cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A023607 a(n) = n * Fibonacci(n+1).

Original entry on oeis.org

0, 1, 4, 9, 20, 40, 78, 147, 272, 495, 890, 1584, 2796, 4901, 8540, 14805, 25552, 43928, 75258, 128535, 218920, 371931, 630454, 1066464, 1800600, 3034825, 5106868, 8580897, 14398412, 24129160, 40388070, 67527579, 112786496, 188195271
Offset: 0

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Author

Keywords

Comments

Convolution of Fibonacci numbers and Lucas numbers.
Central terms of the triangle in A119457 for n>0. - Reinhard Zumkeller, May 20 2006
d/dx(1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + ...) = (1 + 4x + 9x^2 + ...). - Gary W. Adamson, Jun 27 2009
For n > 0: sums of rows of the triangle in A108035. - Reinhard Zumkeller, Oct 08 2012

Crossrefs

First differences of A094584.
Second column of triangle A016095.

Programs

  • Haskell
    a023607 n = a023607_list !! n
    a023607_list = zipWith (*) [0..] $ tail a000045_list
    -- Reinhard Zumkeller, Oct 08 2012
    
  • Maple
    A023607 := proc(n)
        n*combinat[fibonacci](n+1) ;
    end proc:
    seq(A023607(n),n=0..10) ; # R. J. Mathar, Jul 15 2017
  • Mathematica
    Times@@@Thread[{Range[0, 50], Fibonacci[Range[51]]}]  (* Harvey P. Dale, Mar 08 2011 *)
    Table[n*Fibonacci[n + 1], {n, 0, 50}]
  • PARI
    a(n)=n*fibonacci(n+1) \\ Charles R Greathouse IV, Sep 24 2015

Formula

O.g.f.: x(2x+1)/(1-x-x^2)^2. - Len Smiley, Dec 11 2001
a(n) = n*Sum_{k=0..n} binomial(k,n-k). - Paul Barry, Sep 25 2004
a(n) = A215082(2n-2) + A215082(2n-1). - Philippe Deléham, Aug 03 2012
a(n) = Sum_{i=1..n} A000045(i)*A000032(n-i+1). - Vladimir Kruchinin, Nov 08 2013

Extensions

Simpler description from Samuel Lachterman (slachterman(AT)fuse.net), Sep 19 2003
Name improved by T. D. Noe, Mar 08 2011

A094585 Triangle T of all positive differences of distinct Fibonacci numbers; also, triangle of all sums of consecutive distinct Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 5, 8, 10, 11, 8, 13, 16, 18, 19, 13, 21, 26, 29, 31, 32, 21, 34, 42, 47, 50, 52, 53, 34, 55, 68, 76, 81, 84, 86, 87, 55, 89, 110, 123, 131, 136, 139, 141, 142, 89, 144, 178, 199, 212, 220, 225, 228, 230, 231, 144, 233, 288, 322, 343, 356, 364, 369, 372, 374, 375
Offset: 1

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Author

Clark Kimberling, May 13 2004

Keywords

Comments

Row sums = (1,5,14,34,74,...) = A094584. Alternating row sums = (1,1,4,4,12,12,...) given by F(m+1)-1 if m is even and F(m+2)-1 if m is odd. Central numbers = (1,5,16,47,...) = A094586.
Let p(n,x) = Sum_{k=0..n} F(k+1)*x^(n-k) and q(n,x) = x * q(n-1,x)+1, with q(0,x)=1. Then A094585 is the fission of sequence (p(n,x)) by sequence (q(n,x)); see A193842 for the definition of fission. A094585 is the mirror of A193999. - Clark Kimberling, Aug 11 2011

Examples

			Rows 1 to 5:
  1;
  2,  3;
  3,  5,  6;
  5,  8, 10, 11;
  8, 13, 16, 18, 19;
T(5,4) = F(8) - F(4) = 21 - 3 = 18;
T(5,4) = F(6) + F(5) + F(4) + F(3) = 8 + 5 + 3 + 2 = 18.
		

Crossrefs

Programs

  • GAP
    Flat(List([1..11],n->List([1..n],k->Fibonacci(n+3)-Fibonacci(n-k+3)))); # Muniru A Asiru, Apr 28 2019
  • Mathematica
    (* See A193999. *)
    Table[Fibonacci[n+3]-Fibonacci[n+3-k],{n,1,20}, {k,1,n}]//TableForm (* Rigoberto Florez, Oct 03 2019 *)

Formula

T(n, k) = F(n+3) - F(n+3-k) = F(n+1) + F(n) + ... + F(n+2-k), for k=1..n; n >= 1.
G.f.: x*y*(x*y+x+1)/((1-y*x)*(x^2+x-1)*(x^2*y^2+x*y-1)). - Vladimir Kruchinin, Jun 20 2025

A178525 The sum of the costs of all nodes in the Fibonacci tree of order n.

