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.

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A106566 Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, 1, 1, 1, 1, 1, 1, ... ] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ... ] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 5, 3, 1, 0, 14, 14, 9, 4, 1, 0, 42, 42, 28, 14, 5, 1, 0, 132, 132, 90, 48, 20, 6, 1, 0, 429, 429, 297, 165, 75, 27, 7, 1, 0, 1430, 1430, 1001, 572, 275, 110, 35, 8, 1, 0, 4862, 4862, 3432, 2002, 1001, 429, 154, 44, 9, 1
Offset: 0

Views

Author

Philippe Deléham, May 30 2005

Keywords

Comments

Catalan convolution triangle; g.f. for column k: (x*c(x))^k with c(x) g.f. for A000108 (Catalan numbers).
Riordan array (1, xc(x)), where c(x) the g.f. of A000108; inverse of Riordan array (1, x*(1-x)) (see A109466).
Diagonal sums give A132364. - Philippe Deléham, Nov 11 2007

Examples

			Triangle begins:
  1;
  0,   1;
  0,   1,   1;
  0,   2,   2,  1;
  0,   5,   5,  3,  1;
  0,  14,  14,  9,  4,  1;
  0,  42,  42, 28, 14,  5, 1;
  0, 132, 132, 90, 48, 20, 6, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0, 1,
  0, 1, 1,
  0, 1, 1, 1,
  0, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1,
  0, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
		

Crossrefs

The three triangles A059365, A106566 and A099039 are the same except for signs and the leading term.
See also A009766, A033184, A059365 for other versions.
The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.

Programs

  • Magma
    A106566:= func< n,k | n eq 0 select 1 else (k/n)*Binomial(2*n-k-1, n-k) >;
    [A106566(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 06 2021
    
  • Maple
    A106566 := proc(n,k)
        if n = 0 then
            1;
        elif k < 0 or k > n then
            0;
        else
            binomial(2*n-k-1,n-k)*k/n ;
        end if;
    end proc: # R. J. Mathar, Mar 01 2015
  • Mathematica
    T[n_, k_] := Binomial[2n-k-1, n-k]*k/n; T[0, 0] = 1; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 18 2017 *)
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-Sqrt[1-4#])/(2#)&, 11] // Flatten (* Jean-François Alcover, Jul 16 2019 *)
  • PARI
    {T(n, k) = if( k<=0 || k>n, n==0 && k==0, binomial(2*n - k, n) * k/(2*n - k))}; /* Michael Somos, Oct 01 2022 */
  • Sage
    def A106566(n, k): return 1 if (n==0) else (k/n)*binomial(2*n-k-1, n-k)
    flatten([[A106566(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 06 2021
    

Formula

T(n, k) = binomial(2n-k-1, n-k)*k/n for 0 <= k <= n with n > 0; T(0, 0) = 1; T(0, k) = 0 if k > 0.
T(0, 0) = 1; T(n, 0) = 0 if n > 0; T(0, k) = 0 if k > 0; for k > 0 and n > 0: T(n, k) = Sum_{j>=0} T(n-1, k-1+j).
Sum_{j>=0} T(n+j, 2j) = binomial(2n-1, n), n > 0.
Sum_{j>=0} T(n+j, 2j+1) = binomial(2n-2, n-1), n > 0.
Sum_{k>=0} (-1)^(n+k)*T(n, k) = A064310(n). T(n, k) = (-1)^(n+k)*A099039(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = 0,1,2,3,4,5,6,7,8 respectively.
Sum_{k>=0} T(n, k)*x^(n-k) = C(x, n); C(x, n) are the generalized Catalan numbers.
Sum_{j=0..n-k} T(n+k,2*k+j) = A039599(n,k).
Sum_{j>=0} T(n,j)*binomial(j,k) = A039599(n,k).
Sum_{k=0..n} T(n,k)*A000108(k) = A127632(n).
Sum_{k=0..n} T(n,k)*(x+1)^k*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x= 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Aug 25 2007
Sum_{k=0..n} T(n,k)*A000108(k-1) = A121988(n), with A000108(-1)=0. - Philippe Deléham, Aug 27 2007
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 27 2007
T(n,k)*2^(n-k) = A110510(n,k); T(n,k)*3^(n-k) = A110518(n,k). - Philippe Deléham, Nov 11 2007
Sum_{k=0..n} T(n,k)*A000045(k) = A109262(n), A000045: Fibonacci numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A000129(k) = A143464(n), A000129: Pell numbers. - Philippe Deléham, Oct 28 2008
Sum_{k=0..n} T(n,k)*A100335(k) = A002450(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A100334(k) = A001906(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A099322(k) = A015565(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A106233(k) = A003462(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A151821(k+1) = A100320(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A082505(k+1) = A144706(n). - Philippe Deléham, Oct 30 2008
Sum_{k=0..n} T(n,k)*A000045(2k+2) = A026671(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A122367(k) = A026726(n). - Philippe Deléham, Feb 11 2009
Sum_{k=0..n} T(n,k)*A008619(k) = A000958(n+1). - Philippe Deléham, Nov 15 2009
Sum_{k=0..n} T(n,k)*A027941(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
G.f.: Sum_{n>=0, k>=0} T(n, k)*x^k*z^n = 1/(1 - x*z*c(z)) where c(z) the g.f. of A000108. - Michael Somos, Oct 01 2022

