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 1-3 of 3 results.

A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

Original entry on oeis.org

0, 1, 3, 14, 91, 820, 9650, 140601, 2440317, 49109632, 1123595495, 28792920872, 816742025772, 25402428294801, 859492240650847, 31427791175659690, 1234928473553777403, 51893300561135516404, 2322083099525697299278
Offset: 0

Views

Author

Ross La Haye, Dec 14 2004

Keywords

Comments

In what follows a(i,j,k) denotes a three-dimensional array, the terms a(n) are defined as a(n,n,n) in that array. - Joerg Arndt, Jan 03 2021
Previous name was: Three-dimensional array: a(i,j,k) = expansion of x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)), read by a(n,n,n).
a(i,j,k) = the k-th value of the convolution of the Fibonacci numbers (A000045) with the powers of i = Sum_{m=0..k} a(i-1,j,m), both for i = j and i > 0; a(i,j,k) = a(i-1,j,k) + a(j,j,k-1), for i,k > 0; a(i,1,k) = Sum_{m=0..k} a(i-1,0,m), for i > 0. With F = Fibonacci and L = Lucas, then a(1,1,k) = F(k+2) - 1; a(2,1,k) = F(k+3) - 2; a(3,1,k) = L(k+2) - 3; a(4,1,k) = 4*F(k+1) + F(k) - 4; a(1,2,k) = 2^k - F(k+1); a(2,2,k) = 2^(k+1) - F(k+3); a(3,2,k) = 3(2^k - F(k+2)) + F(k); a(4,2,k) = 2^(k+2) - F(k+4) - F(k+2); a(1,3,k) = (3^k + L(k-1))/5, for k > 0; a(2,3,k) = (2 * 3^k - L(k)) /5, for k > 0; a(3,3,k) = (3^(k+1) - L(k+2))/5; a(4,3,k) = (4 * 3^k - L(k+2) - L(k+1))/5, etc..

Examples

			a(1,3,3) = 6 because a(1,3,0) = 0, a(1,3,1) = 1, a(1,3,2) = 2 and 4*2 - 2*1 - 3*0 = 6.

Links

Crossrefs

a(0, j, k) = A000045(k).
a(1, 2, k+1) - a(1, 2, k) = A099036(k).
a(3, 2, k+1) - a(3, 2, k) = A104004(k).
a(4, 2, k+1) - a(4, 2, k) = A027973(k).
a(1, 3, k+1) - a(1, 3, k) = A099159(k).
a(i, 0, k) = A109754(i, k).
a(i, i+1, 3) = A002522(i+1).
a(i, i+1, 4) = A071568(i+1).
a(2^i-2, 0, k+1) = A118654(i, k), for i > 0.
Sequences of the form a(n, 0, k): A000045(k+1) (n=1), A000032(k) (n=2), A000285(k-1) (n=3), A022095(k-1) (n=4), A022096(k-1) (n=5), A022097(k-1) (n=6), A022098(k-1) (n=7), A022099(k-1) (n=8), A022100(k-1) (n=9), A022101(k-1) (n=10), A022102(k-1) (n=11), A022103(k-1) (n=12), A022104(k-1) (n=13), A022105(k-1) (n=14), A022106(k-1) (n=15), A022107(k-1) (n=16), A022108(k-1) (n=17), A022109(k-1) (n=18), A022110(k-1) (n=19), A088209(k-2) (n=k-2), A007502(k) (n=k-1), A094588(k) (n=k).
Sequences of the form a(1, n, k): A000071(k+2) (n=1), A027934(k-1) (n=2), A098703(k) (n=3).
Sequences of the form a(2, n, k): A001911(k) (n=1), A008466(k+1) (n=2), A106517(k-1) (n=3).
Sequences of the form a(3, n, k): A027961(k) (n=1), A094688(k) (n=3).
Sequences of the form a(4, n, k): A053311(k-1) (n=1), A027974(k-1) (n=2).
Cf. A001622, A001891, A001924, A023537, A104161, A106517.

