sums give A016116(n+1). - Yosu Yurramendi, Aug 07 2008 [corrected by Petros Hadjicostas, Nov 04 2017]
Coefficients in expansion of (x + y)^n where x and y anticommute (y x = -x y), that is, q-binomial coefficients when q = -1. - Michael Somos, Feb 16 2009
The sequence of coefficients of a general polynomial recursion that links at w=2 to the Pascal triangle is here w=0. Row sums are {1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, ...}. - Roger L. Bagula and Gary W. Adamson, Dec 04 2009
T(n,k) is the number of palindromic compositions of n+1 with exactly k+1 parts. T(6,4) = 3 because we have the following compositions of n+1=7 with length k+1=5: 1+1+3+1+1, 2+1+1+1+2, 1+2+1+2+1. - Geoffrey Critzer, Mar 15 2014 [corrected by Petros Hadjicostas, Nov 03 2017]
Let P(n,k) be the number of palindromic compositions of n with exactly k parts. MacMahon (1893) was the first to prove that P(n,k) = T(n-1,k-1), where T(n,k) are the numbers in this sequence (see the comment above by G. Critzer). He actually proved that, for 1 <= s <= m, we have P(2*m,2*s) = P(2*m,2*s-1) = P(2*m-1, 2*s-1) = bin(m-1, s-1), but  P(2*m-1, 2*s) = 0. For the current sequence, this can be translated into T(2*m-1, 2*s-1) = T(2*m-1,2*s-2) = T(2*m-2, 2*s-2) = bin(m-1,s-1), but T(2m-2, 2*s-1) = 0 (valid again for 1 <= s <= m). - Petros Hadjicostas, Nov 03 2017
T is the infinite lower triangular matrix for this sequence; define two others, U and V; let U(n,k)=e_k(-1,2,-3,...,(-1)^n n), where e_k is the k-th elementary symmetric polynomial, and let V be the diagonal matrix A057077 (periodic sequence 1,1,-1,-1). Clearly V^-1 = V. Conjecture: U = U^-1, T = U . V, T^-1 = V . U, and |T| = |U|. - George Beck, Dec 16 2017
Let T*(n,k)=T(n,k) except when n is odd and k=(n+1)/2, where T*(n,k) = T(n,k)+2^((n-1)/2). Thus, T*(n,k) is the number of non-isomorphic symmetric stairs with n cells and k steps, i.e., k-1 changes of direction. See A016116. - Christian Barrientos and Sarah Minion, Jul 29 2018

			Triangle starts:
{1},
{1,  1},
{1,  0,  1},
{1,  1,  1,  1},
{1,  0,  2,  0,  1},
{1,  1,  2,  2,  1,  1},
{1,  0,  3,  0,  3,  0,  1},
{1,  1,  3,  3,  3,  3,  1,  1},
{1,  0,  4,  0,  6,  0,  4,  0,  1},
{1,  1,  4,  4,  6,  6,  4,  4,  1,  1},
{1,  0,  5,  0, 10,  0, 10,  0,  5,  0,  1},
{1,  1,  5,  5, 10, 10, 10, 10,  5,  5,  1,  1}
... - _Roger L. Bagula_ and _Gary W. Adamson_, Dec 04 2009

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Nantel Bergeron, Kelvin Chan, Yohana Solomon, Farhad Soltani, and Mike Zabrocki, Quasisymmetric harmonics of the exterior algebra, arXiv:2206.02065 [math.CO], 2022.
E. Burlachenko, Fractal generalized Pascal matrices, arXiv:1612.00970 [math.NT], 2016. See p. 3.
S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757-774. See Table 1 on p. 763.
D. E. Davenport, L. W. Shapiro and L. C. Woodson, The Double Riordan Group, The Electronic Journal of Combinatorics, 18(2) (2012).
M. E. Horn, The Didactical Relevance of the Pauli Pascal Triangle, arXiv:physics/0611277 [physics.ed-ph], 2006. [_Michael Somos_]
F. Al-Kharousi, R. Kehinde, and A. Umar, Combinatorial results for certain semigroups of partial isometries of a finite chain, The Australasian Journal of Combinatorics, Volume 58 (3) (2014), 363-375.
P. A. MacMahon, Memoir on the Theory of the Compositions of Numbers, Phil. Trans. Royal Soc. London A, 184 (1893), 835-901.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A051160, A016116, A034851, A169623, A036355.

