A037027 -id:A037027 - cp's OEIS frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Nov 29 2009

Row sums: A000045(n+1), Fibonacci numbers.
A168561*A007318 = A037027, as lower triangular matrices. Diagonal sums : A077957. - Philippe Deléham, Dec 02 2009
T(n,k) is the number of compositions of n+1 into k+1 odd parts. Example: T(4,2)=3 because we have 5 = 1+1+3 = 1+3+1 = 3+1+1.
Coefficients of monic Fibonacci polynomials (rising powers of x). Ftilde(n, x) = x*Ftilde(n-1, x) + Ftilde(n-2, x), n >=0, Ftilde(-1,x) = 0, Ftilde(0, x) = 1. G.f.: 1/(1 - x*z - z^2). Compare with Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jul 29 2014

			The triangle T(n,k) begins:
n\k 0  1   2   3   4    5    6    7    8    9  10  11  12  13 14 15 ...
0:  1
1:  0  1
2:  1  0   1
3:  0  2   0   1
4:  1  0   3   0   1
5:  0  3   0   4   0    1
6:  1  0   6   0   5    0    1
7:  0  4   0  10   0    6    0    1
8:  1  0  10   0  15    0    7    0    1
9:  0  5   0  20   0   21    0    8    0    1
10: 1  0  15   0  35    0   28    0    9    0   1
11: 0  6   0  35   0   56    0   36    0   10   0   1
12: 1  0  21   0  70    0   84    0   45    0  11   0   1
13: 0  7   0  56   0  126    0  120    0   55   0  12   0   1
14: 1  0  28   0 126    0  210    0  165    0  66   0  13   0  1
15: 0  8   0  84   0  252    0  330    0  220   0  78   0  14  0  1
... reformatted by _Wolfdieter Lang_, Jul 29 2014.
------------------------------------------------------------------------

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - From _N. J. A. Sloane_, May 10 2012
Tom Copeland, Addendum to Elliptic Lie Triad
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A162515 (rows reversed), A112552, A102426 (deflated).

A168561:=proc(n,k) if n-k mod 2 = 0 then binomial((n+k)/2,k) else 0 fi end proc:
seq(seq(A168561(n,k),k=0..n),n=0..12) ; # yields sequence in triangular form

Table[If[EvenQ[n + k], Binomial[(n + k)/2, k], 0], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Apr 16 2017 *)

T(n,k) = if ((n+k) % 2, 0, binomial((n+k)/2,k));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print();); \\ Michel Marcus, Oct 09 2016

Sum_{k=0..n} T(n,k)*x^k = A059841(n), A000045(n+1), A000129(n+1), A006190(n+1), A001076(n+1), A052918(n), A005668(n+1), A054413(n), A041025(n), A099371(n+1), A041041(n), A049666(n+1), A041061(n), A140455(n+1), A041085(n), A154597(n+1), A041113(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 respectively. - Philippe Deléham, Dec 02 2009
T(2n,2k) = A085478(n,k). T(2n+1,2k+1) = A078812(n,k). Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000045(n+1), A006131(n), A015445(n), A168579(n), A122999(n) for x = 0,1,2,3,4,5 respectively. - Philippe Deléham, Dec 02 2009
T(n,k) = binomial((n+k)/2,k) if (n+k) is even; otherwise T(n,k)=0.
G.f.: (1-z^2)/(1-t*z-z^2) if offset is 1.
T(n,k) = T(n-1,k-1) + T(n-2,k), T(0,0) = 1, T(0,1) = 0. - Philippe Deléham, Feb 09 2012
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 09 2012
From R. J. Mathar, Feb 04 2022: (Start)
Sum_{k=0..n} T(n,k)*k = A001629(n+1).
Sum_{k=0..n} T(n,k)*k^2 = 0,1,4,11,... = 2*A055243(n)-A099920(n+1).
Sum_{k=0..n} T(n,k)*k^3 = 0,1,8,29,88,236,... = 12*A055243(n) -6*A001629(n+2) +A001629(n+1)-6*(A001872(n)-2*A001872(n-1)). (End)