Extensions

Formula corrected by Philippe Deléham, Oct 31 2008
Corrected by Philippe Deléham, Sep 17 2009
Corrected by Alois P. Heinz, Aug 02 2012

A291000 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 3, 9, 26, 74, 210, 596, 1692, 4804, 13640, 38728, 109960, 312208, 886448, 2516880, 7146144, 20289952, 57608992, 163568448, 464417728, 1318615104, 3743926400, 10630080640, 30181847168, 85694918912, 243312448256, 690833811712, 1961475291648, 5569190816256
Offset: 0

Views

Author

Clark Kimberling, Aug 22 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).
In the following guide to p-INVERT sequences using s = (1,1,1,1,1,...) = A000012, in some cases t(1,1,1,1,1,...) is a shifted version of the cited sequence:
p(S) t(1,1,1,1,1,...)
1 - S A000079
1 - S^2 A000079
1 - S^3 A024495
1 - S^4 A000749
1 - S^5 A139761
1 - S^6 A290993
1 - S^7 A290994
1 - S^8 A290995
1 - S - S^2 A001906
1 - S - S^3 A116703
1 - S - S^4 A290996
1 - S^3 - S^6 A290997
1 - S^2 - S^3 A095263
1 - S^3 - S^4 A290998
1 - 2 S^2 A052542
1 - 3 S^2 A002605
1 - 4 S^2 A015518
1 - 5 S^2 A163305
1 - 6 S^2 A290999
1 - 7 S^2 A291008
1 - 8 S^2 A291001
(1 - S)^2 A045623
(1 - S)^3 A058396
(1 - S)^4 A062109
(1 - S)^5 A169792
(1 - S)^6 A169793
(1 - S^2)^2 A024007
1 - 2 S - 2 S^2 A052530
1 - 3 S - 2 S^2 A060801
(1 - S)(1 - 2 S) A053581
(1 - 2 S)(1 - 3 S) A291002
(1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S) A291003
(1 - 2 S)^2 A120926
(1 - 3 S)^2 A291004
1 + S - S^2 A000045 (Fibonacci numbers starting with -1)
1 - S - S^2 - S^3 A291000
1 - S - S^2 - S^3 - S^4 A291006
1 - S - S^2 - S^3 - S^4 - S^5 A291007
1 - S^2 - S^4 A290990
(1 - S)(1 - 3 S) A291009
(1 - S)(1 - 2 S)(1 - 3 S) A291010
(1 - S)^2 (1 - 2 S) A291011
(1 - S^2)(1 - 2 S) A291012
(1 - S^2)^3 A291013
(1 - S^3)^2 A291014
1 - S - S^2 + S^3 A045891
1 - 2 S - S^2 + S^3 A291015
1 - 3 S + S^2 A136775
1 - 4 S + S^2 A291016
1 - 5 S + S^2 A291017
1 - 6 S + S^2 A291018
1 - S - S^2 - S^3 + S^4 A291019
1 - S - S^2 - S^3 - S^4 + S^5 A291020
1 - S - S^2 - S^3 + S^4 + S^5 A291021
1 - S - 2 S^2 + 2 S^3 A175658
1 - 3 S^2 + 2 S^3 A291023
(1 - 2 S^2)^2 A291024
(1 - S^3)^3 A291143
(1 - S - S^2)^2 A209917

Crossrefs

Programs

  • Mathematica
    z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291000 *)

Formula

G.f.: (-1 + x - x^2)/(-1 + 4 x - 4 x^2 + 2 x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) for n >= 4.

