A094292 -id:A094292 - cp's OEIS frontend

A049681 a(n) = (Lucas(2*n) - Lucas(n))/2.

0, 1, 2, 7, 20, 56, 152, 407, 1080, 2851, 7502, 19702, 51680, 135461, 354902, 929567, 2434320, 6374236, 16689752, 43697227, 114405500, 299525051, 784179002, 2053027082, 5374926720, 14071792681

Offset: 0

Clark Kimberling

nonn

Create a triangle with T(n,1) = L(n-1) for L a Lucas number and the other side T(n,n) = L(2*(n-1)). Interior elements are defined as T(r,c) = T(r-1,c-1) + T(r-1,c). Half the sum of the terms in row(n)=a(n) for n=1,2,3... - J. M. Bergot, Dec 15 2012

Robert Israel, Table of n, a(n) for n = 0..2380
Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Cf. A000032, A000045, A094292.

List([0..30], n-> (Lucas(1,-1,2*n)[2] - Lucas(1,-1,n)[2])/2 ); # G. C. Greubel, Dec 15 2019

[(Lucas(2*n) - Lucas(n))/2: n in [0..30]]; // G. C. Greubel, Dec 02 2017

Lucas:= n -> combinat:-fibonacci(n+1)+combinat:-fibonacci(n-1):
seq((Lucas(2*n)-Lucas(n))/2,n=0..100); # Robert Israel, Sep 15 2016

Table[(LucasL[2n] - LucasL[n])/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^30); concat([0], Vec(x*(1-2*x+2*x^2)/((1-x-x^2)*(1-3*x+x^2)) )) \\ G. C. Greubel, Dec 02 2017

[(lucas_number2(2*n,1,-1) - lucas_number2(n,1,-1))/2 for n in (0..30)] # G. C. Greubel, Dec 15 2019

G.f.: x*(1-2*x+2*x^2)/( (1-x-x^2)*(1-3*x+x^2) ). - R. J. Mathar, Dec 17 2012
a(n) = Lucas(n)*(Lucas(n) - 1)/2 - (-1)^n = binomial(Lucas(n), 2) - (-1)^n. - Vladimir Reshetnikov, Sep 27 2016
E.g.f.: (1/2)*exp(-2*x/(1+sqrt(5)))*(-1 + exp(x))*(1 + exp(sqrt(5)*x)). - Stefano Spezia, Dec 15 2019

Corrected by Franklin T. Adams-Watters, Oct 25 2006

Corrected by T. D. Noe, Nov 01 2006

A362067 Sum of successive Fibonacci numbers F(n) : a(n) = Sum_{k = 0..n} F(n+k).

Original entry on oeis.org

0, 2, 6, 18, 50, 136, 364, 966, 2550, 6710, 17622, 46224, 121160, 317434, 831430, 2177322, 5701290, 14927768, 39083988, 102327390, 267903350, 701391022, 1836283246, 4807480608, 12586194000, 32951158706, 86267374854, 225851115906, 591286215650, 1548007923880

Offset: 0

Philippe Deléham, Apr 07 2023

			a(n) are the row sums of the triangle T(n,k) (A199334):
  0;
  1, 1;
  1, 2,  3;
  2, 3,  5,  8;
  3, 5,  8, 13, 21;
  5, 8, 13, 21, 34, 55;
  ...
T(n,k) = T(n,k-1) + T(n-1, k-1); T(n,0) = A000045(n).

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

Cf. A000032, A000045, A094292, A199334.

A362067[n_] := Fibonacci[n+1]*(LucasL[n+1] - 1); Array[A362067, 50, 0] (* or *)
LinearRecurrence[{4, -3, -2, 1}, {0, 2, 6, 18}, 50] (* Paolo Xausa, Jun 10 2024 *)

a(n) = 4*a(n-1)-3*a(n-2)-2*a(n-3)+a(n-4), a(0)=0, a(1)=2, a(2)=6, a(3)=18.
G.f.: 2*x*(1-x)/((1-3*x+x^2)*(1-x-x^2)).
a(n) = A000045(2n+2) - A000045(n+1).
a(n) = 2 * A094292(n+1). - Alois P. Heinz, Apr 07 2023
a(n) = A000045(n+1)*(A000032(n+1) - 1). - Paolo Xausa, Jun 10 2024

A327917 Triangle T read by rows: T(k, n) = A(k-n, k) with the array A(k, n) = F(2k+n) = A000045(2k+n), for k >= 0 and n >= 0.

