A054450 -id:A054450 - cp's OEIS frontend

A054453 Triangle of partial row sums of triangle A054450(n,m), n >= m >= 0.

1, 2, 1, 4, 2, 1, 8, 5, 2, 1, 15, 10, 6, 2, 1, 28, 20, 12, 7, 2, 1, 51, 38, 26, 14, 8, 2, 1, 92, 71, 50, 33, 16, 9, 2, 1, 164, 130, 97, 64, 41, 18, 10, 2, 1, 290, 235, 180, 130, 80, 50, 20, 11, 2, 1, 509, 420, 332, 244, 171, 98, 60, 22, 12, 2, 1

Offset: 0

Wolfdieter Lang, Apr 27 2000 and May 08 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is ((1-z^2)*(Fib(z))^2)/(1-x*z/(1-z^2)) Fib(x)=1/(1-x-x^2) = g.f. for A000045(n+1) (Fibonacci numbers without 0).
This is the second member of the family of Riordan-type matrices obtained from the unsigned convolution matrix A049310(n,m) by repeated application of the partial row sums procedure.
The column sequences are A029907, A001629, A054454 for m=0..2.

			{1}; {2,1}; {4,2,1}; {8,5,2,1};...
Fourth row polynomial (n=3): p(3,x)= 8+5*x+2*x^2+x^3

Gregg Musiker, Nick Ovenhouse, and Sylvester W. Zhang, Double Dimers and Super Ptolemy Relations, Séminaire Lotharingien de Combinatoire XX, Proc. 35th Conf. Formal Power, Series and Algebraic Combinatorics (Davis) 2023, Art. #YY. See p. 12.

Cf. A049310, A054450, A000045, A029907, A001629. Row sums: A054455(n).

a(n, m)=sum(A054450(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n

Column m recursion: a(n, m)= sum(a(j-1, m)*|A049310(n-j, 0)|, j=m..n) + A054450(n, m), n >= m >= 0, a(n, m) := 0 if n

G.f. for column m: ((1-x^2)*(Fib(x))^2)*(x/(1-x^2))^m, m >= 0, with Fib(x) G.f. for A000045(n+1).

A099573 Reverse of number triangle A054450.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233

Offset: 0

Paul Barry, Oct 23 2004

			First few rows of the array:
  1, 1, 2, 3,  5,  8, ... (A000045)
  1, 1, 3, 4,  8, 12, ... (A052952)
  1, 1, 4, 5, 12, 17, ... (A054451)
  1, 1, 5, 6, 17, 23, ... (A099571)
  1, 1, 6, 7, 23, 30, ... (A099572)
  ...
Triangle begins as:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 3, 3;
  1, 1, 4, 4,  5;
  1, 1, 5, 5,  8,  8;
  1, 1, 6, 6, 12, 12, 13;
  1, 1, 7, 7, 17, 17, 21, 21;
  1, 1, 8, 8, 23, 23, 33, 33, 34;
  1, 1, 9, 9, 30, 30, 50, 50, 55, 55;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000045, A029907 (row sums), A052951, A052952, A054450, A054451, A052952.

Cf. A099571, A099572, A099574 (diagonal sums), A099575.

[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2)))
flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)

A052952 a(n) = Fibonacci(n+2) - (1-(-1)^n)/2.

Original entry on oeis.org

1, 1, 3, 4, 8, 12, 21, 33, 55, 88, 144, 232, 377, 609, 987, 1596, 2584, 4180, 6765, 10945, 17711, 28656, 46368, 75024, 121393, 196417, 317811, 514228, 832040, 1346268, 2178309, 3524577, 5702887, 9227464, 14930352, 24157816, 39088169, 63245985, 102334155

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Equals row sums of triangle A173284. - Gary W. Adamson, Feb 14 2010
The Kn21 sums (see A180662 for definition) of the 'Races with Ties' triangle A035317 produce this sequence. - Johannes W. Meijer, Jul 20 2011
a(n-1), for n >= 1, gives the number of compositions of n with relative prime parts, and parts not exceeding 2. See the row sums of triangle A030528 where for even n the leading 1 is missing. - Wolfdieter Lang, Jul 27 2023