a051159 n k = a051159_tabl !! n !! k
a051159_row n = a051159_tabl !! n
a051159_tabl = [1] : f [1] [1,1] where
   f us vs = vs : f vs (zipWith (+) ([0,0] ++ us) (us ++ [0,0]))
-- Reinhard Zumkeller, Apr 25 2013

T:= proc(n, k) option remember; `if`(n=0 and k=0, 1,
      `if`(n<0 or k<0, 0, `if`(irem(n, 2)=1 or
       irem(k, 2)=0, T(n-1, k-1) + T(n-1, k), 0)))
    end:
seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, Jul 12 2014

T[ n_, k_] := QBinomial[n, k, -1]; (* Michael Somos, Jun 14 2011; since V7 *)
Clear[p, n, x, a]
w = 0;
p[x, 1] := 1;
p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + w*x + 1)^Floor[n/2]]
a = Table[CoefficientList[p[x, n], x], {n, 1, 12}]
Flatten[a] (* Roger L. Bagula and Gary W. Adamson, Dec 04 2009 *)

{T(n, k) = binomial(n%2, k%2) * binomial(n\2, k\2)};

from math import comb as binomial
def T(n, k): return binomial(n%2, k%2) * binomial(n//2, k//2)
print([T(n, k) for n in range(14) for k in range(n+1)])  # Peter Luschny, Oct 17 2024

@cached_function
def T(n, k):
    if k == 0 or k == n: return 1
    return T(n-1, k-1) + (-1)^k*T(n-1, k)
for n in (0..12): print([T(n, k) for k in (0..n)]) # Peter Luschny, Jul 06 2021

T(n, k) = T(n-1, k-1) + T(n-1, k) if n odd or k even, else 0. T(0, 0) = 1.
T(n, k) = T(n-2, k-2) + T(n-2, k). T(0, 0) = T(1, 0) = T(1, 1) = 1.
Square array made by setting first row/column to 1's (A(i, 0) = A(0, j) = 1); A(1, 1) = 0; A(1, j) = A(1, j-2); A(i, 1) = A(i-2, 1); other entries A(i, j) = A(i-2, j) + A(i, j-2). - Gerald McGarvey, Aug 21 2004
Sum_{k=0..n} k * T(n,k) = A093968(n); A093968 = S-D transform of A001477. - Philippe Deléham, Aug 02 2006
Equals 2*A034851 - A007318. - Gary W. Adamson, Dec 31 2007. [Corrected by Yosu Yurramendi, Aug 07 2008]
A051160(n, k) = (-1)^floor(k/2) * T(n, k).
Sum_{k = 0..n} T(n,k)*x^k = A000012(n), A016116(n+1), A056487(n), A136859(n+2) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Mar 11 2014
G.f.: (1+x+x*y)/(1-x^2-y^2*x^2). - Philippe Deléham, Mar 11 2014
For n,k >= 1, T(n, k) = 0 when n odd and k even; otherwise, T(n, k) = binomial(floor((n-1)/2), floor((k-1)/2)). - Christian Barrientos, Mar 14 2020
From Werner Schulte, Jun 25 2021: (Start)
T(n,k) = T(n-1,k-1) + (-1)^k * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0.
Matrix inverse is T^-1(n,k) = (-1)^((n-k)*(n+k+1)/2) * T(n,k) for 0 <= k <= n. (End)
From Peter Bala, Aug 08 2021: (Start)
Double Riordan array ( 1/(1 - x); x/(1 + x), x/(1 - x) ) in the notation of Davenport et al.
G.f. for column 2*n: (1 + x)*x^(2*n)/(1 - x^2)^(n+1); G.f. for column 2*n+1: x^(2*n+1)/(1 - x^2)^(n+1)
Row polynomials: R(2*n,x) = (1 + x^2)^n; R(2*n+1,x) = (1 + x)*(1 + x^2)^n.
The infinitesimal generator of this triangle has the sequence [1, 0, 1, 0, 1, 0, ...] on the main subdiagonal, the sequence [1, 1, 2, 2, 3, 3, 4, 4, ...] on the diagonal immediately below and zeros elsewhere.
Let T denote this lower triangular array. Then T^a, for a in C, is the double Riordan array ( (1 + a*x)/(1 - a*x^2); x/(1 + a*x), (1 + a*x)/(1 - a*x^2) ) with o.g.f. (1 + x*(a + y))/(1 - x^2*(a + y^2)) =  1 + (a + y)*x + (a + y^2)*x^2 + (a^2 + a*y + a*y^2 + y^3)*x^3 + (a^2 + 2*a*y^2 + y^4)*x^4 + ....
The (2*n)-th row polynomial of T^a is (a + y^2)^n; The (2*n+1)-th row polynomial of T^a is (a + y)*(a + y^2)^n. (End)

