A081057 -id:A081057 - cp's OEIS frontend

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

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

A014335 Exponential convolution of Fibonacci numbers with themselves (divided by 2).

Original entry on oeis.org

0, 0, 1, 3, 11, 35, 115, 371, 1203, 3891, 12595, 40755, 131891, 426803, 1381171, 4469555, 14463795, 46805811, 151466803, 490156851, 1586180915, 5132989235, 16610702131, 53753361203, 173949530931, 562912506675, 1821623137075, 5894896300851, 19076285150003

Offset: 0

N. J. A. Sloane

It can be noticed that A014335/A011782 is an "autosequence", that is a sequence which is identical to its inverse binomial transform, except for alternating signs. - Jean-François Alcover, Jun 15 2016

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

Cf. A000032, A000045, A014334, A081057, A203579.
Cf. (partial sums of) A063727.
Column k=2 of A346415.

[(2^n*Lucas(n)-2)/10: n in [0..40]]; // G. C. Greubel, Jan 06 2023

a[0]:=0:a[1]:=0:for n from 2 to 50 do a[n]:=2*a[n-1]+4*a[n-2]+1 od: seq(a[n], n=0..29); # Zerinvary Lajos, Dec 14 2008
# second Maple program:
a:= n-> (<<0|1|0>, <0|0|1>, <-4|2|3>>^n)[1,3]:
seq(a(n), n=0..30);  # Alois P. Heinz, Oct 04 2016

LinearRecurrence[{3,2,-4}, {0,0,1}, 41] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
Table[(2^n LucasL[n] - 2)/10, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 06 2016 *)

[(2^n*lucas_number2(n,1,-1) -2)/10 for n in range(41)] # G. C. Greubel, Jan 06 2023

a(n) = A014334(n)/2.
G.f.: x^2/((1-x)*(1-2*x-4*x^2)). - Vladeta Jovovic, Mar 05 2003
E.g.f.: exp(x)*(cosh(sqrt(5)*x)-1)/5. - Vladeta Jovovic, Sep 01 2004
From Benoit Cloitre, Sep 25 2004: (Start)
a(n+1) = Sum_{i=0..n} A000045(i)*2^(i-1).
a(n) = (1/5)*(2^(n-1)*A000032(n) - 1). (End)
a(n) = 2*a(n-1) + 4*a(n-2) + 1, a(0)=0; a(1)=0. - Zerinvary Lajos, Dec 14 2008
G.f.: G(0)*x^2/(2*(1-x)^2), where G(k)= 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = (A203579(n) - 2)/5. - Vladimir Reshetnikov, Oct 06 2016

cp's OEIS Frontend

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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