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.

A014445 Even Fibonacci numbers; or, Fibonacci(3*n).

Original entry on oeis.org

0, 2, 8, 34, 144, 610, 2584, 10946, 46368, 196418, 832040, 3524578, 14930352, 63245986, 267914296, 1134903170, 4807526976, 20365011074, 86267571272, 365435296162, 1548008755920, 6557470319842, 27777890035288, 117669030460994, 498454011879264, 2111485077978050
Offset: 0

Views

Author

Mohammad K. Azarian

Keywords

Comments

a(n) = 3^n*b(n;2/3) = -b(n;-2), but we have 3^n*a(n;2/3) = F(3n+1) = A033887 and a(n;-2) = F(3n-1) = A015448, where a(n;d) and b(n;d), n=0,1,...,d, denote the so-called delta-Fibonacci numbers (the argument "d" of a(n;d) and b(n;d) is abbreviation of the symbol "delta") defined by the following equivalent relations: (1 + d*((sqrt(5) - 1)/2))^n = a(n;d) + b(n;d)*((sqrt(5) - 1)/2) equiv. a(0;d)=1, b(0;d)=0, a(n+1;d) = a(n;d) + d*b(n;d), b(n+1;d) = d*a(n;d) + (1-d)b(n;d) equiv. a(0;d)=a(1;d)=1, b(0;1)=0, b(1;d)=d, and x(n+2;d) + (d-2)*x(n+1;d) + (1-d-d^2)*x(n;d) = 0 for every n=0,1,...,d, and x=a,b equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k-1)*(-d)^k, and b(n;d) = Sum_{k=0..n} C(n,k)*(-1)^(k-1)*F(k)*d^k equiv. a(n;d) = Sum_{k=0..n} C(n,k)*F(k+1)*(1-d)^(n-k)*d^k, and b(n;d) = Sum_{k=1..n} C(n;k)*F(k)*(1-d)^(n-k)*d^k. The sequences a(n;d) and b(n;d) for special values d are connected with many known sequences: A000045, A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A163073 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012
For any odd k, Fibonacci(k*n) = sqrt(Fibonacci((k-1)*n) * Fibonacci((k+1)*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012
The ratio of consecutive terms approaches the continued fraction 4 + 1/(4 + 1/(4 +...)) = A098317. - Hal M. Switkay, Jul 05 2020

Examples

			G.f. = 2*x + 8*x^2 + 34*x^3 + 144*x^4 + 610*x^5 + 2584*x^6 + 10946*x^7 + ...

References

  • Arthur T. Benjamin and Jennifer J. Quinn,, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 232.

Links

Crossrefs

Cf. A000032, A000045, A001076, A049310, A111125.
Cf. A001519, A001906, A015448, A020699, A033887, A033889, A074872, A081567, A081568, A081569, A081574, A081575, A098317, A163073.
First differences of A099919.
Third column of array A102310.