New name using a formula of the author by Peter Luschny, Oct 17 2024

A074829 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Fibonacci number.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 4, 4, 3, 5, 7, 8, 7, 5, 8, 12, 15, 15, 12, 8, 13, 20, 27, 30, 27, 20, 13, 21, 33, 47, 57, 57, 47, 33, 21, 34, 54, 80, 104, 114, 104, 80, 54, 34, 55, 88, 134, 184, 218, 218, 184, 134, 88, 55, 89, 143, 222, 318, 402, 436, 402, 318, 222, 143, 89

Offset: 1

Joseph L. Pe, Sep 30 2002

			The first and second Fibonacci numbers are 1, 1, so the first and second rows of the triangle are 1; 1 1; respectively. The third row of the triangle begins and ends with the third Fibonacci number, 2 and the middle term is the sum of the contiguous two terms in the second row, i.e., 1 + 1 = 2, so the third row is 2 2 2.
Triangle begins:
   1;
   1,  1;
   2,  2,  2;
   3,  4,  4,   3;
   5,  7,  8,   7,   5;
   8, 12, 15,  15,  12,   8;
  13, 20, 27,  30,  27,  20, 13;
  21, 33, 47,  57,  57,  47, 33, 21;
  34, 54, 80, 104, 114, 104, 80, 54, 34;
  ...
Formatted as a symmetric triangle:
                           1;
                        1,    1;
                     2,    2,    2;
                  3,    4,    4,    3;
               5,    7,    8,    7,    5;
            8,   12,   15,   15,   12,    8;
        13,   20,   27,   30,   27,   20,   13;
     21,   33,   47,   57,   57,   47,   33,   21;
  34,   54,   80,  104,  114,  104,   80,   54,   34;

Reinhard Zumkeller, Rows n = 1..120 of table, flattened
Hebert Pérez-Rosés, Asymptotic Analysis of Central Binomiacci Numbers, arXiv:2503.17462 [math.CO], 2025.
Index entries for triangles and arrays related to Pascal's triangle

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A105809, A108617, A109906, A111006, A114197, A162741, A228074.

Cf. A074878 (row sums).

T:= function(n,k)
    if k=1 then return Fibonacci(n);
    elif k=n then return Fibonacci(n);
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([1..15], n-> List([1..n], k-> T(n,k) ))); # G. C. Greubel, Jul 12 2019

a074829 n k = a074829_tabl !! (n-1) !! (k-1)
a074829_row n = a074829_tabl !! (n-1)
a074829_tabl = map fst $ iterate
   (\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [u]))) ([1], [1,1])
-- Reinhard Zumkeller, Aug 15 2013

A074829 := proc(n,k)
    option remember ;
    if k=1 or k=n then
        combinat[fibonacci](n) ;
    else
        procname(n-1,k-1)+procname(n-1,k) ;
    end if;
end proc:
seq(seq(A074829(n,k),k=1..n),n=1..12) ; # R. J. Mathar, Mar 31 2025

T[n_, 1]:= Fibonacci[n]; T[n_, n_]:= Fibonacci[n]; T[n_, k_]:= T[n-1, k-1] + T[n-1, k]; Table[T[n, k], {n, 1, 12}, {k, 1, n}]//Flatten (* G. C. Greubel, Jul 12 2019 *)

T(n,k) = if(k==1 || k==n, fibonacci(n), T(n-1,k-1) + T(n-1,k));
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 12 2019

def T(n, k):
    if (k==1 or k==n): return fibonacci(n)
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 12 2019

More terms from Philippe Deléham, Sep 20 2006

Data error in 7th row fixed by Reinhard Zumkeller, Aug 15 2013

A162741 Fibonacci-Pascal triangle; same as Pascal triangle, but beginning another Pascal triangle to the right of each row starting at row 2.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 4, 3, 2, 1, 1, 1, 4, 7, 7, 5, 3, 2, 1, 1, 1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1, 1, 7, 22, 41, 51, 46, 33, 21, 13, 8, 5, 3, 2, 1, 1, 1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1