A049660 a(n) = Fibonacci(6*n)/8.

Original entry on oeis.org

0, 1, 18, 323, 5796, 104005, 1866294, 33489287, 600940872, 10783446409, 193501094490, 3472236254411, 62306751484908, 1118049290473933, 20062580477045886, 360008399296352015, 6460088606857290384, 115921586524134874897, 2080128468827570457762
Offset: 0

Views

Author

Keywords

Comments

For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 18's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,17}. - Milan Janjic, Jan 25 2015
10*a(n)^2 = Tri(4)*S(n-1, 18)^2 is the triangular number Tri((T(n, 9) - 1)/2), with Tri, S and T given in A000217, A049310 and A053120. This is instance k = 4 of the k-family of identities given in a comment on A001109. - Wolfdieter Lang, Feb 01 2016
Possible solutions for y in Pell equation x^2 - 80*y^2 = 1. The values for x are given in A023039. - Herbert Kociemba, Jun 05 2022

Examples

			a(3) = F(6 * 3) / 8 = F(18) / 8 = 2584 / 8 = 323. - _Indranil Ghosh_, Feb 06 2017
		

Crossrefs

Column m=6 of array A028412.
Partial sums of A007805.

Programs

Formula

G.f.: x/(1 - 18*x + x^2).
a(n) = A134492(n)/8.
a(n) ~ (1/40)*sqrt(5)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all terms k of the sequence, 80*k^2 + 1 is a square. Limit_{n->oo} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 14 2002
a(n) = S(n-1, 18) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n))/(8*sqrt(5)).
a(n) = sqrt((A023039(n)^2 - 1)/80) (cf. Richardson comment).
a(n) = 18*a(n-1) - a(n-2). - Gregory V. Richardson, Oct 14 2002
a(n) = A001076(2n)/4.
a(n) = 17*(a(n-1) + a(n-2)) - a(n-3) = 19*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, May 26 2007
a(n+1) = Sum_{k=0..n} A101950(n,k)*17^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = (1/2)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = (2/9)*(2 + sqrt(5)). - Peter Bala, Dec 23 2012
a(n) = (1/32)*(F(6*n + 3) - F(6*n - 3)).
Sum_{n>=1} 1/(4*a(n) + 1/(4*a(n))) = 1/4. Compare with A001906 and A049670. - Peter Bala, Nov 29 2013
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -G(x)*G(-x), where G(x) = Sum_{n >= 1} A001076(n)*x^n.
1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = (1 + F(x))*(1 + F(-x)) = (1 + 2*x*G(x))*(1 - 2*x*G(-x)), where F(x) = Sum_{n >= 1} Fibonacci(3*n + 3)*x^n.
1 + 7*Sum_{n >= 1} a(n)*x^(2*n) = (1 + G(x))*(1 + G(-x)) = (1 + 7*G(x))*(1 + 7*G(-x)).
1 + 12*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*G(x))*(1 + 2*G(-x)) = (1 + 6*G(x))*(1 + 6*G(-x)) = (1 + A(x))*(1 + A(-x)), where A(x) = Sum_{n >= 1} Fibonacci(3*n)*x^n is the o.g.f for A014445.
1 + 15*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 5*G(x))*(1 + 5*G(-x)) = (1 + 3*G(x))*(1 + 3*G(-x)) = H(x)*H(-x), where H(x) = Sum_{n >= 0} A155179(n)*x^n.
1 + 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 4*G(x))*(1 + 4*G(-x)) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n - 1)*(-x)^n) = (1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*x^n)*(1 + 2* Sum_{n >= 1} Fibonacci(3*n + 1)*(-x)^n).
1 + 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} Lucas(3*n)*x^n)*(1 + Sum_{n >= 1} Lucas(3*n)*(-x)^n).
1 - 5*Sum_{n >= 1} a(n)*x^(2*n) = (1 + Sum_{n >= 1} A001077(n+1)*x^n)*(1 + Sum_{n >= 1} A001077(n+1)*(-x)^n).
1 - 9*Sum_{n >= 1} a(n)*x^(2*n) = (1 - G(x))*(1 - G(-x)) = (1 + 9*G(x))*(1 + 9*G(-x)).
1 - 16*Sum_{n >= 1} a(n)*x^(2*n) = (1 + 2*Sum_{n >= 1} A099843(n)*x^n)*(1 + 2*Sum_{n >= 1} A099843(n)*(-x)^n).
1 - 20*Sum_{n >= 1} a(n)*x^(2*n) = (1 - 2*G(x))*(1 - 2*G(-x)) = (1 + 10*G(x))*(1 + 10*G(-x)).
(End)

