A027926 -id:A027926 - cp's OEIS frontend

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

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

A026268 Triangle, T(n, k): T(n,k) = 1 for n < 3, T(3,1) = T(3,2) = T(3,3) = 2, T(n,0) = 1, T(n,1) = n-1, T(n,n) = T(n-1,n-2) + T(n-1,n-1), otherwise T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 3, 5, 6, 4, 1, 4, 9, 14, 15, 10, 1, 5, 14, 27, 38, 39, 25, 1, 6, 20, 46, 79, 104, 102, 64, 1, 7, 27, 72, 145, 229, 285, 270, 166, 1, 8, 35, 106, 244, 446, 659, 784, 721, 436, 1, 9, 44, 149, 385, 796, 1349, 1889, 2164, 1941, 1157, 1, 10, 54, 202, 578, 1330, 2530, 4034, 5402, 5994, 5262, 3098

Offset: 0

Clark Kimberling

a(n) = number of strings s(0)..s(n) such that s(n) = n-k, where s(0) = 0, s(1) = 1, |s(i)-s(i-1)| <= 1 for i >= 2; |s(2)-s(1)| = 1, and |s(3)-s(2)| = 1 if s(2) = 1.

			Triangle begins as:
  1;
  1, 1;
  1, 1,  1;
  1, 2,  2,   2;
  1, 3,  5,   6,   4;
  1, 4,  9,  14,  15,  10;
  1, 5, 14,  27,  38,  39,   25;
  1, 6, 20,  46,  79, 104,  102,   64;
  1, 7, 27,  72, 145, 229,  285,  270,  166;
  1, 8, 35, 106, 244, 446,  659,  784,  721,  436;
  1, 9, 44, 149, 385, 796, 1349, 1889, 2164, 1941, 1157;

Clark Kimberling, Rows 0..100, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A026269, A026270, A026288, A026289, A026290, A026291, A026292, A026293.
Cf. A026294, A026295, A026296, A026297, A026299, A026519, A026536, A026552.
Cf. A026584, A027926, A212342.

f:= func< n | n eq 2 select 1 else (n^2 -n -2)/2 >;
function T(n,k) // T = A026268
  if k eq 0 or n lt 3 then return 1;
  elif k eq 1 then return n-1;
  elif k eq 2 then return f(n);
  elif k eq n then return T(n-1, n-2) + T(n-1, n-1);
  else return T(n-1, k-2) + T(n-1, k-1) + T(n-1, k);
  end if; return T;
end function;
[T(n,k): k in [0..n], n in [0..14]]; // G. C. Greubel, Sep 24 2022

T[n_, k_]:= T[n, k]= If[n<3 || k==0, 1, If[k==1, n-1, If[k==2, (n^2-n-2)/2 + Boole[n==2], If[k==n, T[n-1, n-2] +T[n-1, n-1], T[n-1, k-2] + T[n-1, k-1] + T[n -1, k] ]]]];
Table[T[n, k], {n,0,14}, {k,0,n}]//Flatten (* corrected by G. C. Greubel, Sep 24 2022 *)

def T(n,k): # T = A026268
    if n<3 or k==0: return 1
    elif k==1: return n-1
    elif k==2: return (n^2 -n -2)//2 + int(n==2)
    elif k==n: return T(n-1, n-2) + T(n-1, n-1)
    else: return T(n-1, k-2) + T(n-1, k-1) + T(n-1, k)
flatten([[T(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Sep 24 2022

From G. C. Greubel, Sep 24 2022: (Start)
T(n, 1) = A000027(n-1), n >= 1.
T(n, 2) = A212342(n-1), n >= 2.
T(n, n-1) = A026270(n), n >= 2.
T(n, n-2) = A026288(n), n >= 2.
T(n, n-3) = A026289(n), n >= 3.
T(n, n-4) = A026290(n), n >= 4.
T(n, n) = A026269(n), n >= 2.
T(n, floor(n/2)) = A026297(n), n >= 0.
T(2*n, n) = A026292(n).
T(2*n, n-1) = A026295(n), n >= 1.
T(2*n, n+1) = A026296(n), n >= 1.
T(2*n-1, n-1) = A026291(n), n >= 2.
T(3*n, n) = A026293(n), n >= 0.
T(4*n, n) = A026294(n), n >= 0.
Sum_{k=0..n} T(n, k) = A026299(n-1), n >= 3.(End)

Updated by Clark Kimberling, Aug 29 2014

Indices of b-file corrected by Sidney Cadot, Jan 06 2023.

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

A027927 Number of plane regions after drawing (in general position) a convex n-gon and all its diagonals.

Original entry on oeis.org

1, 2, 5, 12, 26, 51, 92, 155, 247, 376, 551, 782, 1080, 1457, 1926, 2501, 3197, 4030, 5017, 6176, 7526, 9087, 10880, 12927, 15251, 17876, 20827, 24130, 27812, 31901, 36426, 41417, 46905, 52922, 59501, 66676, 74482, 82955, 92132, 102051, 112751, 124272, 136655, 149942, 164176, 179401

Offset: 2

Clark Kimberling

For n>=1, a(n+1) is the number of Grassmannian permutations that avoid a pattern, sigma, where sigma is a pattern of size 5 with exactly one descent. - Jessica A. Tomasko, Nov 15 2022

			a(2)=1 (segment traced twice has only exterior).