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 12*x^5 + 21*x^6 + 33*x^7 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1023
K. Kuhapatanakul, On the Sums of Reciprocal Generalized Fibonacci Numbers, J. Int. Seq. 16 (2013) #13.7.1, eq (1).
Steven Linton, James Propp, Tom Roby, and Julian West, Equivalence Classes of Permutations under Various Relations Generated by Constrained Transpositions, Journal of Integer Sequences, Vol. 15 (2012), #12.9.1.
H. Ohtsuka and S. Nakamura, On the sum of reciprocal sums of Fibonacci numbers, Fibonacci Quart. 46/47 (2008/2009), 153-159.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).

Cf. A062114, A173284, A059841, A014217, A180662, A035317.
Partial sums of A008346, first differences of A129696.
Cf. also A000032, A000045, A030528.
Cf. A001595, A054450, A066983, A074331, A168309.

List([0..40], n-> Fibonacci(n+2) -(1-(-1)^n)/2); # G. C. Greubel, Jul 10 2019

a052952 n = a052952_list !! n
a052952_list = 1 : 1 : zipWith (+)
   a059841_list (zipWith (+) a052952_list $ tail a052952_list)
-- Reinhard Zumkeller, Jan 06 2012

[Fibonacci(n+2)-(1-(-1)^n)/2: n in [0..40]]; // Vincenzo Librandi, Dec 02 2016

A052952 :=proc(n)
    option remember;
    local t1;
    if n <= 1 then
        return 1 ;
    fi:
    if n mod 2 = 1 then
        t1:=0
    else
        t1:=1;
    fi:
    procname(n-1)+procname(n-2)+t1;
end proc;
seq(A052952(n), n=0..40) ; # N. J. A. Sloane, May 25 2008

Table[Fibonacci[n+2] -(1-(-1)^n)/2, {n, 0, 40}] (* Vincenzo Librandi, Dec 02 2016 *)
Sum[(-1)^k*Fibonacci[Range[2,41], 1-k], {k,0,1}] (* G. C. Greubel, Oct 21 2019 *)
CoefficientList[Series[1/((1-x-x^2)*(1-x^2)),{x,0,40}],x] (* Harvey P. Dale, Sep 12 2020 *)