Extensions

Chebyshev and other comments from Wolfdieter Lang, Nov 08 2002

A001090 a(n) = 8*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 63, 496, 3905, 30744, 242047, 1905632, 15003009, 118118440, 929944511, 7321437648, 57641556673, 453811015736, 3572846569215, 28128961537984, 221458845734657, 1743541804339272, 13726875588979519, 108071462907496880, 850844827670995521, 6698687158460467288
Offset: 0

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Author

Keywords

Comments

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 15*y^2 = 1; the corresponding values of x are in A001091. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 02 2014]
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 8's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,7}. - Milan Janjic, Jan 25 2015
From Klaus Purath, Jul 25 2024: (Start)
For any three consecutive terms (x, y, z) y^2 - x*z = 1 always applies.
a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 9t(k-1) - 9t(k-2) + t(k-3) or t(k) = 8t(k-1) - t(k-2) without regard to initial values.
In particular, if the recurrence (t) of the form (9,-9,1) has the initial values t(0) = 1, t(1) = 2, t(2) = 9, a(n) = t(n) - 1 applies. (End)

Examples

			G.f. = x + 8*x^2 + 63*x^3 + 496*x^4 + 3905*x^5 + 30744*x^6 + 242047*x^7 + ...
		

References

  • Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equals one-third A136325.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), this sequence (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

Programs

  • GAP
    m:=4;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Magma
    I:=[0,1]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017
    
  • Maple
    A001090:=1/(1-8*z+z**2); # Simon Plouffe in his 1992 dissertation
    seq( simplify(ChebyshevU(n-1, 4)), n=0..20); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Table[GegenbauerC[n-1, 1, 4], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
    LinearRecurrence[{8,-1},{0,1},30] (* Harvey P. Dale, Aug 29 2012 *)
    a[n_]:= ChebyshevU[n-1, 4]; (* Michael Somos, May 28 2014 *)
    CoefficientList[Series[x/(1-8*x+x^2), {x,0,20}], x] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    {a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 15}; /* Michael Somos, Apr 05 2008 */
    
  • PARI
    {a(n) = polchebyshev(n-1, 2, 4)}; /* Michael Somos, May 28 2014 */
    
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(x/(1-8*x-x^2))) \\ G. C. Greubel, Dec 20 2017
    
  • SageMath
    [lucas_number1(n,8,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008
    
  • SageMath
    [chebyshev_U(n-1,4) for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

15*a(n)^2 - A001091(n)^2 = -1.
a(n) = sqrt((A001091(n)^2 - 1)/15).
a(n) = S(2*n-1, sqrt(10))/sqrt(10) = S(n-1, 8); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310, with S(-1, x) := 0.
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ((4+sqrt(15))^n - (4-sqrt(15))^n)/(2*sqrt(15)).
G.f.: x/(1-8*x+x^2). (End)
Limit_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
[A070997(n-1), a(n)] = [1,6; 1,7]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(-n) = -a(n). - Michael Somos, Apr 05 2008
a(n+1) = Sum_{k=0..n} A101950(n,k)*7^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(15)).
Product_{n >= 2} (1 - 1/a(n)) = (1/8)*(3 + sqrt(15)).
(End)
a(n) = A136325(n)/3. - Greg Dresden, Sep 12 2019
E.g.f.: exp(4*x)*sinh(sqrt(15)*x)/sqrt(15). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*6^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*10^k. - Peter Bala, Jul 17 2023

Extensions

More terms from Wolfdieter Lang, Aug 02 2000

A109466 Riordan array (1, x(1-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Aug 28 2005

Keywords

Comments

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

Examples

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)
		

Crossrefs

Cf. A026729 (unsigned version), A000108, A030528, A124644.

