A228074 -id:A228074 - cp's OEIS frontend

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

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

A014166 Apply partial sum operator 4 times to Fibonacci numbers.

Original entry on oeis.org

0, 1, 5, 16, 41, 92, 189, 365, 674, 1204, 2098, 3588, 6050, 10093, 16703, 27476, 44995, 73440, 119575, 194345, 315460, 511576, 829060, 1342936, 2174596, 3520457, 5698329, 9222440, 14924829, 24151764, 39081553

Offset: 0

N. J. A. Sloane

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Ligia Loretta Cristea, Ivica Martinjak, and Igor Urbiha, Hyperfibonacci Sequences and Polytopic Numbers, arXiv:1606.06228 [math.CO], 2016.
Index entries for linear recurrences with constant coefficients, signature (5,-9,6,1,-3,1).

Cf. A000045, A228074, A136431.

Right-hand column 8 of triangle A011794.

List([0..30], n-> Fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6); # G. C. Greubel, Sep 06 2019

[Fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6: n in [0..30]]; // G. C. Greubel, Sep 06 2019

with(combinat); seq(fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6, n = 0..30); # G. C. Greubel, Sep 06 2019

Nest[Accumulate, Fibonacci[Range[0, 30]], 4] (* Jean-François Alcover, Jan 08 2019 *)

a(n)=fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6 \\ Charles R Greathouse IV, Jun 11 2015

[fibonacci(n+8)-(n^3+12*n^2+59*n+126)/6 for n in (0..30)] # G. C. Greubel, Sep 06 2019

a(n) = Fibonacci(n+8) - (n^3 +12*n^2 +59*n +126)/6.

G.f.: x/((1-x)^4*(1-x-x^2)).

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

A014162 Apply partial sum operator thrice to Fibonacci numbers.

Original entry on oeis.org

0, 1, 4, 11, 25, 51, 97, 176, 309, 530, 894, 1490, 2462, 4043, 6610, 10773, 17519, 28445, 46135, 74770, 121115, 196116, 317484, 513876, 831660, 1345861, 2177872, 3524111, 5702389, 9226935, 14929789

Offset: 0

N. J. A. Sloane

With offset 4, number of 132-avoiding two-stack sortable permutations which contain exactly one subsequence of type 51234.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Ligia Loretta Cristea, Ivica Martinjak, and Igor Urbiha, Hyperfibonacci Sequences and Polytopic Numbers, arXiv:1606.06228 [math.CO], 2016.
E. S. Egge and T. Mansour, 132-avoiding Two-stack Sortable Permutations, Fibonacci Numbers, and Pell Numbers, arXiv:math/0205206 [math.CO], 2002.
T. Langley, J. Liese, and J. Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011) # 11.4.2.
Index entries for linear recurrences with constant coefficients, signature (4,-5,1,2,-1).

Cf. A000045, A001924, A228074.

Right-hand column 6 of triangle A011794.

List([0..40], n-> Fibonacci(n+6) - (n^2 + 7*n + 16)/2); # G. C. Greubel, Sep 05 2019

[Fibonacci(n+6) - (n^2 + 7*n + 16)/2: n in [0..40]]; // G. C. Greubel, Sep 05 2019

with(combinat); seq(fibonacci(n+6)-(n^2+7*n+16)*(1/2), n = 0..40); # G. C. Greubel, Sep 05 2019

Nest[Accumulate,Fibonacci[Range[0,30]],3] (* or *) LinearRecurrence[{4,-5,1,2,-1},{0,1,4,11,25},40] (* Harvey P. Dale, Aug 19 2017 *)

a(n)=fibonacci(n+6)-n*(n+7)/2-8 \\ Charles R Greathouse IV, Jun 11 2015

[fibonacci(n+6) - (n^2 + 7*n + 16)/2 for n in (0..40)] # G. C. Greubel, Sep 05 2019

a(n) = Sum_{k=0..n} A000045(n-k)*k*(k+1)/2. - Benoit Cloitre, Jan 06 2003
G.f.: x/((1-x)^3*(1-x-x^2)).
From Paul Barry, Oct 07 2004: (Start)
a(n-2) = Sum_{k=0..floor(n/2)} binomial(n-k, k+3).
a(n-2) = Sum_{k=0..n} binomial(k, n-k+3). (End)
Convolution of A000045 and A000217 (Fibonacci and triangular numbers). - Ross La Haye, Nov 08 2004
a(n) = Fibonacci(n+6) - (n^2 + 7*n + 16)/2.