Vincenzo Librandi, Table of n, a(n) for n = 2..10000
Lapo Cioni and Luca Ferrari, Enumerative Results on the Schröder Pattern Poset, In: Dennunzio A., Formenti E., Manzoni L., Porreca A. (eds) Cellular Automata and Discrete Complex Systems, AUTOMATA 2017, Lecture Notes in Computer Science, vol 10248.
Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees, Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227. - From _N. J. A. Sloane_, Feb 01 2013
J. B. Gil and J. Tomasko, Restricted Grassmannian permutations, ECA 2:4 (2022) Article S4PP6.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A006522 (does not count exterior of n-gon).

Cf. A000045, A000217, A004006, A027926, A228074.

List([2..50], n-> (n^4 -6*n^3 +23*n^2 -42*n +48)/24); # G. C. Greubel, Sep 06 2019

[(n^4 -6*n^3 +23*n^2 -42*n +48)/24: n in [2..50]]; // G. C. Greubel, Sep 06 2019

seq((n^4 -6*n^3 +23*n^2 -42*n +48)/24, n=2..50); # G. C. Greubel, Sep 06 2019

LinearRecurrence[{5,-10,10,-5,1 }, {1,2,5,12,26}, 50] (* Vincenzo Librandi, Feb 01 2012 *)
S[n_] :=n*(n+1)/2; Table[S[S[n]+2]/3, {n, 0, 50}] (* Waldemar Puszkarz, Jan 22 2016 *)

a(n)=n*(n^3-6*n^2+23*n-42)/24+2 \\ Charles R Greathouse IV, Jan 31 2012

[(n^4 -6*n^3 +23*n^2 -42*n +48)/24 for n in (2..50)] # G. C. Greubel, Sep 06 2019

a(n) = T(n, 2*n-4), T given by A027926.
a(n) = 1 + binomial(n, 4) + binomial(n-1, 2) = (n^4 - 6*n^3 + 23*n^2 - 42*n + 48)/24.
G.f.: x^2*(1 -3*x +5*x^2 -3*x^3 +x^4)/(1-x)^5. - Colin Barker, Jan 31 2012
a(n) = (1/6)*A152950(n-1)*A152948(n). - Bruno Berselli, Jan 31 2012
a(n) = A000217(A000217(n-2)+2)/3, a(n+1) - a(n) = A004006(n-1) for n > 2. - Waldemar Puszkarz, Jan 22 2016 [Adjusted for offset by Peter Munn, Jan 10 2023]
a(n) = 1 + Sum {i=3..5} binomial(n-1, i-1). - Jessica A. Tomasko, Nov 15 2022

New name from Len Smiley, Oct 19 2001

A027994 a(n) = (F(2n+3) - F(n))/2 where F() = Fibonacci numbers A000045.

Original entry on oeis.org

1, 2, 6, 16, 43, 114, 301, 792, 2080, 5456, 14301, 37468, 98137, 256998, 672946, 1761984, 4613239, 12078110, 31621701, 82787980, 216743836, 567446112, 1485598681, 3889356696, 10182482353, 26658108074, 69791870526, 182717549872, 478360854115, 1252365133866, 3278734743901, 8583839415648, 22472784017272

Offset: 0

Clark Kimberling

Substituting x*(1-x)/(1-2x) into x^2/(1-x^2) yields x^2*(g.f. of sequence).
The number of (s(0), s(1), ..., s(n+1)) such that 0 < s(i) < 5 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n+1, s(0) = 2, s(n+1) = 3. - Herbert Kociemba, Jun 02 2004
Diagonal sums of triangle in A125171. - Philippe Deléham, Jan 14 2014

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Cf. A000667, A059216, A059219, A059502, A027926.

[(Fibonacci(2*n+3)-Fibonacci(n))/2 : n in [0..40]]; // Vincenzo Librandi, Jan 01 2025

Table[(Fibonacci[2n+3]-Fibonacci[n])/2,{n,0,30}] (* or *) LinearRecurrence[{4,-3,-2,1},{1,2,6,16},30] (* Harvey P. Dale, Apr 28 2022 *)

PARI
```
a(n)=(fibonacci(2*n+3)-fibonacci(n))/2
```

G.f.: (1-x)^2/((1-x-x^2)*(1-3*x+x^2)). - Floor van Lamoen and N. J. A. Sloane, Jan 21 2001
a(n) = Sum_{k=0..n} T(n, k)*T(n, n+k), T given by A027926.
a(n) = 2*a(n-1) + Sum_{m < n-1} a(m) + F(n-1) = A059512(n+2) - F(n) where F(n) is the n-th Fibonacci number (A000045). - Floor van Lamoen, Jan 21 2001
a(n) = (2/5)*Sum_{k=1..4} sin(2*Pi*k/5)*sin(3*Pi*k/5)*(1+2*cos(Pi*k/5))^(n+1). - Herbert Kociemba, Jun 02 2004
a(-1-2n) = A056014(2n), a(-2n) = A005207(2n-1).
E.g.f.: exp(3*x/2)*cosh(sqrt(5)*x/2) + exp(x/2)*(2*exp(x) - 1)*sinh(sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Jan 01 2025