Programs

  • Magma
    /* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Formula

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A006139 n*a(n) = 2*(2*n-1)*a(n-1) + 4*(n-1)*a(n-2) with a(0) = 1.

Original entry on oeis.org

1, 2, 8, 32, 136, 592, 2624, 11776, 53344, 243392, 1116928, 5149696, 23835904, 110690816, 515483648, 2406449152, 11258054144, 52767312896, 247736643584, 1164829376512, 5484233814016, 25852072517632, 121997903495168
Offset: 0

Views

Author

Keywords

Comments

a(n) = number of Delannoy paths (A001850) from (0,0) to (n,n) in which every Northeast step is immediately preceded by an East step. - David Callan, Mar 14 2004
The Hankel transform (see A001906 for definition) of this sequence is A036442 : 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
In general, 1/sqrt(1-4*r*x-4*r*x^2) has e.g.f. exp(2rx)BesselI(0,2r*sqrt((r+1)/r)x), a(n) = Sum_{k=0..n} C(2k,k)*C(k,n-k)*r^k, gives the central coefficient of (1+(2r)x+r(r+1)x^2) and is the (2r)-th binomial transform of 1/sqrt(1-8*C(n+1,2)x^2). - Paul Barry, Apr 28 2005
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the H and U steps can have two colors. - N-E. Fahssi, Feb 05 2008
Self-convolution of a(n)/2^n gives Pell numbers A000129(n+1). - Vladimir Reshetnikov, Oct 10 2016
This sequence gives the integer part of an integral approximation to Pi, and also appears in Frits Beukers's "A Rational Approach to Pi" (cf. Links, Example). Despite quality M ~ 0.9058... reported by Beukers, measurements between n = 10000 and 30000 lead to a contentious quality estimate, M ~ 0.79..., at the 99% confidence level. In "Searching for Apéry-Style Miracles" Doron Zeilberger Quotes that M = 0.79119792... and also gives a closed form. The same rational approximation to Pi also follows from time integration on a quartic Hamiltonian surface, 2*H=(q^2+p^2)*(1-4*q*(q-p)). - Bradley Klee, Jul 19 2018, updated Mar 17 2019
Diagonal of rational function 1/(1 - (x + y + x*y^2)). - Gheorghe Coserea, Aug 06 2018

Examples

			G.f. = 1 + 2*x + 8*x^2 + 32*x^3 + 136*x^4 + 592*x^5 + 2624*x^6 + 11776*x^7 + ...
J_3 = Integral_{y=0..Pi/4} 4*(4*(sin(y)-cos(y))*sin(y))^3*dy = 32*Pi - (304/3), |J_3| < 1. - _Bradley Klee_, Jul 19 2018
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

First column of A110446. A higher-quality Pi approximation: A123178.

Programs

  • GAP
    a:=[1,2];; for n in [3..25] do a[n]:=1/(n-1)*(2*(2*n-3)*a[n-1]+4*(n-2)*a[n-2]); od; a; # Muniru A Asiru, Aug 06 2018
  • Maple
    seq(add(binomial(2*k, k)*binomial(k, n-k), k=0..n), n=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
    A006139 := n -> 2^n*hypergeom([-n/2, 1/2-n/2], [1], 2):
    seq(simplify(A006139(n)), n=0..29); # Peter Luschny, Sep 18 2014
  • Mathematica
    Table[SeriesCoefficient[1/(1-4x-4x^2)^(1/2),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 05 2012 *)
    Table[Abs[LegendreP[n, I]] 2^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 22 2015 *)
    Table[Sum[Binomial[2*k, k]*Binomial[k, n - k], {k,0,n}], {n,0,50}] (* G. C. Greubel, Feb 28 2017 *)
    a[n_] := If[n == 0, 1, Coefficient[(1 + 2 x + 2 x^2)^n, x^n]] (* Emanuele Munarini, Aug 04 2017 *)
    CoefficientList[Series[1/Sqrt[(-4 x^2 - 4 x + 1)], {x, 0, 24}], x] (* Robert G. Wilson v, Jul 28 2018 *)
  • Maxima
    a(n) := coeff(expand((1+2*x+2*x^2)^n),x,n);
    makelist(a(n),n,0,12); /* Emanuele Munarini, Aug 04 2017 */
    