Offset: 1

Mark Dols, Jul 12 2009, Jul 19 2009

Intertwined Pascal-triangles;

the first five rows seen as numbers in decimal representation: row(n) = 110*row(n-1) + 1. - corrected by Reinhard Zumkeller, Jul 16 2013

			.                                           1
.                                       1,  1, 1
.                                   1,  2,  2, 1, 1
.                               1,  3,  4,  3, 2, 1, 1
.                           1,  4,  7,  7,  5, 3, 2, 1, 1
.                       1,  5, 11, 14, 12,  8, 5, 3, 2, 1, 1
.                   1,  6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1,1
.               1,  7, 22, 41, 51, 46, 33, 21,13, 8, 5, 3, 2,1,1
.           1,  8, 29, 63, 92, 97, 79, 54, 34,21,13, 8, 5, 3,2,1,1
.       1,  9, 37, 92,155,189,176,133, 88, 55,34,21,13, 8, 5,3,2,1,1
.    1,10, 46,129,247,344,365,309,221,143, 89,55,34,21,13, 8,5,3,2,1,1
. 1,11,56,175,376,591,709,674,530,364,232,144,89,55,34,21,13,8,5,3,2,1,1 .

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. See Table 3.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A005408 (row length), A000225 (row sums), A000045 (central terms), A007318, A136431.
Cf. A021113. - Mark Dols, Jul 18 2009
Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A228074.

a162741 n k = a162741_tabf !! (n-1) !! (k-1)
a162741_row n = a162741_tabf !! (n-1)
a162741_tabf = iterate
   (\row -> zipWith (+) ([0] ++ row ++ [0]) (row ++ [0,1])) [1]
-- Reinhard Zumkeller, Jul 16 2013

T[, 1] = 1; T[n, k_] /; k == 2*n-2 || k == 2*n-1 = 1; T[n_, k_] := T[n, k] = T[n-1, k-1] + T[n-1, k]; Table[T[n, k], {n, 1, 9}, {k, 1, 2*n-1}] // Flatten (* Jean-François Alcover, Oct 30 2017, after Reinhard Zumkeller *)

T(n,k) = T(n-1,k-1) + T(n-1,k), T(n,1)=1 and for n>1: T(n,2*n-2) = T(n,2*n-1)=1. - Reinhard Zumkeller, Jul 16 2013

A105809 Riordan array (1/(1 - x - x^2), x/(1 - x)).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 5, 7, 7, 4, 1, 8, 12, 14, 11, 5, 1, 13, 20, 26, 25, 16, 6, 1, 21, 33, 46, 51, 41, 22, 7, 1, 34, 54, 79, 97, 92, 63, 29, 8, 1, 55, 88, 133, 176, 189, 155, 92, 37, 9, 1, 89, 143, 221, 309, 365, 344, 247, 129, 46, 10, 1, 144, 232, 364, 530, 674, 709, 591

Offset: 0

Paul Barry, May 04 2005

Previous name was: A Fibonacci-Pascal matrix.
From Wolfdieter Lang, Oct 04 2014: (Start)
In the column k of this triangle (without leading zeros) is the k-fold iterated partial sums of the Fibonacci numbers, starting with 1. A000045(n+1), A000071(n+3), A001924(n+1), A014162(n+1), A014166(n+1), ..., n >= 0. See the Riordan property.
For a combinatorial interpretation of these iterated partial sums see the H. Belbachir and A. Belkhir link. There table 1 shows in the rows these columns. In their notation (with r = k) f^(k)(n) = T(k, n+k).
The A-sequence of this Riordan triangle is [1, 1] (see the recurrence for T(n, k), k >= 1, given in the formula section). The Z-sequence is A165326 = [1, repeat(1, -1)]. See the W. Lang link under A006232 for Riordan A- and Z-sequences. (End)