A123736 Triangle T(n,k) = Sum_{j=0..k/2} binomial(n-j-1,k-2*j), read by rows.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 3, 4, 3, 2, 1, 1, 0, 1, 4, 7, 7, 5, 3, 2, 1, 1, 0, 1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 0, 1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1, 0, 1, 7, 22, 41, 51, 46, 33, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1

Offset: 1

Roger L. Bagula, Nov 14 2006

Row sums give: A000225

			The triangle starts in row n=1 with columns 0 <= k < 2*n:
  1, 0;
  1, 1,  1,  0;
  1, 2,  2,  1,  1,  0;
  1, 3,  4,  3,  2,  1,  1,  0;
  1, 4,  7,  7,  5,  3,  2,  1,  1,  0;
  1, 5, 11, 14, 12,  8,  5,  3,  2,  1,  1, 0;
  1, 6, 16, 25, 26, 20, 13,  8,  5,  3,  2, 1, 1, 0;
  1, 7, 22, 41, 51, 46, 33, 21, 13,  8,  5, 3, 2, 1, 1, 0;
  1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0;

G. C. Greubel, Rows n = 1..100 of triangle, flattened
Eric Weisstein's World of Mathematics, Steenrod Algebra

Cf. A136431 (antidiagonals), A027926 (row-reversed), A004006 (column m=3)

Flat(List([1..10], n-> List([0..2*n-1], k-> Sum([0..Int(k/2)], j-> Binomial(n-j-1, k-2*j) )))); # G. C. Greubel, Sep 05 2019

[&+[Binomial(n-j-1, k-2*j): j in [0..Floor(k/2)]]: k in [0..2*n-1], n in [1..10]]; // G. C. Greubel, Sep 05 2019

seq(seq(sum(binomial(n-j-1, k-2*j), j=0..floor(k/2)), k=0..2*n-1), n=1..10); # G. C. Greubel, Sep 05 2019

Table[Sum[Binomial[n-j-1, k-2*j], {j,0,Floor[k/2]}], {n, 10}, {k, 0, 2*n-1}]//Flatten (* modified by G. C. Greubel, Sep 05 2019 *)

T(n,k) = sum(j=0, k\2, binomial(n-j-1, k-2*j));
for(n=1,10, for(k=0,2*n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 05 2019

[[sum(binomial(n-j-1, k-2*j) for j in (0..floor(k/2))) for k in (0..2*n-1)] for n in (1..10)] # G. C. Greubel, Sep 05 2019

A246648 Numbers k such that 2*k + 1 divides 2^(k+1) - 1.

Original entry on oeis.org

0, 1, 7, 127, 227, 647, 1351, 1907, 3239, 4607, 5219, 5975, 11447, 13159, 13919, 21527, 22049, 23759, 23939, 24839, 30959, 31283, 31583, 31967, 32767, 37223, 46091, 46511, 47267, 60479, 65663, 66527, 78539, 78599, 81727, 82799, 84311, 98405, 102671, 103967

Offset: 1

Clark Kimberling, Sep 01 2014

These are the numbers k such that mean of the k-th row of the triangle at A027926 is an integer.

Numbers k such that 2*k + 1 divides 2^k + k. - Thomas Ordowski, Jun 04 2024

			The sum of the numbers row 7 of the triangular array at A027926 is 2^8 - 1 = 255, and the number of numbers in row 7 is 15, and 255/15 = 17; thus 7 is in this sequence, and 17 is in A246649.

Robert Israel, Table of n, a(n) for n = 1..2000

Cf. A246637, A246649.

filter:= k -> 2 &^ (k+1) - 1 mod (2*k+1) = 0:
select(filter, [$0..2*10^5]); # Robert Israel, Jan 10 2020

z = 140000; u = Select[Range[0, z], IntegerQ[(2^(# + 1) - 1)/(2 # + 1)] &]   (* A246648 *)
v = Table[(2^(u[[k]] + 1) - 1)/(2 u[[k]] + 1), {k, 1, 6}] (* A246649 *)

Edited and offset changed by Robert Israel, Jan 10 2020

cp's OEIS Frontend

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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A026268 Triangle, T(n, k): T(n,k) = 1 for n < 3, T(3,1) = T(3,2) = T(3,3) = 2, T(n,0) = 1, T(n,1) = n-1, T(n,n) = T(n-1,n-2) + T(n-1,n-1), otherwise T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k), read by rows.

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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).

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A027927 Number of plane regions after drawing (in general position) a convex n-gon and all its diagonals.

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