  • PARI
    for(n=0,30,t=polcoeff((1+2*x+2*x^2)^n,n,x); print1(t","))
    
  • PARI
    for(n=0,25, print1(sum(k=0,n, binomial(2*k,k)*binomial(k,n-k)), ", ")) \\ G. C. Greubel, Feb 28 2017
    
  • PARI
    {a(n) = (-2*I)^n * pollegendre(n, I)}; /* Michael Somos, Aug 04 2018 */
    

Formula

a(n) = Sum_{k=0..n} C(2*k, k)*C(k, n-k). - Detlef Pauly (dettodet(AT)yahoo.de), Nov 08 2001
G.f.: 1/(1-4x-4x^2)^(1/2); also, a(n) is the central coefficient of (1+2x+2x^2)^n. - Paul D. Hanna, Jun 01 2003
Inverse binomial transform of central Delannoy numbers A001850. - David Callan, Mar 14 2004
E.g.f.: exp(2*x) * BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..floor(n/2)} C(n,2k) * C(2k,k) * 2^(n-k). - Paul Barry, Sep 19 2006
a(n) ~ 2^(n - 3/4) * (1 + sqrt(2))^(n + 1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Oct 05 2012, simplified Jan 31 2023
G.f.: 1/(1 - 2*x*(1+x)*Q(0)), where Q(k)= 1 + (4*k+1)*x*(1+x)/(k+1 - x*(1+x)*(2*k+2)*(4*k+3)/(2*x*(1+x)*(4*k+3)+(2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
a(n) = 2^n*hypergeom([-n/2, 1/2-n/2], [1], 2). - Peter Luschny, Sep 18 2014
0 = a(n)*(+16*a(n+1) + 24*a(n+2) - 8*a(n+3)) + a(n+1)*(+8*a(n+1) + 16*a(n+2) - 6*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Oct 13 2016
It appears that Pi/2 = Sum_{n >= 1} (-1)^(n-1)*4^n/(n*a(n-1)*a(n)). - Peter Bala, Feb 20 2017
G.f.: G(x) = (1/(2*Pi))*Integral_{y=0..2*Pi} 1/(1-x*(4*(sin(y)-cos(y))*sin(y)))*dy, also satisfies: (2+4*x)*G(x)-(1-4*x-4*x^2)*G'(x)=0. - Bradley Klee, Jul 19 2018
a(n) = Sum_{k=0..n} (1-i)^k * (1+i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit. - Seiichi Manyama, Aug 29 2025

A088305 a(0) = 1, a(n) = Fibonacci(2*n). It has the property that a(n) = 1*a(n-1) + 2*a(n-2) + 3*a(n-3) + 4*a(n-4) + ...

Original entry on oeis.org

1, 1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040, 2178309, 5702887, 14930352, 39088169, 102334155, 267914296, 701408733, 1836311903, 4807526976, 12586269025, 32951280099, 86267571272, 225851433717
Offset: 0

