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.

Previous Showing 11-13 of 13 results.

A138134 a(n) = Sum_{i=0..n} Fibonacci(5*i).

Original entry on oeis.org

0, 5, 60, 670, 7435, 82460, 914500, 10141965, 112476120, 1247379290, 13833648315, 153417510760, 1701426266680, 18869106444245, 209261597153380, 2320746675131430, 25737475023599115, 285432971934721700, 3165500166305537820
Offset: 0

Views

Author

Gary Detlefs, Dec 07 2010

Keywords

Comments

Partial sums of A102312.
Other sequences in the OEIS related to the sum of Fibonacci(k*n) (although not defined as such) are:
k = 1: A000071 = Fibonacci(n) - 1 (delete leading 0);
k = 2: A027941 = Fibonacci(2n+1) - 1;
k = 3: A099919 = (Fibonacci(3n+2) - 1)/2;
k = 4: A058038 = Fibonacci(2n)*Fibonacci(2n+2);
k = 6: A053606 = (Fibonacci(6n+3) - 2)/4.
These sequences appear to be second order linear inhomogeneous sequences of the form: a(0) = 0, a(1) = Fibonacci(k), a(n) = L(k)*a(n-1) + (-1)^(k+1)*a(n-2) + Fibonacci(k), where L(n) = A000032(n), n > 1.
The Koshy reference gives the closed form:
Sum_{i=0..n} Fibonacci(k*i) = (Fibonacci(n*k+k) - (-1)^k*Fibonacci(n*k) - Fibonacci(k))/(L(k) - (-1)^k - 1).

References

  • Thomas Koshy; Fibonacci and Lucas numbers with applications, Wiley,2001, p. 86.

Links

Crossrefs

Cf. A000071, A027941, A099919, A058038, A102312, A053606.
Cf. also A000032, A000045.

Programs

  • Maple
    with(combinat):fs5:=n-> sum(fibonacci(5*k),k=0..n):
seq(fs5(n),n=0..18)
  • PARI
    a(n)=(fibonacci(5*n+5)+fibonacci(5*n)-5)/11 \\ Charles R Greathouse IV, Jun 11 2015

Formula

G.f.: 5*x/((x - 1)*(x^2 + 11*x - 1)). - R. J. Mathar, Dec 09 2010
a(n) = 11*a(n) + a(n-1) + 5, n > 1.
a(n) = 12*a(n-1) - 10*a(n-2) - a(n-3), n > 2.
a(n) = 1/11*(Fibonacci(5*n+5) + Fibonacci(5n) - 5).

A081079 a(n) = Lucas(4*n+2) - 3 = 5*Fibonacci(2*n)*Fibonacci(2*n+2).

Original entry on oeis.org

0, 15, 120, 840, 5775, 39600, 271440, 1860495, 12752040, 87403800, 599074575, 4106118240, 28143753120, 192900153615, 1322157322200, 9062201101800, 62113250390415, 425730551631120, 2918000611027440, 20000273725560975
Offset: 0

Views

Author

R. K. Guy, Mar 04 2003

Keywords

References

  • Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Links

Crossrefs

Cf. A000032 (Lucas numbers), A000045 (Fibonacci numbers), A058038, A092521, A156088.

Programs

  • Magma
    A081079:= func< n | Lucas(4*n+2) -3 >; // G. C. Greubel, Jun 12 2025
  • Maple
    luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,luc(4*n+2)-3) od: # James Sellers, Mar 05 2003
  • Mathematica
    LinearRecurrence[{8,-8,1}, {0,15,120}, 20] (* Jean-François Alcover, Nov 29 2023 *)
  • SageMath
    def A081079(n): return lucas_number2(4*n+2,1,-1) -3 # G. C. Greubel, Jun 12 2025

Formula

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
From R. J. Mathar, Sep 03 2010: (Start)
a(n) = 15*A092521(n) = 5*A058038(n).
G.f.: 15*x/((1-x)*(1-7*x+x^2)). (End)
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = 15*(-1)^n*A156088(n).
E.g.f.: exp(7*x/2)*( 3*cosh(3*sqrt(5)*x/2) - sqrt(5)*sinh(3*sqrt(5)*x/2) ) - 3*exp(x). (End)

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181
Offset: 1

Views

Author

Yuriy Sibirmovsky, Sep 12 2016

Keywords

Comments

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

Examples

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Links

Crossrefs

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Programs

  • Mathematica
    Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid
  • PARI
    T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Formula

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.
Previous Showing 11-13 of 13 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.