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

A053295 Partial sums of A053739.

Original entry on oeis.org

1, 7, 29, 92, 247, 591, 1300, 2683, 5270, 9955, 18228, 32551, 56967, 98086, 166681, 280271, 467301, 773906, 1274856, 2091266, 3419252, 5576298, 9076280, 14750858, 23945893, 38839257, 62955061, 101995694

Offset: 0

Barry E. Williams, Mar 04 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 189, 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-20,29,-20,1,8,-5,1).

Cf. A053739, A014166 and A000045.
Right-hand column 12 of triangle A011794.
Cf. A228074.

[(&+[Binomial(n+6-j, n-2*j): j in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, May 24 2018

Table[Sum[Binomial[n+6-j, n-2*j], {j, 0, Floor[n/2]}], {n, 0, 50}] (* G. C. Greubel, May 24 2018 *)

for(n=0, 30, print1(sum(j=0, floor(n/2), binomial(n+6-j, n-2*j)), ", ")) \\ G. C. Greubel, May 24 2018

a(n) = Sum_{i=0..floor(n/2)} C(n+6-i, n-2i), n >= 0.
a(n) = a(n-1) + a(n-2) + C(n+5,5); n >= 0; a(-1)=0.
G.f.: -1 / ( (x^2 + x - 1)*(x-1)^6 ). - R. J. Mathar, Oct 10 2014

A053739 Partial sums of A014166.

Original entry on oeis.org

1, 6, 22, 63, 155, 344, 709, 1383, 2587, 4685, 8273, 14323, 24416, 41119, 68595, 113590, 187030, 306605, 500950, 816410, 1327986, 2157046, 3499982, 5674578, 9195035, 14893364, 24115804, 39040633, 63192397, 102273950, 165512723, 267839033, 433410661, 701315739, 1134800215

Offset: 0

Barry E. Williams, Feb 13 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.
Index entries for linear recurrences with constant coefficients, signature (6,-14,15,-5,-4,4,-1).

Cf. A014166 and A000045.
Right-hand column 10 of triangle A011794.
Cf. A228074.

List([0..35], n-> Fibonacci(n+11)-(n^4+22*n^3+203*n^2+974*n + 2112)/24); # G. C. Greubel, Sep 06 2019

[Fibonacci(n+11) - (n^4+22*n^3+203*n^2+974*n+2112)/24: n in [0..35]]; // G. C. Greubel, Sep 06 2019

with(combinat); seq(fibonacci(n+11)-(n^4 + 22*n^3 + 203*n^2 + 974*n + 2112)/4!, n = 0..35); # G. C. Greubel, Sep 06 2019

Table[Fibonacci[n+11] -(n^4+22*n^3+203*n^2+974*n+2112)/4!, {n,0,35}] (* G. C. Greubel, Sep 06 2019 *)

vector(35, n, m=n-1; fibonacci(n+10) - (m^4+22*m^3+203*m^2+974*m +2112)/4!) \\ G. C. Greubel, Sep 06 2019

[fibonacci(n+11) - (n^4+22*n^3+203*n^2+974*n+2112)/24 for n in (0..35)] # G. C. Greubel, Sep 06 2019

a(n) = Sum_{i=0..floor(n/2)} binomial(n+5-i, n-2*i) for n >= 0.
a(n) = a(n-1) + a(n-2) + C(n+4,4); n >= 0; a(-1)=0.
G.f.: 1/((1-x-x^2)*(1-x)^5). - R. J. Mathar, May 22 2013
a(n) = Fibonacci(n+11) - (n^4 + 22*n^3 + 203*n^2 + 974*n + 2112)/4!. - G. C. Greubel, Sep 06 2019

Terms a(28) onward added by G. C. Greubel, Sep 06 2019

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Haskell

Maple

Mathematica

PARI

Sage

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

Extensions

A014166 Apply partial sum operator 4 times to Fibonacci numbers.

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

A111006 Another version of Fibonacci-Pascal triangle A037027.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Formula

A114197 A Pascal-Fibonacci triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A014162 Apply partial sum operator thrice to Fibonacci numbers.

Views

Author

Keywords