Programs

Formula

a(n) = Sum_{k=0..n} binomial(n, k)*F(k)*2^k. - Benoit Cloitre, Oct 25 2003
From Lekraj Beedassy, Jun 11 2004: (Start)
a(n) = 4*a(n-1) + a(n-2), with a(-1) = 2, a(0) = 0.
a(n) = 2*A001076(n).
a(n) = (F(n+1))^3 + (F(n))^3 - (F(n-1))^3. (End)
a(n) = Sum_{k=0..floor((n-1)/2)} C(n, 2*k+1)*5^k*2^(n-2*k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004
a(n) = Sum_{k=0..n} F(n+k)*binomial(n, k). - Benoit Cloitre, May 15 2005
O.g.f.: 2*x/(1 - 4*x - x^2). - R. J. Mathar, Mar 06 2008
a(n) = second binomial transform of (2,4,10,20,50,100,250). This is 2* (1,2,5,10,25,50,125) or 5^n (offset 0): *2 for the odd numbers or *4 for the even. The sequences are interpolated. Also a(n) = 2*((2+sqrt(5))^n - (2-sqrt(5))^n)/sqrt(20). - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
a(n) = 3*F(n-1)*F(n)*F(n+1) + 2*F(n)^3, F(n)=A000045(n). - Gary Detlefs, Dec 23 2010
a(n) = (-1)^n*3*F(n) + 5*F(n)^3, n >= 0. See the D. Jennings formula given in a comment on A111125, where also the reference is given. - Wolfdieter Lang, Aug 31 2012
With L(n) a Lucas number, F(3*n) = F(n)*(L(2*n) + (-1)^n) = (L(3*n+1) + L(3*n-1))/5 starting at n=1. - J. M. Bergot, Oct 25 2012
a(n) = sqrt(Fibonacci(2*n)*Fibonacci(4*n) + Fibonacci(n)^2). - Gary Detlefs, Dec 28 2012
For n > 0, a(n) = 5*F(n-1)*F(n)*F(n+1) - 2*F(n)*(-1)^n. - J. M. Bergot, Dec 10 2015
a(n) = -(-1)^n * a(-n) for all n in Z. - Michael Somos, Nov 15 2018
a(n) = (5*Fibonacci(n)^3 + Fibonacci(n)*Lucas(n)^2)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022
a(n) = 2*i^(n-1)*S(n-1,-4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(-1, x) = 0. From the simplified trisection formula. - Gary Detlefs and Wolfdieter Lang, Mar 04 2023
E.g.f.: 2*exp(2*x)*sinh(sqrt(5)*x)/sqrt(5). - Stefano Spezia, Jun 03 2024
a(n) = 2*F(n) + 3*Sum_{k=0..n-1} F(3*k)*F(n-k). - Yomna Bakr and Greg Dresden, Jun 10 2024

A094440 Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 8, 6, 4, 5, 15, 20, 10, 5, 8, 30, 45, 40, 15, 6, 13, 56, 105, 105, 70, 21, 7, 21, 104, 224, 280, 210, 112, 28, 8, 34, 189, 468, 672, 630, 378, 168, 36, 9, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 89, 605, 1870, 3465, 4290, 3696, 2310, 990, 330, 55, 11
Offset: 1

Views

Author

Clark Kimberling, May 03 2004

Keywords

Comments

Row sums yield the even-subscripted Fibonacci numbers (A001906).
Row n shows the coefficients of the numerator of the n-th derivative of c(n)/(x^2+x-1), where c(n) = ((-1)^(n + 1))/n!; see the Mathematica program. - Clark Kimberling, Oct 22 2019

Examples

			Triangle starts:
   1;
   1,  2;
   2,  3,   3;
   3,  8,   6,   4;
   5, 15,  20,  10,  5;
   8, 30,  45,  40, 15,  6;
  13, 56, 105, 105, 70, 21, 7;
  ...
T(4,3) = F(2)*C(4,2) = 1*6 = 6.

Links

Crossrefs

Cf. A000032, A000045, A001906, A030191, A081574, A094435, A132148.
Cf. A094435, A094436, A094437, A094438, A094439, A094441, A094442, A094443, A094444.
Cf. A367208, A367209, A367210, A367211.

Programs

  • GAP
    Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k-1)* Fibonacci(n-k+1) ))); # G. C. Greubel, Oct 30 2019
  • Magma
    /* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 15 2017
  • Maple
    with(combinat): T:=(n,k)->binomial(n,k-1)*fibonacci(n+1-k): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # Emeric Deutsch
  • Mathematica
    Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 14 2016 *)
(* Next program outputs polynomials having coefficients T(n,k) *)
g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[1/(1 - x - x^2), {x, n}]]]
Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* Clark Kimberling, Oct 22 2019 *)
  • PARI
    T(n,k) = binomial(n,k-1)*fibonacci(n-k+1);
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019
  • Sage
    [[binomial(n,k-1)*fibonacci(n-k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Oct 30 2019