Original entry on oeis.org

0, 0, 3, 8, 22, 49, 104, 208, 403, 760, 1406, 2561, 4608, 8208, 14499, 25432, 44342, 76913, 132808, 228416, 391475, 668840, 1139518, 1936513, 3283392, 5555424, 9381699, 15815528, 26618518, 44733745, 75073256, 125827696, 210642643
Offset: 0

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Author

Emeric Deutsch, Jun 15 2010

Keywords

Comments

A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node. In a Fibonacci tree the cost of a left (right) edge is defined to be 1 (2). The cost of a node in a Fibonacci tree is defined to be the sum of the costs of the edges that form the path from the root to this node.
A178525 is the 1-sequence of reduction of the odd number sequence (2n-1) by x^2 -> x+1; as such it is related to 0-sequence of this reduction, A192304. See A192232 for definition of "k-sequence of reduction of [sequence] by [substitution]". - Clark Kimberling, Jun 27 2011

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Crossrefs

Programs

  • GAP
    List([0..40], n -> 3 +(2*n-3)*Fibonacci(n-1) +(2*n-5)*Fibonacci(n)); # G. C. Greubel, Jan 30 2019
  • Magma
    [3 +(2*n-3)*Fibonacci(n-1) +(2*n-5)*Fibonacci(n): n in [0..40]]; // G. C. Greubel, Jan 30 2019
    
  • Maple
    with(combinat): seq(3+(2*n-3)*fibonacci(n-1)+(2*n-5)*fibonacci(n), n = 0 .. 32);
  • Mathematica
    Table[3 +(2*n-3)*Fibonacci[n-1] +(2*n-5)*Fibonacci[n], {n,0,40}] (* G. C. Greubel, Jan 30 2019 *)
  • PARI
    a(n) = 3+(2*n-3)*fibonacci(n-1) + (2*n-5)*fibonacci(n); \\ Michel Marcus, Jan 21 2019
    
  • Sage
    [3 +(2*n-3)*fibonacci(n-1) +(2*n-5)*fibonacci(n) for n in range(40)] # G. C. Greubel, Jan 30 2019
    

Formula

a(n) = 3 + (2*n-3)*F(n-1) + (2*n-5)*F(n), where F(k)=A000045(k) are the Fibonacci numbers.
a(n) = a(n-1) + a(n-2) + 2*F(n+1) + 2*F(n-1) - 3 (n>=2), F(0)=0, F(1)=0.
G.f.: z^2*(3-z+z^2)/((1-z)*(1-z-z^2)^2).

A104796 Triangle read by rows: T(n,k) = (n+1-k)*Fibonacci(n+2-k), for n>=1, 1<=k<=n.

Original entry on oeis.org

1, 4, 1, 9, 4, 1, 20, 9, 4, 1, 40, 20, 9, 4, 1, 78, 40, 20, 9, 4, 1, 147, 78, 40, 20, 9, 4, 1, 272, 147, 78, 40, 20, 9, 4, 1, 495, 272, 147, 78, 40, 20, 9, 4, 1, 890, 495, 272, 147, 78, 40, 20, 9, 4, 1, 1584, 890, 495, 272, 147, 78, 40, 20, 9, 4, 1, 2796, 1584, 890, 495, 272
Offset: 1

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Author

Gary W. Adamson, Mar 26 2005

Keywords

Comments

The first column is A023607 (without the leading zero).

Examples

			Rows 1,2,3,4,5,6 and columns 1,2,3,4,5,6 of the triangle are:
1;
4, 1;
9, 4, 1;
20, 9, 4, 1;
40, 20, 9, 4, 1;
78, 40, 20, 9, 4, 1;
...
Row 3 for example is 3*F(4), 2*F(3), 1*F(2) = 3*3, 2*2, 1*1 = 9, 4, 1.
Row 4 is 4*F(5), 3*F(4), 2*F(3), 1*F(2) = 4*5, 3*3, 2*2, 1*1 = 20, 9, 4, 1.
Reading the rows backwards gives an initial segment of the terms of A023607 (but without the initial zero).
		

Crossrefs

Row sums are in A094584.

Programs

  • Mathematica
    Table[(n+1-k)Fibonacci[n+2-k],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 24 2020 *)
    Module[{nn=20,c},c=LinearRecurrence[{2,1,-2,-1},{1,4,9,20},nn];Table[ Reverse[ Take[c,n]],{n,nn}]]//Flatten (* Harvey P. Dale, Sep 25 2020 *)

Extensions

Edited by Ralf Stephan, Apr 05 2009
Entry revised by N. J. A. Sloane, Sep 23 2020

A120940 Alternating sum of the Fibonacci numbers multiplied by their (combinatorial) indices.

Original entry on oeis.org

0, 1, 3, 6, 14, 26, 52, 95, 177, 318, 572, 1012, 1784, 3117, 5423, 9382, 16170, 27758, 47500, 81035, 137885, 234046, 396408, 670056, 1130544, 1904281, 3202587, 5378310, 9020102, 15109058, 25279012, 42248567, 70537929, 117657342, 196076468, 326485852
Offset: 0

Views

Author

Marcello M. Herreshoff (m(AT)marcello.gotdns.com), Jul 18 2006

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(2*z^2 + z)/((z + 1)*(z^2 + z - 1)^2), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 08 2011 *)
    LinearRecurrence[{1,3,-1,-3,-1},{0,1,3,6,14},40] (* Harvey P. Dale, Apr 21 2018 *)
  • PARI
    concat(0, Vec(x*(1 + 2*x) / ((1 + x)*(1 - x - x^2)^2) + O(x^40))) \\ Colin Barker, Apr 03 2019

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*k*f(k) also, when n >= 3, a(n) = nf(n-1) + f(n-3) + (-1)^n where f(n) = F(n+1).
a(n) = (-1)^n+A000045(n)-A001629(n+2)+3*A001629(n+1). - R. J. Mathar, Jul 11 2011 [Corrected by Alan Michael Gómez Calderón, Jul 23 2025]
From Colin Barker, Apr 03 2019: (Start)
G.f.: x*(1 + 2*x) / ((1 + x)*(1 - x - x^2)^2).
a(n) = a(n-1) + 3*a(n-2) - a(n-3) - 3*a(n-4) - a(n-5) for n>4.
(End)
Showing 1-5 of 5 results.