Typo in name corrected (1(1-x^2) changed to 1/(1-x^2)) by Wolfdieter Lang, Nov 20 2010

A036355 Fibonacci-Pascal triangle read by rows.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 5, 5, 3, 5, 10, 14, 10, 5, 8, 20, 32, 32, 20, 8, 13, 38, 71, 84, 71, 38, 13, 21, 71, 149, 207, 207, 149, 71, 21, 34, 130, 304, 478, 556, 478, 304, 130, 34, 55, 235, 604, 1060, 1390, 1390, 1060, 604, 235, 55, 89, 420, 1177, 2272, 3310, 3736, 3310, 2272, 1177, 420, 89

Offset: 0

Floor van Lamoen, Dec 28 1998

T(n,k) is the number of lattice paths from (0,0) to (n-k,k) using steps (1,0),(2,0),(0,1),(0,2). - Joerg Arndt, Jun 30 2011, corrected by Greg Dresden, Aug 25 2020
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins
   1;
   1,   1;
   2,   2,   2;
   3,   5,   5,    3;
   5,  10,  14,   10,    5;
   8,  20,  32,   32,   20,    8;
  13,  38,  71,   84,   71,   38,   13;
  21,  71, 149,  207,  207,  149,   71,  21;
  34, 130, 304,  478,  556,  478,  304, 130,  34;
  55, 235, 604, 1060, 1390, 1390, 1060, 604, 235, 55;
with indices
  T(0,0);
  T(1,0),  T(1,1);
  T(2,0),  T(2,1),  T(2,2);
  T(3,0),  T(3,1),  T(3,2),  T(3,3);
  T(4,0),  T(4,1),  T(4,2),  T(4,3),  T(4,4);
For example, T(4,2) = 14 and there are 14 lattice paths from (0,0) to (4-2,2) = (2,2) using steps (1,0),(2,0),(0,1),(0,2). - _Greg Dresden_, Aug 25 2020

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

Row sums form sequence A002605. T(n, 0) forms the Fibonacci sequence (A000045). T(n, 1) forms sequence A001629.
Derived sequences: A036681, A036682, A036683, A036684, A036692 (central terms).
Cf. A007318, A051159, A091533, A228196, A228576.
Some other Fibonacci-Pascal triangles: A027926, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074.

a036355 n k = a036355_tabl !! n !! k
a036355_row n = a036355_tabl !! n
a036355_tabl = [1] : f [1] [1,1] where
   f us vs = vs : f vs (zipWith (+)
                       (zipWith (+) ([0,0] ++ us) (us ++ [0,0]))
                       (zipWith (+) ([0] ++ vs) (vs ++ [0])))
-- Reinhard Zumkeller, Apr 23 2013

nmax = 11; t[n_, m_] := t[n, m] = tp[n-1, m-1] + tp[n-2, m-2] + tp[n-1, m] + tp[n-2, m]; tp[n_, m_] /; 0 <= m <= n && n >= 0 := t[n, m]; tp[n_, m_] = 0; t[0, 0] = 1; Flatten[ Table[t[n, m], {n, 0, nmax}, {m, 0, n}]] (* Jean-François Alcover, Nov 09 2011, after formula *)

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

T(n, m) = T'(n-1, m-1)+T'(n-2, m-2)+T'(n-1, m)+T'(n-2, m), where T'(n, m) = T(n, m) if 0<=m<=n and n >= 0 and T'(n, m)=0 otherwise. Initial term T(0, 0)=1.