PARI
```
{a(n) = fibonacci(n+2) - n%2};
```

[fibonacci(n+2) -(1-(-1)^n)/2 for n in (0..40)] # G. C. Greubel, Jul 10 2019

G.f.: 1/((1-x-x^2)*(1-x^2)).
a(n) = A074331(n+1).
a(n) = A054450(n+1, 1) (second column of triangle).
a(n) = 2*a(n-2) + a(n-3) + 1, with a(0)=1, a(1)=1, a(2)=3.
a(n) = Sum_{alpha=RootOf(-1+z+z^2)} (3+alpha)*alpha^(-1-n)/3 - Sum_{beta=RootOf(-1+z^2)} beta^(-1-n)/2.
a(2*k) = Sum_{j=0..k} F(2*j+1) = F(2*(k+1)) for k >= 0; a(2*k-1) = Sum_{j=0..k} F(2*j) = F(2*k+1)-1 for k >= 1 (F = A000045, Fibonacci numbers).
a(n) = a(n-1) + a(n-2) + (1+(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+1, k). - Paul Barry, Oct 23 2004
a(n) = floor(phi^(n+2) / sqrt(5)), where phi is the golden ratio: phi = (1+sqrt(5))/2. - Reinhard Zumkeller, Apr 19 2005
a(n) = Fibonacci(n+1) + a(n-2) with n>1, a(0)=a(1)=1. - Zerinvary Lajos, Mar 17 2008
a(n) = floor(Fibonacci(n+3)^2/Fibonacci(n+4)). - Gary Detlefs, Nov 29 2010
a(n) = (A001595(n+3) - A066983(n+4))/2. - Gary Detlefs, Dec 19 2010
a(4*n) = F(4*n+2); a(4*n+1) = F(4*n+3) - 1; a(4*n+2) = F(4*n+4); a(4*n+3) = F(4*n+5) - 1. - Johannes W. Meijer, Jul 20 2011
a(n+1) = a(n) + a(n-1) + A059841(n+1). - Reinhard Zumkeller, Jan 06 2012
a(n) = floor(|F((1+i)*(n+2))|), n >= 0, with the complex Fibonacci function F: C -> C, z -> F(z) with F(z) := (exp(log(phi)*z) - exp(i*Pi*z)*exp(-log(phi)*z))/(2*phi-1) with the modulus |z|, the imaginary unit i and the golden section phi:=(1+sqrt(5))/2. A Conjecture: For F(z) see, e.g., the T. Koshy reference. ch. 45, p. 523, where F is called f, given in A000045. - Wolfdieter Lang, Jul 24 2012
5*a(n) = (L(n+3)-1)*(L(n+4)+3) -14 -Sum_{k=0..n} L(k+1)*L(k+5) = (L(n+3)-1)*(L(n+4)+3) -L(2*n+7) +A168309(n), where L=A000032. - J. M. Bergot, Jun 13 2014
a(n) = floor(phi*Fibonacci(n+1)), where phi is the golden section. - Michel Dekking, Dec 02 2016
a(n) = -(-1)^n * a(-4-n) for all n in Z. - Michael Somos, Dec 03 2016
a(n) = Sum_{k=0..n} Sum_{i=0..n} C(n-k-1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = floor(1/(Sum_{k>=n+4} 1/Fibonacci(k))) [Ohtsuka and Nakamura]. - Michel Marcus, Aug 09 2018
a(n) = floor(abs(chebyshevU(n/2, 3/2))). - Federico Provvedi, Feb 23 2022
E.g.f.: exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - sinh(x). - Stefano Spezia, Mar 09 2024

Additional formulas and more terms from Wolfdieter Lang, May 02 2000

Better description from Olivier Gérard, Jun 05 2001

A029907 a(n+1) = a(n) + a(n-1) + Fibonacci(n), with a(0) = 0 and a(1) = 1.

Original entry on oeis.org

0, 1, 2, 4, 8, 15, 28, 51, 92, 164, 290, 509, 888, 1541, 2662, 4580, 7852, 13419, 22868, 38871, 65920, 111556, 188422, 317689, 534768, 898825, 1508618, 2528836, 4233872, 7080519, 11828620, 19741179, 32916068, 54835556, 91276202, 151814645, 252318312

Offset: 0

N. J. A. Sloane

Number of matchings of the fan graph on n vertices, n>0 (a fan is the join of the path graph with one extra vertex).
a(n+1) gives row sums of A054450. - Paul Barry, Oct 23 2004
Number of parts in all compositions of n into odd parts. Example: a(5)=15 because the compositions 5, 311, 131, 113, and 11111 have a total of 1+3+3+3+5=15 parts.
a(n-1) is the number of compositions of n that contain one even part; for example, a(5-1)=a(4)=8 counts the compositions 1112, 1121, 1211, 14, 2111, 23, 32, 41. - Joerg Arndt, May 21 2013

			a(4)=8 because matchings of fan graph with edges {OA,OB,OC,AB,AC} are: {},{OA},{OB},{OC},{AB},{AC},{OA,BC},{OC,AB}.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Jean-Luc Baril and Nathanaël Hassler, Intervals in a family of Fibonacci lattices, Univ. de Bourgogne (France, 2024). See p. 5.
David Broadhurst, Zig-zag resistance and OEIS sequence A029907, Apr 11 2024
Jia Huang, Compositions with restricted parts, arXiv:1812.11010 [math.CO], 2018.
Mengmeng Liu and Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000032, A000045, A001629, A010049, A054450, A152163, A240847.