			The triangle T(n,k) begins:
n\k   0   1   2    3    4    5    6    7    8   9  10 11 12 13 ...
0:    1
1:    1   1
2:    2   2   1
3:    3   4   3    1
4:    5   7   7    4    1
5:    8  12  14   11    5    1
6:   13  20  26   25   16    6    1
7:   21  33  46   51   41   22    7    1
8:   34  54  79   97   92   63   29    8    1
9:   55  88 133  176  189  155   92   37    9   1
10:  89 143 221  309  365  344  247  129   46  10   1
11: 144 232 364  530  674  709  591  376  175  56  11  1
12: 233 376 596  894 1204 1383 1300  967  551 231  67 12  1
13: 377 609 972 1490 2098 2587 2683 2267 1518 782 298 79 13  1
... reformatted and extended - _Wolfdieter Lang_, Oct 03 2014
------------------------------------------------------------------
Recurrence from Z-sequence (see a comment above): 8 = T(0,5) = (+1)*5 + (+1)*7 + (-1)*7 + (+1)*4 + (-1)*1 = 8. - _Wolfdieter Lang_, Oct 04 2014

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Anton Vladimirovich Shutov, On some analogue of the Gelfond problem for Zeckendorf representations, Chebyshevskii Sbornik (2024) Vol. 25, No. 5, 195-215. See p. 215. (In Russian)
Index entries for triangles and arrays related to Pascal's triangle

Cf. A165326 (Z-sequence), A027934 (row sums), A010049(n+1) (antidiagonal sums), A212804 (alternating row sums), inverse is A105810.

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A109906, A111006, A114197, A162741, A228074.

a105809 n k = a105809_tabl !! n !! k
a105809_row n = a105809_tabl !! n
a105809_tabl = map fst $ iterate
   (\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [0]))) ([1], [1,1])
-- Reinhard Zumkeller, Aug 15 2013

T := (n,k) -> `if`(n=0,1,binomial(n,k)*hypergeom([1,k/2-n/2,k/2-n/2+1/2], [k+1,-n], -4)); for n from 0 to 13 do seq(simplify(T(n,k)),k=0..n) od; # Peter Luschny, Oct 10 2014

T[n_, k_] := Sum[Binomial[n-j, k+j], {j, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

Riordan array (1/(1-x-x^2), x/(1-x)).
Triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0) = A000045(n+1);
T(n, m) = T(n-1, m-1) + T(n-1, m).
T(n, k) = Sum_{j=0..n} binomial(j, n+k-j). - Paul Barry, Oct 23 2006
G.f. of row polynomials Sum_{k=0..n} T(n, k)*x^k is (1 - z)/((1 - z - z^2)*(1 - (1 + x)*z)) (Riordan property). - Wolfdieter Lang, Oct 04 2014
T(n, k) = binomial(n, k)*hypergeom([1, k/2 - n/2, k/2 - n/2 + 1/2],[k + 1, -n], -4) for n > 0. - Peter Luschny, Oct 10 2014
From Wolfdieter Lang, Feb 13 2025: (Start)
Array A(k, n) = Sum_{j=0..n} F(j+1)*binomial(k-1+n-j, k-1), k >= 0, n >= 0, with F = A000045, (from Riordan triangle k-th convolution in columns without leading 0s).
A(k, n) = F(n+1+2*k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n+1+j, j), (from iteration of partial sums).
Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(n-j+1) * binomial(j-1, k-1), 0 <= k <= n.
T(n, k) = F(n+1+k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n - (k-1-j), j). (End)
T(n, k) =  A027926(n, n+k), for 0 <= k <= n. - Wolfdieter Lang, Mar 08 2025

Use first formula as a more descriptive name, Joerg Arndt, Jun 08 2021

A111006 Another version of Fibonacci-Pascal triangle A037027.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 0, 2, 3, 0, 0, 1, 5, 5, 0, 0, 0, 3, 10, 8, 0, 0, 0, 1, 9, 20, 13, 0, 0, 0, 0, 4, 22, 38, 21, 0, 0, 0, 0, 1, 14, 51, 71, 34, 0, 0, 0, 0, 0, 5, 40, 111, 130, 55, 0, 0, 0, 0, 0, 1, 20, 105, 233, 235, 89, 0, 0, 0, 0, 0, 0, 6, 65, 256, 474, 420, 144

Offset: 0

Philippe Deléham, Oct 02 2005

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Row sums are the Jacobsthal numbers A001045(n+1) and column sums form Pell numbers A000129.
Maximal column entries: A038149 = {1, 1, 2, 5, 10, 22, ...}.
T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, ...).
Triangle read by rows: T(n,n-k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)). - Philippe Deléham, Feb 17 2014
Diagonal sums are A013979(n). - Philippe Deléham, Feb 17 2014
T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and 1 X 2 tiles. - Emeric Deutsch, Aug 14 2014