Programs

  • Magma
    A101220:= func< n | (&+[n^k*Fibonacci(n-k): k in [0..n]]) >;
[A101220(n): n in [0..30]]; // G. C. Greubel, Jun 01 2025
  • Mathematica
    Join[{0}, Table[Sum[Fibonacci[n-k]*n^k, {k, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 03 2021 *)
  • PARI
    a(n)=sum(k=0,n,fibonacci(n-k)*n^k) \\ Joerg Arndt, Jan 03 2021
  • SageMath
    def A101220(n): return sum(n^k*fibonacci(n-k) for k in range(n+1))
print([A101220(n) for n in range(31)]) # G. C. Greubel, Jun 01 2025

Formula

a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1; a(i, j, k) = ((j+1)*a(i, j, k-1)) - ((j-1)*a(i, j, k-2)) - (j*a(i, j, k-3)), for k > 2.
a(i, j, k) = Fibonacci(k) + i*a(j, j, k-1), for i, k > 0.
a(i, j, k) = (Phi^k - (-Phi)^-k + i(((j^k - Phi^k) / (j - Phi)) - ((j^k - (-Phi)^-k) / (j - (-Phi)^-1)))) / sqrt(5), where Phi denotes the golden mean/ratio (A001622).
i^k = a(i-1, i, k) + a(i-2, i, k+1).
A104161(k) = Sum_{m=0..k} a(k-m, 0, m).
a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1, a(i, j, 3) = i*(j+1) + 2; a(i, j, k) = (j+2)*a(i, j, k-1) - 2*j*a(i, j, k-2) - a(i, j, k-3) + j*a(i, j, k-4), for k > 3. a(i, j, 0) = 0, a(i, j, 1) = 1; a(i, j, k) = a(i, j, k-1) + a(i, j, k-2) + i * j^(k-2), for k > 1.
G.f.: x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)).
a(n, n, n) = Sum_{k=0..n} Fibonacci(n-k) * n^k. - Ross La Haye, Jan 14 2006
Sum_{m=0..k} binomial(k,m)*(i-1)^m = a(i-1,i,k) + a(i-2,i,k+1), for i > 1. - Ross La Haye, May 29 2006
From Ross La Haye, Jun 03 2006: (Start)
a(3, 3, k+1) - a(3, 3, k) = A106517(k).
a(1, 1, k) = A001924(k) - A001924(k-1), for k > 0.
a(2, 1, k) = A001891(k) - A001891(k-1), for k > 0.
a(3, 1, k) = A023537(k) - A023537(k-1), for k > 0.
Sum_{j=0..i+1} a(i-j+1, 0, j) - Sum_{j=0..i} a(i-j, 0, j) = A001595(i). (End)
a(i,j,k) = a(j,j,k) + (i-j)*a(j,j,k-1), for k > 0.
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jan 03 2021

Extensions

New name from Joerg Arndt, Jan 03 2021

A052964 Expansion of (1-x)/((1-2x)(1+x-x^2)).

Original entry on oeis.org

1, 0, 3, 1, 10, 7, 35, 36, 127, 165, 474, 715, 1807, 3004, 6995, 12393, 27370, 50559, 107883, 204820, 427351, 826045, 1698458, 3321891, 6765175, 13333932, 26985675, 53457121, 107746282, 214146295, 430470899, 857417220, 1720537327
Offset: 0

Views

Author

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

Keywords

Comments

Number of walks of length n+1 between two adjacent vertices in the cycle graph C_5. Example: a(2)=3 because in the cycle ABCDE we have three walks of length 3 between A and B: ABAB, ABCB and AEAB. - Emeric Deutsch, Apr 01 2004
In general a(n,m)=2^n/m*Sum(k,0,m-1,Cos(2Pi*k/m)^(n+1)) gives number of walks of length n between two adjacent vertices in the cycle graph C_m. Here we have m=5. - Herbert Kociemba, May 31 2004
Counts walks of length n at the vertex of degree 3 of the graph with adjacency matrix A=[0,1,1,1;1,0,0,0;1,0,0,0;1,0,0,1]. Binomial transform is (L(n-2)+2*3^n)/5, or A099159. - Paul Barry, Oct 01 2004
Also, the cogrowth sequence for the 10-element dihedral group D5; that is a(n) is the number of words of length n+2 that reduce to the identity using the presentation . - Sean A. Irvine, Nov 04 2024

Links

Programs

  • Maple
    spec := [S,{S=Sequence(Prod(Union(Prod(Sequence(Z),Z),Z,Z),Z))},unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
  • Mathematica
    CoefficientList[Series[(1-x)/((1-2x)(1+x-x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{1,3,-2},{1,0,3},40] (* Harvey P. Dale, Jun 03 2019 *)