Formula

From Peter Bala, Aug 17 2007: (Start)
With an offset of 0, the row polynomials F(n,x) = Sum_{k = 0..n} C(n,k)* Fibonacci(n-k)*x^k satisfy F(n,x)*L(n,x) = F(2*n,x), where L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k.
Other identities and formulas include:
F(n+1,x)^2 - F(n,x)*F(n+2,x) = (x^2 + x - 1)^n;
Sum_{k = 0..n} C(n,k)*F(n-k,x)*L(k,x) = (2^n)*F(n,x);
F(n,2*x) = Sum_{k = 0..n} C(n,k)*F(n-k,x)*x^k;
F(n,3*x) = Sum_{k = 0..n} C(n,k)*F(n-k,2*x)*x^k, etc.
The sequence {F(n,r)}n>=1 gives the r-th binomial transform of the Fibonacci numbers: r = 1 gives A001906, r = 2 gives A030191, r = 3 gives A099453, r = 4 gives A081574, r = 5 gives A081575.
F(n,1/phi) = (-1)^(n-1)*F(n,-phi) = sqrt(5)^(n-1) for n >= 1, where phi = (1 + sqrt(5))/2.
The polynomials F(n,-x) satisfy a Riemann hypothesis: the zeros of F(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
G.f.: t/(1 - (2*x + 1)*t + (x^2 + x - 1)*t^2) = t + (1 + 2*x)*t^2 + (2 + 3*x + 3*x^2)*t^3 + (3 + 8*x + 6*x^2 + 4*x^3)*t^4 + ... . (End)
From Peter Bala, Jun 29 2016: (Start)
Working with an offset of 0, the n-th row polynomial F(n,x) = 1/sqrt(5)*( (x + phi)^n - (x - 1/phi)^n ), where phi = (1 + sqrt(5))/2.
d/dx(F(n,x)) = n*F(n-1,x).
F(-n,x) = -F(n,x)/(x^2 + x - 1)^n.
F(n,x - 1) = (-1)^(n-1)*F(n,-x).
F(n,x) is a divisibility sequence of polynomials, that is, if n divides m then F(n,x) divides F(m,x) in the polynomial ring Z[x]. (End)
From G. C. Greubel, Oct 30 2019: (Start)
Sum_{k = 1..n} T(n,k) = Fibonacci(2*n).
Sum_{k = 1..n} (-1)^k * T(n,k) = (-1)^n * Fibonacci(n). (End)
From Clark Kimberling, Oct 30 2019: (Start)
F(n,x) is a strong divisibility sequence of polynomials in Z[x]; that is,
gcd(F(x,h),F(x,k)) = F(x,gcd(h,k)) for h,k >= 1. Thus, if x is an integer, then F(n,x) is a strong divisibility sequence of integers; e.g., for x=3, we have A099453. (End)
Let p(n) denote the polynomial F(x,n). Then p(n) = k(b^n - c^n), where k = -1/sqrt(5), b = (1/2)(2x + 1 - sqrt(5)), c = (1/2)(2x + 1 + sqrt(5)), and for n >=3, p(n) = u*p(n - 1) + v*p(n - 2), where u = 1 + 2 x, v = 1 - x - x^2. - Clark Kimberling, Nov 11 2023

Extensions

Error in expansion of generating function corrected by Peter Bala, Sep 24 2008

A289785 p-INVERT of the (5^n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 7, 48, 325, 2183, 14588, 97161, 645719, 4285240, 28411789, 188257719, 1246893028, 8256349457, 54659946215, 361825274112, 2394939574997, 15851402375719, 104912178457996, 694343294142105, 4595323060281271, 30412598132972936, 201274210714545437
Offset: 0

Views

Author

Clark Kimberling, Aug 10 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).
See A289780 for a guide to related sequences.

Links

Crossrefs

Cf. A000351, A289780.

Programs

  • Mathematica
    z = 60; s = x/(1 - 5*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000351 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289785 *)
  • PARI
    Vec(x*(1 - 4*x) / (1 - 11*x + 29*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

Formula

G.f.: (1 - 4 x)/(1 - 11 x + 29 x^2).
a(n) = 11*a(n-1) - 29*a(n-2).
a(n) = (2^(-n-1)*((11-sqrt(5))^(n+1)*(-7+2*sqrt(5)) + (11+sqrt(5))^(n+1)*(7+2*sqrt(5)))) / (29*sqrt(5)). - Colin Barker, Aug 11 2017
a(n) = A081575(n+1)-4*A081575(n). - R. J. Mathar, Jul 08 2022

A081576 Square array of binomial transforms of Fibonacci numbers, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 5, 8, 3, 0, 1, 7, 20, 21, 5, 0, 1, 9, 38, 75, 55, 8, 0, 1, 11, 62, 189, 275, 144, 13, 0, 1, 13, 92, 387, 905, 1000, 377, 21, 0, 1, 15, 128, 693, 2305, 4256, 3625, 987, 34, 0, 1, 17, 170, 1131, 4955, 13392, 19837, 13125, 2584, 55
Offset: 0

Views

Author

Paul Barry, Mar 22 2003

Keywords

Comments

Array rows are solutions of the recurrence a(n) = (2*k+1)*a(n-1) - A028387(k-1)*a(n-2) where a(0) = 0 and a(1) = 1.