a029907 n = a029907_list !! n
a029907_list = 0 : 1 : zipWith (+) (tail a000045_list)
                      (zipWith (+) (tail a029907_list) a029907_list)
-- Reinhard Zumkeller, Nov 01 2013

[((n+4)*Fibonacci(n)+2*n*Fibonacci(n-1))/5: n in [0..40]]; // Vincenzo Librandi, Feb 25 2018

with(combinat); A029907 := proc(n) options remember; if n <= 1 then n else procname(n-1)+procname(n-2)+fibonacci(n-1); fi; end;

CoefficientList[Series[x(1-x^2)/(1-x-x^2)^2, {x, 0, 37}], x] (* or *)
a[n_]:= a[n]= a[n-1] +a[n-2] +Fibonacci[n-1]; a[0]=0; a[1]=1; Array[a, 37] (* or *)
LinearRecurrence[{2,1,-2,-1}, {0,1,2,4}, 38] (* Robert G. Wilson v, Jun 22 2014 *)

alias(F,fibonacci); a(n)=((n+4)*F(n)+2*n*F(n-1))/5;

def A029907(n): return (1/5)*(n*lucas_number2(n, 1, -1) + 4*fibonacci(n))
[A029907(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022

G.f.: x*(1-x^2)/(1-x-x^2)^2.
a(n) = ((n+4)*Fibonacci(n) + 2*n*Fibonacci(n-1))/5.
a(n+1) = Sum_{k=0..n} Sum_{j=0..floor(k/2)} binomial(n-j, j). - Paul Barry, Oct 23 2004
a(n) = A010049(n+1) + A152163(n+1). - R. J. Mathar, Dec 10 2011
a(n) = F(n) + Sum_{k=1..n-1} F(k)*F(n-k), where F=Fibonacci. - Reinhard Zumkeller, Nov 01 2013
a(n) = (1/5)*(n*A000032(n) + 4*A000045(n)). - G. C. Greubel, Apr 06 2022
a(n) = A001629(n+1) - A001629(n-1), where A001629 is the first convolution of the Fibonacci numbers. - Gregory L. Simay, Aug 30 2022
E.g.f.: exp(x/2)*(5*x*cosh(sqrt(5)*x/2) + sqrt(5)*(5*x + 8)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023

Additional formula from Wolfdieter Lang, May 02 2000

Additional comments from Michael Somos, Jul 23 2002

A054452 Partial sums of A027941(n-1) with a(-1) = 0.

Original entry on oeis.org

0, 0, 1, 5, 17, 50, 138, 370, 979, 2575, 6755, 17700, 46356, 121380, 317797, 832025, 2178293, 5702870, 14930334, 39088150, 102334135, 267914275, 701408711, 1836311880, 4807526952, 12586269000, 32951280073, 86267571245, 225851433689, 591286729850

Offset: 0

Wolfdieter Lang, Apr 27 2000

Colin Barker, Table of n, a(n) for n = 0..1000
László Németh, Pascal pyramid in the space H^2 x R, arXiv:1701.06022 [math.CO], 2017 (5th line of Table 1 is a(n-2)).
A. Shriki and O. Liba, Polygons with Fibonacci Number Coordinates: Problem B-1167, Fib. Quart. 54,2 May 2016, p. 180-181.
Index entries for linear recurrences with constant coefficients, signature (5,-8,5,-1).

Cf. A027941, A054451, A001906, A052952.

I:=[0,0,1,5]; [n le 4 select I[n] else 5*Self(n-1)-8*Self(n-2)+5*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Mar 26 2015

a[0]:=0: a[1]:=1: for n from 2 to 50 do a[n] := 3*a[n-1]-a[n-2] od: seq(a[n]-n, n=0..27); # Zerinvary Lajos, Mar 20 2008
with(combinat): seq(fibonacci(2*n)-n, n=0..27); # Zerinvary Lajos, Jun 19 2008
g:=z/(1-3*z+z^2): gser:=series(g, z=0, 43): seq(abs(coeff(gser, z, n)-n), n=0..27); # Zerinvary Lajos, Mar 22 2009