G.f.: 1/(1-(1+y)*x-(1+y^2)*x^2). - Vladeta Jovovic, Oct 11 2003

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

A155161 A Fibonacci convolution triangle: Riordan array (1, x/(1 - x - x^2)). Triangle T(n,k), 0 <= k <= n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 3, 5, 3, 1, 0, 5, 10, 9, 4, 1, 0, 8, 20, 22, 14, 5, 1, 0, 13, 38, 51, 40, 20, 6, 1, 0, 21, 71, 111, 105, 65, 27, 7, 1, 0, 34, 130, 233, 256, 190, 98, 35, 8, 1, 0, 55, 235, 474, 594, 511, 315, 140, 44, 9, 1, 0, 89, 420, 942, 1324, 1295, 924, 490, 192, 54, 10, 1

Offset: 0

Philippe Deléham, Jan 21 2009

			Triangle begins:
[0] 1;
[1] 0,  1;
[2] 0,  1,   1;
[3] 0,  2,   2,   1;
[4] 0,  3,   5,   3,   1;
[5] 0,  5,  10,   9,   4,   1;
[6] 0,  8,  20,  22,  14,   5,  1;
[7] 0, 13,  38,  51,  40,  20,  6,  1;
[8] 0, 21,  71, 111, 105,  65, 27,  7, 1;
[9] 0, 34, 130, 233, 256, 190, 98, 35, 8, 1.

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.

Row sums are in A215928.
Central terms: T(2*n,n) = A213684(n) for n > 0.
Cf. A000045, A037027, A122542, A059283, A321620.

a155161 n k = a155161_tabl !! n !! k
a155161_row n = a155161_tabl !! n
a155161_tabl = [1] : [0,1] : f [0] [0,1] where
   f us vs = ws : f vs ws where
     ws = zipWith (+) (us ++ [0,0]) $ zipWith (+) ([0] ++ vs) (vs ++ [0])
-- Reinhard Zumkeller, Apr 17 2013

T := (n, k) -> binomial(n-1, k-1)*hypergeom([-(n-k)/2, -(n-k-1)/2], [1-n], -4):
seq(seq(simplify(T(n, k)), k = 0..n), n = 0..11); # Peter Luschny, May 23 2021
# Uses function PMatrix from A357368.
PMatrix(10, n -> combinat:-fibonacci(n)); # Peter Luschny, Oct 07 2022

CoefficientList[#, y]& /@ CoefficientList[(1-x-x^2)/(1-x-x^2-x*y)+O[x]^12, x] // Flatten (* Jean-François Alcover, Mar 01 2019 *)
(* Generates the triangle without the leading '1' (rows are rearranged). *)
(* Function RiordanSquare defined in A321620. *)
RiordanSquare[x/(1 - x - x^2), 11] // Flatten  (* Peter Luschny, Feb 27 2021 *)

M(n,k):=pochhammer(n,k)/k!;
create_list(sum(M(k,i)*binomial(i,n-i-k),i,0,n-k),n,0,8,k,0,n); /* Emanuele Munarini, Mar 15 2011 */

T(n, k) given by [0,1,1,-1,0,0,0,...] DELTA [1,0,0,0,...] where DELTA is the operator defined in A084938.
a(n,k) = Sum_{i=0..n-k} M(k,i)*binomial(i,n-i-k), where M(n,k) = n(n+1)(n+2)...(n+k-1)/k!. - Emanuele Munarini, Mar 15 2011
Recurrence: a(n+2,k+1) = a(n+1,k+1) + a(n+1,k) + a(n,k+1). - Emanuele Munarini, Mar 15 2011
G.f.: (1-x-x^2)/(1-x-x^2-x*y). - Philippe Deléham, Feb 08 2012
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000129(n) (n > 0), A052991(n), A155179(n), A155181(n), A155195(n), A155196(n), A155197(n), A155198(n), A155199(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Feb 08 2012
T(n, k) = binomial(n-1, k-1)*hypergeom([-(n-k)/2, -(n-k-1)/2], [1-n], -4). - Peter Luschny, May 23 2021

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

A001872 Convolved Fibonacci numbers.