			Triangle begins:
  1;
  0, 1;
  0, 1, 2;
  0, 0, 2, 3;
  0, 0, 1, 5,  5;
  0, 0, 0, 3, 10,  8;
  0, 0, 0, 1,  9, 20, 13;
  0, 0, 0, 0,  4, 22, 38,  21;
  0, 0, 0, 0,  1, 14, 51,  71,  34;
  0, 0, 0, 0,  0,  5, 40, 111, 130,  55;
  0, 0, 0, 0,  0,  1, 20, 105, 233, 235,  89;
  0, 0, 0, 0,  0,  0,  6,  65, 256, 474, 420, 144;

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000045, A000129, A001045, A037027, A038112, A038149, A084938, A128100 (reversed version).

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A114197, A162741, A228074.

a111006 n k = a111006_tabl !! n !! k
a111006_row n = a111006_tabl !! n
a111006_tabl =  map fst $ iterate (\(us, vs) ->
   (vs, zipWith (+) (zipWith (+) ([0] ++ us ++ [0]) ([0,0] ++ us))
                    ([0] ++ vs))) ([1], [0,1])
-- Reinhard Zumkeller, Aug 15 2013

T(0, 0) = 1, T(n, k) = 0 for k < 0 or for n < k, T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-2, k-2).
T(n, k) = A037027(k, n-k). T(n, n) = A000045(n+1). T(3n, 2n) = (n+1)*A001002(n+1) = A038112(n).
G.f.: 1/(1-yx(1-x)-x^2*y^2). - Paul Barry, Oct 04 2005
Sum_{k=0..n} x^k*T(n,k) = (-1)^n*A053524(n+1), (-1)^n*A083858(n+1), (-1)^n*A002605(n), A033999(n), A000007(n), A001045(n+1), A083099(n) for x = -4, -3, -2, -1, 0, 1, 2 respectively. - Philippe Deléham, Dec 02 2006
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 respectively. - Philippe Deléham, Feb 17 2014

A114197 A Pascal-Fibonacci triangle.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 31, 21, 6, 1, 1, 7, 31, 61, 61, 31, 7, 1, 1, 8, 43, 106, 142, 106, 43, 8, 1, 1, 9, 57, 169, 286, 286, 169, 57, 9, 1, 1, 10, 73, 253, 520, 659, 520, 253, 73, 10, 1

Offset: 0

Paul Barry, Nov 16 2005

T(2n,n) is A114198. Row sums are A114199. Row sums of inverse are 0^n.

			Triangle begins
  1;
  1,   1;
  1,   2,   1;
  1,   3,   3,   1;
  1,   4,   7,   4,   1;
  1,   5,  13,  13,   5,   1;
  1,   6,  21,  31,  21,   6,   1;
  1,   7,  31,  61,  61,  31,   7,   1;
  1,   8,  43, 106, 142, 106,  43,   8,   1;

Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A114197, A162741, A228074.

As a number triangle, T(n,k) = Sum_{j=0..n-k} C(n-k, j)C(k, j)F(j);
As a number triangle, T(n,k) = Sum_{j=0..n} C(n-k, n-j)C(k, j-k)F(j-k);
As a number triangle, T(n,k) = Sum_{j=0..n} C(k, j)C(n-k, n-j)F(k-j) if k <= n, 0 otherwise.
As a square array, T(n,k) = Sum_{j=0..n} C(n, j)C(k, j)F(j);
As a square array, T(n,k) = Sum_{j=0..n+k} C(n, n+k-j)C(k, j-k)F(j-k);
Column k has g.f.: (Sum_{j=0..k} C(k, j)F(j+1)(x/(1-x))^j)*x^k/(1-x);
G.f.: -((x^2-x)*y-x+1)/((x^4+x^3-x^2)*y^2+(x^3-3*x^2+2*x)*y-x^2+2*x-1). - Vladimir Kruchinin, Jan 15 2018

A109906 A triangle based on A000045 and Pascal's triangle: T(n,m) = Fibonacci(n-m+1) * Fibonacci(m+1) * binomial(n,m).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 6, 6, 3, 5, 12, 24, 12, 5, 8, 25, 60, 60, 25, 8, 13, 48, 150, 180, 150, 48, 13, 21, 91, 336, 525, 525, 336, 91, 21, 34, 168, 728, 1344, 1750, 1344, 728, 168, 34, 55, 306, 1512, 3276, 5040, 5040, 3276, 1512, 306, 55, 89, 550, 3060, 7560, 13650, 16128, 13650, 7560, 3060, 550, 89

Offset: 0

Roger L. Bagula and Gary W. Adamson, Aug 24 2008

Row sums give A081057.