CoefficientList[Series[x^2 / ((1 - x)^2 (1 - 3 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 26 2015 *)

makelist(sum(fib(k)*binomial(n+1,k+2),k,0,n),n,0,20); /* Vladimir Kruchinin, Oct 21 2016 */

concat(vector(2), Vec(x^2/((1-x)^2*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Jan 28 2017

[(lucas_number1(n, 3, 1)-lucas_number1(n, 2, 1)) for n in range(1, 28)]# Zerinvary Lajos, Mar 13 2009

a(n) = +5*a(n-1) -8*a(n-2) +5*a(n-3) -1*a(n-4).
G.f.: x^2/((1-x)^2*(1-3*x+x^2)).
a(n) = Sum_{k=0..n} A027941(k-1) = F(2*n)-n = A054450(2*n-1, 2) = A054451(2*n-3).
G.f.: x^2*Fibe(x)/(1-x)^2, with Fibe(x) := 1/(1-3*x+x^2) = g.f. A001906(n+1) (Fibonacci numbers F(2(n+1))).
Fourth diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = Max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
a(n) = Sum_{k=0..n-2} binomial(2*n-k-1, k). - Johannes W. Meijer, Aug 12 2013
a(n) = Sum_{i=1..n-1} Sum_{j=1..n-1} binomial(i+j, i-j). - Wesley Ivan Hurt, Mar 25 2015
a(n) = Sum_{k=0..n} (binomial(n+1,k+2)*Fibonacci(k)). - Vladimir Kruchinin, Oct 21 2016
a(n) = (-((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n)/sqrt(5) - n. - Colin Barker, Jan 28 2017

More terms from James Sellers, Apr 28 2000

a(0) added by Arkadiusz Wesolowski, Jun 07 2011

A099572 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).

Original entry on oeis.org

1, 1, 6, 7, 23, 30, 73, 103, 211, 314, 581, 895, 1560, 2455, 4135, 6590, 10890, 17480, 28590, 46070, 74946, 121016, 196326, 317342, 514123, 831465, 1346148, 2177613, 3524441, 5702054, 9227311, 14929365, 24157645, 39087010, 63245795, 102332805

Offset: 0

Paul Barry, Oct 23 2004

Fifth column of triangle A054450. In general Sum_{k=0..floor(n/2)} binomial(n-k+r, k), r>=0, will have g.f. 1/((1-x^2)^r*(1-x-x^2)) and for r>0, a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+r-1, r-1)*(1+(-1)^k)/2.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,5,-4,-10,6,10,-4,-5,1,1).

Cf. A000045, A054450, A054451, A052952, A099571.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x^2)^4*(1-x-x^2)) )); // G. C. Greubel, Jul 25 2022

Table[Fibonacci(n+5) +(-1)^n*(n^3+9*n^2+35*n+33)/96 -(n^3+21*n^2+155*n+417)/96, {n,0,40}] (* G. C. Greubel, Jul 25 2022 *)

[fibonacci(n+5) + (-1)^n*(n^3+9*n^2+35*n+33)/96 - (n^3+21*n^2+155*n + 417)/96 for n in (0..40)] # G. C. Greubel, Jul 25 2022

G.f.: 1/((1-x^2)^4*(1-x-x^2)). - corrected by R. J. Mathar, Feb 20 2011
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+3, 3)*((1+(-1)^k)/2).
a(n) = Fibonacci(n+5) + (-1)^n*(n^3 + 9*n^2 + 35*n + 33)/96 - (n^3 + 21*n^2 + 155*n + 417)/96. - G. C. Greubel, Jul 25 2022

A099571 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+3, k).

Original entry on oeis.org

1, 1, 5, 6, 17, 23, 50, 73, 138, 211, 370, 581, 979, 1560, 2575, 4135, 6755, 10890, 17700, 28590, 46356, 74946, 121380, 196326, 317797, 514123, 832025, 1346148, 2178293, 3524441, 5702870, 9227311, 14930334, 24157645, 39088150, 63245795

Offset: 0

Paul Barry, Oct 23 2004

Fourth column of triangle A054450.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-3,-6,3,4,-1,-1).

