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

A221719 a(n) = 3*2^n - Fibonacci(n+3) - 1.

Original entry on oeis.org

0, 2, 6, 15, 34, 74, 157, 328, 678, 1391, 2838, 5766, 11677, 23588, 47554, 95719, 192426, 386450, 775485, 1555152, 3117070, 6245087, 12507886, 25044430, 50135229, 100345484, 200812362, 401821143, 803960098, 1608434426, 3217700893, 6436748056, 12875674422, 25754873423, 51515449734, 103040126934, 206095184221, 412214526260, 824468140690
Offset: 0

Views

Author

N. J. A. Sloane, Jan 31 2013

Keywords

Comments

Number of 3-strand braids with n crossings, divided by 2.

References

  • Paul K. Stockmeyer, Personal communication, Jan 12 2013

Links

Crossrefs

Cf. A000045, A083329, A104004.

Programs

  • Magma
    A221719:= func< n | 3*2^n - Fibonacci(n+3) - 1 >; // G. C. Greubel, Jun 05 2025
  • Mathematica
    LinearRecurrence[{4,-4,-1,2},{0,2,6,15},40] (* Harvey P. Dale, Aug 25 2015 *)
A221719[n_]:= 3*2^n -Fibonacci[n+3] -1; (* G. C. Greubel, Jun 05 2025 *)
  • PARI
    concat(0, Vec(x*(x^2+2*x-2)/((x-1)*(2*x-1)*(x^2+x-1)) + O(x^100))) \\ Colin Barker, Jun 07 2015
  • SageMath
    def A221719(n): return 3*2**n - fibonacci(n+3) - 1 # G. C. Greubel, Jun 05 2025

Formula

From Colin Barker, Jun 07 2015: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4) for n>3.
G.f.: x*(2-2*x-x^2) / ((1-x)*(1-2*x)*(1-x-x^2)). (End)
a(n) = -1 + 3*2^n + ( (2-sqrt(5))*((1-sqrt(5))/2)^n - (2+sqrt(5))*((1+sqrt(5))/2)^n )/sqrt(5). - Colin Barker, Nov 03 2016
From G. C. Greubel, Jun 05 2025: (Start)
a(n) = A083329(n+1) - A000045(n+3).
a(n) = A104004(n) - 1.
E.g.f.: 3*exp(2*x) - exp(x) - (2/sqrt(5))*exp(x/2)*( 2*sinh(sqrt(5)*x/2) + sqrt(5)*cosh(sqrt(5)*x/2) ). (End)

A104487 a(n+3) = 6a(n+2) - 10a(n+1) + 3a(n); a(0) = 1, a(1) = 4, a(2) = 14.

Original entry on oeis.org

1, 4, 14, 47, 154, 496, 1577, 4964, 15502, 48103, 148490, 456416, 1397905, 4268740, 13002638, 39522143, 119912698, 363262672, 1099015481, 3321204260, 10026858766, 30246156439, 91171963754, 274650794432, 826923598369
Offset: 0

Views

Author

Creighton Dement, Apr 19 2005

Keywords

Comments

Binomial transform of A104004.
If another a(0)=0 is added in front, also the binomial transform of A027934.

Links

Programs

  • Magma
    [3^(n+1) - Fibonacci(2*n+3): n in [0..30]]; // Vincenzo Librandi, Apr 21 2011
  • Mathematica
    LinearRecurrence[{6,-10,3},{1,4,14},30] (* Harvey P. Dale, May 07 2017 *)

Formula

G.f.: (2*x-1)/((3*x-1)*(x^2-3*x+1)). Define c = (3+sqrt(5))/2 and d = (3-sqrt(5))/2. Then a(n) = 3^(n+1) - ((2*sqrt(5)/5)+1)*c^n + ((2*sqrt(5)/5)-1)*d^n = 3^(n+1) - Fibonacci(2n+3). - Ralf Stephan, May 20 2007

Extensions

Comment concerning the binomial transforms corrected by R. J. Mathar, Oct 26 2009
Showing 1-3 of 3 results.

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

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