			Triangle T(n,k) begins:
   1;
   1,   1;
   2,   2,    2;
   3,   6,    6,    3;
   5,  12,   24,   12,     5;
   8,  25,   60,   60,    25,     8;
  13,  48,  150,  180,   150,    48,    13;
  21,  91,  336,  525,   525,   336,    91,   21;
  34, 168,  728, 1344,  1750,  1344,   728,  168,   34;
  55, 306, 1512, 3276,  5040,  5040,  3276, 1512,  306,  55;
  89, 550, 3060, 7560, 13650, 16128, 13650, 7560, 3060, 550, 89;
  ...

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
Peter McCalla, Asamoah Nkwanta, Catalan and Motzkin Integral Representations, arXiv:1901.07092 [math.NT], 2019.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A141611, A141617, A000045, A081057.

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A111006, A114197, A162741, A228074.

a109906 n k = a109906_tabl !! n !! k
a109906_row n = a109906_tabl !! n
a109906_tabl = zipWith (zipWith (*)) a058071_tabl a007318_tabl
-- Reinhard Zumkeller, Aug 15 2013

f:= n-> combinat[fibonacci](n+1):
T:= (n, k)-> binomial(n, k)*f(k)*f(n-k):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 26 2023

Clear[t, n, m] t[n_, m_] := Fibonacci[(n - m + 1)]*Fibonacci[(m + 1)]*Binomial[n, m]; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%]

T(n,m) = Fibonacci(n-m+1)*Fibonacci(m+1)*binomial(n,m).

T(n,k) = A058071(n,k) * A007318(n,k). - Reinhard Zumkeller, Aug 15 2013

Offset changed by Reinhard Zumkeller, Aug 15 2013

A091533 Triangle read by rows, related to Pascal's triangle, starting with rows 1; 1,1.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 3, 7, 7, 3, 5, 15, 21, 15, 5, 8, 30, 53, 53, 30, 8, 13, 58, 124, 157, 124, 58, 13, 21, 109, 273, 417, 417, 273, 109, 21, 34, 201, 577, 1029, 1239, 1029, 577, 201, 34, 55, 365, 1181, 2405, 3375, 3375, 2405, 1181, 365, 55, 89, 655, 2358, 5393, 8625, 10047, 8625, 5393, 2358, 655, 89

Offset: 0

Christian G. Bower, Jan 19 2004

T(n,k) is the number of lattice paths from (0,0) to (k,n-k) using steps (1,0),(2,0),(0,1),(0,2),(1,1). - Seiichi Manyama, Apr 26 2025.

			This triangle begins:
   1;
   1,   1;
   2,   3,    2;
   3,   7,    7,    3;
   5,  15,   21,   15,    5;
   8,  30,   53,   53,   30,     8;
  13,  58,  124,  157,  124,    58,   13;
  21, 109,  273,  417,  417,   273,  109,   21;
  34, 201,  577, 1029, 1239,  1029,  577,  201,   34;
  55, 365, 1181, 2405, 3375,  3375, 2405, 1181,  365,  55;
  89, 655, 2358, 5393, 8625, 10047, 8625, 5393, 2358, 655, 89;
  ...

Seiichi Manyama, Rows n = 0..139, flattened

Row sums: A015518(n+1). Columns 0-1: A000045(n+1), A023610(n-1).
Cf. A090174, A212338 (column 2), A192364 (central terms).
Cf. A036355.