Examples

			Square array begins as:
  0, 1,  1,   2,    3,    5,     8, ... A000045;
  0, 1,  3,   8,   21,   55,   144, ... A001906;
  0, 1,  5,  20,   75,  275,  1000, ... A030191;
  0, 1,  7,  38,  189,  905,  4256, ... A099453;
  0, 1,  9,  62,  387, 2305, 13392, ... A081574;
  0, 1, 11,  92,  693, 4955, 34408, ... A081575;
  0, 1, 13, 128, 1131, 9455, 76544, ...
The antidiagonal triangle begins as:
  0;
  0, 1;
  0, 1,  1;
  0, 1,  3,  2;
  0, 1,  5,  8,   3;
  0, 1,  7, 20,  21,   5;
  0, 1,  9, 38,  75,  55,   8;
  0, 1, 11, 62, 189, 275, 144, 13;

Links

Crossrefs

Array row n: A000045 (n=0), A001906 (n=1), A030191 (n=2), A099453 (n=3), A081574 (n=4), A081575 (n=5).
Array columns k: A005408 (k=3), A077588 (k=4).
Cf. A028387, A081573, A081574, A081575.

Programs

  • Magma
    A081576:= func< n,k | (&+[Binomial(k,j)*Fibonacci(j)*(n-k)^(k-j): j in [0..k]]) >;
[A081576(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021
  • Mathematica
    T[n_, k_]:= If[n==0, Fibonacci[k], Sum[Binomial[k, j]*Fibonacci[j]*n^(k-j), {j, 0, k}]]; Table[T[n-k, k], {n,0,12}, {k,0,n}] //Flatten (* G. C. Greubel, May 26 2021 *)
  • Sage
    def A081576(n,k): return sum( binomial(k,j)*fibonacci(j)*(n-k)^(k-j) for j in (0..k) )
flatten([[A081576(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Formula

Rows are successive binomial transforms of F(n).
T(n, k) = ( ( (2*n + 1 + sqrt(5))/2 )^k - ( (2*n + 1 - sqrt(5))/2 )^k )/sqrt(5).
From G. C. Greubel, May 26 2021: (Start)
T(n, k) = Sum_{j=0..k} binomial(k,j)*Fibonacci(j)*n^(k-j) with T(0, k) = Fibonacci(k) (square array).
T(n, k) = Sum_{j=0..k} binomial(k,j)*Fibonacci(j)*(n-k)^(k-j) (antidiagonal triangle). (End)

A171731 Triangle T : T(n,k)= binomial(n,k)*Fibonacci(n-k)= A007318(n,k)*A000045(n-k).

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 2, 3, 3, 0, 3, 8, 6, 4, 0, 5, 15, 20, 10, 5, 0, 8, 30, 45, 40, 15, 6, 0, 13, 56, 105, 105, 70, 21, 7, 0, 21, 104, 224, 280, 210, 112, 28, 8, 0, 34, 189, 468, 672, 630, 378, 168, 36, 9, 0, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 16 2009

Keywords

Comments

Diagonal sums : A112576.
Essentially the same as A094440. - Peter Bala, Jan 06 2015

Examples

			Triangle begins :
0 ;
1,0 ;
1,2,0 ;
2,3,3,0 ;
3,8,6,4,0 ;
5,15,20,10,5,0 ;
...

Crossrefs

Cf. A000045, A007318, A094440.

Programs

  • Mathematica
    Flatten[Table[Binomial[n,k]Fibonacci[n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jan 16 2013 *)

Formula

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000045(n), A001906(n), A093131(n), A099453(n-1), A081574(n), A081575(n) for x = 0,1,2,3,4,5 respectively. Sum_{k, 0<=k<=n} T(n,k)*2^(n-k) = A014445(n).
Showing 1-5 of 5 results.

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