Original entry on oeis.org

1, 4, 14, 40, 105, 256, 594, 1324, 2860, 6020, 12402, 25088, 49963, 98160, 190570, 366108, 696787, 1315072, 2463300, 4582600, 8472280, 15574520, 28481220, 51833600, 93914325, 169457708, 304597382, 545556512, 973877245, 1733053440, 3075011478

Offset: 0

N. J. A. Sloane and Simon Plouffe

6*a(n) is the number of ways to tile a strip of length n+6 with squares and three colors of dominos, where we must have the same number of dominos of each color. - Greg Dresden and Jiachen Weng, Aug 14 2025

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 101.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
Daniel Birmajer, Juan Gil and Michael D. Weiner, Linear recurrence sequences and their convolutions via Bell polynomials, arXiv:1405.7727 [math.CO], 2014.
Daniel Birmajer, Juan Gil and Michael D. Weiner, Linear Recurrence Sequences and Their Convolutions via Bell Polynomials, Journal of Integer Sequences, 18 (2015), #15.1.2.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Verner E. Hoggatt, Jr. and Marjorie Bicknell-Johnson, Fibonacci convolution sequences, Fib. Quart., 15 (1977), 117-122.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Toufik Mansour, Generalization of some identities involving the Fibonacci numbers, arXiv:math/0301157 [math.CO], 2003.
Pieter Moree, Convoluted convolved Fibonacci numbers, arXiv:math/0311205 [math.CO], 2003.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 18.
Index entries for linear recurrences with constant coefficients, signature (4,-2,-8,5,8,-2,-4,-1).

[(n+5)*(n+3)*(4*(n+1)*Fibonacci(n+2)+3*(n+2)*Fibonacci(n+1))/150: n in [0..30]]; // Vincenzo Librandi, Nov 19 2014

a := n-> (Matrix(8, (i,j)-> if (i=j-1) then 1 elif j=1 then [4,-2,-8,5, 8,-2,-4,-1][i] else 0 fi)^n)[1,1]; seq (a(n), n=0..29);  # Alois P. Heinz, Aug 15 2008

CoefficientList[Series[1/(1 - x - x^2)^4, {x, 0, 100}], x] (* Stefan Steinerberger, Apr 15 2006 *)

Vec( 1/(1 - x - x^2)^4 + O(x^66) )  \\ Joerg Arndt, May 12 2014

G.f.: 1/(1 - x - x^2)^4.
a(n) = A037027(n+3, 3) (Fibonacci convolution triangle).
a(n) = (n+5)*(n+3)*(4*(n+1)*F(n+2)+3*(n+2)*F(n+1))/150, F(n)=A000045(n). - Wolfdieter Lang, Apr 12 2000
For n > 3, a(n-3) = Sum_{h+i+j+k=n} F(h)*F(i)*F(j)*F(k). - Benoit Cloitre, Nov 01 2002
a(n) = F'''(n+3, 1)/6, i.e., 1/6 times the 3rd derivative of the (n+3)th Fibonacci polynomial evaluated at 1. - T. D. Noe, Jan 18 2006
a(n) = (((-i)^n)/3!)*(d^3/dx^3)S(n+3,x)|A049310%20for%20the%20S-polynomials.%20-%20_Wolfdieter%20Lang">{x=i}, where i is the imaginary unit. Third derivative of Chebyshev S(n+3,x) polynomial evaluated at x=i multiplied by ((-i)^(n-3))/3!. See A049310 for the S-polynomials. - _Wolfdieter Lang, Apr 04 2007
a(n) = Sum_{i=ceiling(n/2)..n} (i+1)*(i+2)*(i+3)*binomial(i,n-i)/6. - Vladimir Kruchinin, Apr 26 2011
Recurrence: a(n) = 4*a(n-1) - 2*a(n-2) - 8*a(n-3) + 5*a(n-4) + 8*a(n-5) - 2*a(n-6) - 4*a(n-7) - a(n-8). - Fung Lam, May 11 2014
n*a(n) - (n+3)*a(n-1) - (n+6)*a(n-2) = 0, n > 1. - Michael D. Weiner, Nov 18 2014

cp's OEIS Frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Python

Sage

Formula

Extensions

A168561 Riordan array (1/(1-x^2), x/(1-x^2)). Unsigned version of A049310.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A036355 Fibonacci-Pascal triangle read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Formula

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