Cf. A000045, A052952, A054450, A054451, A099572.

[Fibonacci(n+4) +(-1)^n*(n^2+4*n+7)/16 -(n^2+12*n+39)/16: n in [0..40]]; // G. C. Greubel, Jul 25 2022

Table[Sum[Binomial[n-k+3,k],{k,0,Floor[n/2]}],{n,0,40}] (* or *) LinearRecurrence[{1,4,-3,-6,3,4,-1,-1},{1,1,5,6,17,23,50,73},40] (* Harvey P. Dale, Jun 04 2021 *)

[fibonacci(n+4) +(-1)^n*(n^2+4*n+7)/16 -(n^2+12*n+39)/16 for n in (0..40)] # G. C. Greubel, Jul 25 2022

G.f.: 1/((1-x^2)^3*(1-x-x^2)).
a(n) = a(n-1) + 4*a(n-2) - 3*a(n-3) - 6*a(n-4) + 3*a(n-5) + 4*a(n-6) - a(n-7) - a(n-8);
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2 +2, 2)*(1+(-1)^k)/2.
a(n) = Fibonacci(n+4) + (-1)^n*(n^2 + 4*n + 7)/16 - (n^2 + 12*n + 39)/16. - G. C. Greubel, Jul 25 2022

A079284 Diagonal sums of triangle A008949.

Original entry on oeis.org

1, 1, 3, 4, 9, 13, 26, 39, 73, 112, 201, 313, 546, 859, 1469, 2328, 3925, 6253, 10434, 16687, 27633, 44320, 72977, 117297, 192322, 309619, 506037, 815656, 1329885, 2145541, 3491810, 5637351, 9161929, 14799280, 24026745, 38826025, 62983842, 101809867, 165055853, 266865720

Offset: 0

Paul Barry, Feb 08 2003

a(2n) - a(2n-1) = Fibonacci(2n+1).

Diagonal sums of triangle A054450. - Paul Barry, Oct 23 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Donatella Merlini, Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
Index entries for linear recurrences with constant coefficients, signature (1,3,-2,-2).

Cf. A000045, A005672, A008949, A061667, A079284, A102517.

[Fibonacci(n+3)-2^Floor((n+1)/2): n in [0..40]]; // Vincenzo Librandi, Aug 05 2013

with (combinat):a[0]:=0:a[1]:=1:a[2]:=1:for n from 2 to 50 do a[n]:=fibonacci(n-1)+2*a[n-2] od: seq(a[n], n=1..31); # Zerinvary Lajos, Mar 17 2008

CoefficientList[Series[(1 - x^2) / ((1 - x - x^2) (1 - 2 x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 05 2013 *)
LinearRecurrence[{1,3,-2,-2},{1,1,3,4},40] (* Harvey P. Dale, Nov 30 2018 *)

a(n) = Sum_{j=0..floor(n/2)} Sum_{i=0..j} binomial(n-j, i).
a(n) = Fibonacci(n+3) - 2^floor((n+1)/2). - Vladeta Jovovic, Feb 12 2003
G.f.: (1-x^2)/((1-x-x^2)(1-2x^2)). - Paul Barry, Jan 13 2005

Showing 1-9 of 9 results.

cp's OEIS Frontend

A054451 Third column of triangle A054450 (partial row sums of unsigned Chebyshev triangle A049310).

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A054453 Triangle of partial row sums of triangle A054450(n,m), n >= m >= 0.

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A099573 Reverse of number triangle A054450.

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A052952 a(n) = Fibonacci(n+2) - (1-(-1)^n)/2.

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A029907 a(n+1) = a(n) + a(n-1) + Fibonacci(n), with a(0) = 0 and a(1) = 1.

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A054452 Partial sums of A027941(n-1) with a(-1) = 0.

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