Formula

G.f.: -(-1+x)/(1-x-3*x^2+2*x^3)
Recurrence: {a(1)=0, a(0)=1, a(2)=3, 2*a(n)-3*a(n+1)-a(n+2)+a(n+3)=0}
Sum(-1/25*(-1-11*_alpha+6*_alpha^2)*_alpha^(-1-n), _alpha=RootOf(1-_Z-3*_Z^2+2*_Z^3))
a(n-1)=2^n/5*Sum(k, 0, 4, Cos(2Pi*k/5)^(n+1)), n>=1 - Herbert Kociemba, May 31 2004
a(n)=((sqrt(5)-1)/2)^n(3/10-sqrt(5)/10)+((-sqrt(5)-1)/2)^n(3/10+sqrt(5)/10)+2^(n+1)/5 - Paul Barry, Oct 01 2004
a(n) = (2^(n+1) + A000032(n+2)*(-1)^n)/5 - Ross La Haye, May 31 2006
a(n) = |A084179(n+1)|-|A084179(n)|. - R. J. Mathar, Feb 27 2019

A098703 a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.

Original entry on oeis.org

0, 1, 2, 6, 17, 50, 148, 441, 1318, 3946, 11825, 35454, 106328, 318929, 956698, 2869950, 8609617, 25828474, 77484812, 232453449, 697358750, 2092073666, 6276216817, 18828643686, 56485920112, 169457742625, 508373199218, 1525119551286
Offset: 0

Views

Author

Ross La Haye, Oct 27 2004

Keywords

Comments

Sums of antidiagonals of A090888.
Partial sums of A099159.
Form an array with m(0,n) = A000045(n), the Fibonacci numbers, and m(i,j) = Sum_{kJ. M. Bergot, May 27 2013

Examples

			a(2) = 2 because 3^2 = 9, Luc(1) = 1 and (9 + 1) / 5 = 2.

Links

Crossrefs

Cf. A000032, A000045, A000244, A000285, A001622, A001654, A022383, A052964, A084178, A090888, A094688, A099159.

Programs

  • Magma
    I:=[0,1,2]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-3*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 18 2018
  • Mathematica
    f[n_] := (3^n + Fibonacci[n] + Fibonacci[n - 2])/5; Table[ f[n], {n, 0, 27}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{4, -2, -3}, {0, 1, 2}, 30] (* Jean-François Alcover, Feb 17 2018 *)
  • SageMath
    def A098703(n): return (3**n + lucas_number2(n-1,1,-1))//5
print([A098703(n) for n in range(21)]) # G. C. Greubel, Jun 02 2025

Formula

a(n) = (((1 + sqrt(5))^n - (1 - sqrt(5))^n) / (2^n*sqrt(5))) + ((3^n - (((1 + sqrt(5)) / 2)^(n+1) + ((1 - sqrt(5)) / 2)^(n+1))) / 5).
a(n) = (3^n + (((1 + sqrt(5)) / 2)^(n-1) + ((1 - sqrt(5)) / 2)^(n-1))) / 5.
a(n) = (3^n + A000032(n-1))/5 = A000045(n) + (3^n - A000032(n+1))/5.
a(n) = (3^n + A000045(n) + A000045(n-2))/5.
a(n) = (3^n + 4*A000045(n) - A000045(n+2))/5.
a(n) = Sum_{k=0...n-1} (A000045(k)*3^(n-k-1) - A000045(k-2)*2^(n-k-1)).
a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
a(n) = A000045(n) + A094688(n-1).
a(n) = 3^1 * a(n-1) - A000045(n-3), for n > 2.
a(n) = 3^2 * a(n-2) - A000285(n-4), for n > 3.
a(n) = 3^3 * a(n-3) - A022383(n-5), for n > 4.
Limit_{n -> oo} a(n) / a(n-1) = 3.
From Ross La Haye, Dec 21 2004: (Start)
a(n) = A101220(1,3,n).
Binomial transform of unsigned A084178.
Binomial transform of signed A084178 gives the Fibonacci oblongs (A001654). (End)
G.f.: x*(1-2*x)/((1-3*x)*(1-x-x^2)). - Ross La Haye, Aug 09 2005
a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3^(n-2) for n > 1. - Ross La Haye, Aug 20 2005
Binomial transform of A052964 beginning {0,1,0,3,1,10,...}. - Ross La Haye, May 31 2006

Extensions

More terms from Robert G. Wilson v, Nov 04 2004
More terms from Ross La Haye, Dec 21 2004
Showing 1-3 of 3 results.

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