Original entry on oeis.org

0, 1, 1, 3, 2, 1, 8, 5, 3, 2, 21, 13, 8, 5, 3, 55, 34, 21, 13, 8, 5, 144, 89, 55, 34, 21, 13, 8, 377, 233, 144, 89, 55, 34, 21, 13, 987, 610, 377, 233, 144, 89, 55, 34, 21, 2584, 1597, 987, 610, 377, 233, 144, 89, 55, 34, 6765, 4181, 2584, 1597, 987, 610, 377, 233, 144, 89, 55

Offset: 0

Wolfdieter Lang, Oct 06 2019

This is the row reversed triangle A199334.
This is the analog of the array and the triangle A327916, where the positive odd numbers instead of the Fibonacci numbers are used.
The array A arises from the following Pascal-type triangles PF(k), for k >= 0, based on A000045 (Fibonacci). For example, the Pascal type triangle PF(k), for k = 3 is
  0   1   1   2
    1   2   3
      3   5
        8
For k -> infinity the left-aligned row sequences build the array A(k, n), with k >= 0 and n >= 0, that is, A(k, n) = F(2*k + n). See the example section for the first rows of A.
The first column sequence of A, {F(2*n) = A001906(n)}{n>=0}, is the binomial transform of the first (k=0) row sequence of A, {F(n)}{n>=0}.
The triangle T is the array A read by upwards antidiagonals.

			The Array A(k, n) begins:
k\n  0  1   2   3   4   5 ...
-----------------------------
0:   0  1   1   2   3   5 ...   F(n)
1:   1  2   3   5   8  13 ...   F(n+2)
2:   3  5   8  13  21  34 ...   F(n+4)
3:   8 13  21  34  55  89 ...   F(n+6)
4:  21 34  55  89 144 233 ...   F(n+8)
5:  55 89 144 233 377 610 ...   F(n+10)
...
---------------------------------------
The triangle T(k, n) begins:
k\n     0    1    2    3   4   5   6   7   8  9 10 ...
------------------------------------------------------
0:      0
1:      1    1
2:      3    2    1
3:      8    5    3    2
4:     21   13    8    5   3
5:     55   34   21   13   8   5
6:    144   89   55   34  21  13   8
7:    377  233  144   89  55  34  21  13
8:    987  610  377  233 144  89  55  34  21
9:   2584 1597  987  610 377 233 144  89  55 34
10:  6765 4181 2584 1597 987 610 377 233 144 89 55
...

Cf. A000045, A199334 (row reversed), A001906, A327916.
Column sequences of T (no leading zeros) and A: from the shifted Fibonacci bisection {F(2*k) = A001906(k)} for even n, and {F(2*k+1) = A001519(k+1)}, for odd n.
Row sums: 2*A094292(n+1) = F(2*(n+1)) - F(n+1), n >= 0.
Alternating row sums: 2*A164267(n-1), n >= 0, with 0 for n = 0.

A(k, n) = Sum_{j=0..k} binomial(k, j)*F(n+j) = F(2*k+n), for k >= 0 and n >= 0.
T(k, n) = A(k - n, n)  = F(2*k - n), for k >= 0 and n = 0..k, with the Fibonacci numbers F = A000045.
Recurrence: T(k,0) = F(2*k), k >= 0, T(k, n) = T(k, n-1) - T(k-1, n-1), k >= 1, n = 1..k, and T(k, n) = 0 if k < n.
O.g.f. for row polynomials R(n, x) = Sum_{n=0..k} T(k, n)*x^n:
G(x, z) = Sum_{n=0} R(n, x)*z^n = z*(1 + x - 2*x*z)/((1 - 3*z + z^2)*(1 - x*z - (x*z)^2)).
T(k, 0) =  Sum_{n=0..k} binomial(k,n)*T(n, n), k >= 0 (binomial transform).

cp's OEIS Frontend

A049681 a(n) = (Lucas(2*n) - Lucas(n))/2.

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A362067 Sum of successive Fibonacci numbers F(n) : a(n) = Sum_{k = 0..n} F(n+k).

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A327917 Triangle T read by rows: T(k, n) = A(k-n, k) with the array A(k, n) = F(2k+n) = A000045(2k+n), for k >= 0 and n >= 0.

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cp's OEIS Frontend

A049681 a(n) = (Lucas(2*n) - Lucas(n))/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A362067 Sum of successive Fibonacci numbers F(n) : a(n) = Sum_{k = 0..n} F(n+k).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A327917 Triangle T read by rows: T(k, n) = A(k-n, k) with the array A(k, n) = F(2*k+n) = A000045(2*k+n), for k >= 0 and n >= 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A327917 Triangle T read by rows: T(k, n) = A(k-n, k) with the array A(k, n) = F(2k+n) = A000045(2k+n), for k >= 0 and n >= 0.