T:= proc(n, k) option remember; `if`(k<0 or k>n, 0,
     `if`(n<1, 1, add(add(T(n-i, k-j), j=0..i), i=1..2)))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jan 14 2022

A091533[-2, n2_] = 0; A091533[n1_, -2] = 0; A091533[-1, n2_] = 0; A091533[n1_, -1] = 0; A091533[0, 0] = 1; A091533[n1_, n2_] := A091533[n1, n2] = A091533[n1 - 1, n2] + A091533[n1, n2 - 1] + A091533[n1 - 1, n2 - 1] + A091533[n1 - 2, n2] + A091533[n1, n2 - 2]; Table[A091533[x - y, y], {x, 0, 9}, {y, 0, x}] // Flatten (* Robert P. P. McKone, Jan 14 2022 *)

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) + T(n-2, k-2) for n >= 2, k >= 0, with initial conditions specified by first two rows.
G.f.: A(x, y) = 1/(1-x-x*y-x^2-x^2*y-x^2*y^2).
Sum_{k = 0..n} T(n,k)*x^k = A000045(n+1), A015518(n+1), A015524(n+1), A200069(n+1) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 30 2013
Sum_{k = 0..floor(n/2)} T(n-k,k) = (-1)^n*A079926(n). - Philippe Deléham, Oct 30 2013

A191354 Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2), and (2,1).

Original entry on oeis.org

1, 1, 3, 9, 25, 75, 227, 693, 2139, 6645, 20757, 65139, 205189, 648427, 2054775, 6526841, 20775357, 66251247, 211617131, 676930325, 2168252571, 6953348149, 22322825865, 71735559255, 230735316795, 742773456825, 2392949225565, 7714727440755, 24888317247705, 80341227688095

Offset: 0

Joerg Arndt, Jun 30 2011

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.

Cf. A001850, A026641, A036355, A137644, A192364, A192365, A192369.

m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( 1/Sqrt(1-2*x-3*x^2-4*x^3) )); // G. C. Greubel, Feb 18 2019

a[n_]:= Sum[Binomial[2k, k]*Sum[Binomial[j, n-k-j]*Binomial[k, j]*2^(j-k) *3^(-n+k+2j)*4^(n-k-2j), {j, 0, k}], {k, 0, n}];
Array[a, 30, 0] (* Jean-François Alcover, Jul 21 2018, after Vladimir Kruchinin *)
CoefficientList[Series[1/Sqrt[1-2*x-3*x^2-4*x^3], {x, 0, 30}], x] (* G. C. Greubel, Feb 18 2019 *)

a(n):=sum(binomial(2*k,k) * sum(binomial(j,n-k-j) * 2^(j-k) * binomial(k,j) * 3^(-n+k+2*j) * 4^(n-k-2*j),j,0,k),k,0,n); /* Vladimir Kruchinin, Feb 27 2016 */

/* same as in A092566 but use */
steps=[[1,0], [1,1], [1,2], [2,1]];
/* Joerg Arndt, Jun 30 2011 */

my(x='x+O('x^30)); Vec(1/sqrt(1-2*x-3*x^2-4*x^3)) \\ G. C. Greubel, Feb 18 2019

(1/sqrt(1-2*x-3*x^2-4*x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019

G.f.: 1/sqrt(1-2*x-3*x^2-4*x^3). - Mark van Hoeij, Apr 16 2013
G.f.: Q(0), where Q(k) = 1 + x*(2+3*x+4*x^2)*(4*k+1)/( 4*k+2 - x*(2+3*x+4*x^2)*(4*k+2)*(4*k+3)/(x*(2+3*x+4*x^2)*(4*k+3) + 4*(k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 14 2013
a(n) = Sum_{k=0..n} (binomial(2*k,k) * Sum_{j=0..k} (binomial(j,n-k-j) *binomial(k,j)*2^(j-k)*3^(-n+k+2*j)*4^(n-k-2*j))). - Vladimir Kruchinin, Feb 27 2016
D-finite with recurrence: +(n)*a(n) +(-2*n+1)*a(n-1) +3*(-n+1)*a(n-2) +2*(-2*n+3)*a(n-3)=0. - R. J. Mathar, Jan 14 2020

cp's OEIS Frontend

A051159 Triangle read by rows: T(n, k) = binomial(n mod 2, k mod 2) * binomial(n div 2, k div 2), where 'div' denotes integer division.

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A074829 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Fibonacci number.

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A162741 Fibonacci-Pascal triangle; same as Pascal triangle, but beginning another Pascal triangle to the right of each row starting at row 2.

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A105809 Riordan array (1/(1 - x - x^2), x/(1 - x)).

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A111006 Another version of Fibonacci-Pascal triangle A037027.

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A114197 A Pascal-Fibonacci triangle.

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A109906 A triangle based on A000045 and Pascal's triangle: T(n,m) = Fibonacci(n-m+1) * Fibonacci(m+1) * binomial(n,m).