Views

Author

Miklos Kristof, Nov 05 2003

Keywords

Comments

Number of compositions of n into one sort of 1's, two sorts of 2's, ..., k sorts of k's, ... - Joerg Arndt, Jun 21 2011
Also the number of spanning trees of a graph formed by joining a single vertex to all vertices of a path on n-1 vertices. - Edward Scheinerman (ers(AT)jhu.edu), Feb 28 2007
Row sums of triangle A128908. - Philippe Deléham, Nov 21 2007
Let P = the partial sum operator, A000012: (1; 1,1; 1,1,1; ...) and A153463 = M, the partial sum & shift operator. It appears that beginning with any randomly taken sequence S(n), iterates of the operations M * S(n), -> M * ANS, -> P * ANS, etc. (or starting with P) will rapidly converge upon a two-sequence limit cycle of (1, 2, 5, 13, 34, ...) and (1, 1, 3, 8, 21, ...). - Gary W. Adamson, Dec 27 2008
Eigensequence of triangle A004736. - Paul Barry, Nov 03 2010
a(n) = the sum of the products of all compositions of n.
Number of nonisomorphic graded posets with 0 and uniform Hasse graph of rank n, with exactly 2 elements of each rank level above 0.(Uniform used in the sense of Retakh, Serconek and Wilson. Graded poset is being used in Stanley's sense that every maximal chain has the same length n.) - David Nacin, Feb 26 2012
a(n) is the top left entry in the n-th power of the 3 X 3 matrix [1, 1, 1; 1, 1, 1; 1, 0, 1] or of the 3 X 3 matrix [1, 1, 1; 1, 1, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

Examples

			a(5) = 55 = 1*21 + 2*8 + 3*3 + 4*1 + 5*1 = 21 + 16 + 9 + 4 + 5.
a(3) = 8 because if we multiply the compositions of three:
3; 2,1; 1,2; 1,1,1, we get 3,2,2,1 respectively, which sums to eight.
		

References

  • R. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Crossrefs

Apart from initial term, same as A001906.

Programs

  • Magma
    [1] cat [Fibonacci(2*n): n in [1..40]]; // G. C. Greubel, Dec 16 2022
    
  • Maple
    H := (n, a, b) -> hypergeom([a - n/2, b - n/2], [1 - n], -4):
    a := n -> `if`(n = 0, 1, H(2*n, 1, 1/2)):
    seq(simplify(a(n)), n=0..28); # Peter Luschny, Sep 03 2019
    # third Maple program:
    a:= proc(n) option remember; `if`(n=0, 1,
          add(a(n-i)*i, i=1..n))
        end:
    seq(a(n), n=0..36);  # Alois P. Heinz, Feb 09 2021
  • Mathematica
    f[list_]:=Apply[Times,list]; Table[Total[Map[f, Level[Map[Permutations, Partitions[n]], {2}]]], {n, 0, 20}]
    CoefficientList[Series[(1 - 2 x + x^2)/(1 - 3 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 16 2014 *)
    Join[{1}, Fibonacci[2*Range[40]]] (* G. C. Greubel, Dec 16 2022 *)
  • PARI
    N=66;  x='x+O('x^N);
    Vec( 1/( 1 - sum(k=1,N, k*x^k ) ) )
    /* Joerg Arndt, Sep 30 2012 */
    
  • Python
    def a(n, adict={0:1, 1:1, 2:3}):
        if n in adict:
            return adict[n]
        adict[n]=3*a(n-1)-a(n-2)
        return adict[n]
    # David Nacin, Mar 04 2012
    
  • SageMath
    def A088305(n): return 1 if (n==0) else fibonacci(2*n)
    [A088305(n) for n in range(41)] # G. C. Greubel, Dec 16 2022

Formula

a(n) = 1*a(n-1) + 2*a(n-2) + 3*a(n-3) + 4*a(n-4) + ...
G.f.: (1 -2*x + x^2)/(1 - 3*x + x^2) = 1 + x/(1 - 3*x + x^2) (see Agarwal (2000), p. 1424).
G.f.: 1/(1 - Sum_{k >= 1} k*x^k). - Joerg Arndt, Jun 21 2011
G.f.: Sum_{n >= 0} q^n / (1 - q)^(2*n). - Joerg Arndt, Dec 09 2012
a(0) = 1, a(n) = (h^(2*n) - h^(-2*n))/sqrt(5), where h = (1+sqrt(5))/2.
a(0) = 1, a(1) = 1, a(2) = 3, a(n+1) = 3*a(n) - a(n-1) for n >= 2. - Philippe Deléham, Nov 21 2007
a(n) = (((3 + sqrt(5))/2)^n - ((3 - sqrt(5))/2)^n)/sqrt(5). - Geoffrey Critzer, Sep 23 2008
F(2n) = 1*F(2n-2) + 2*F(2n-4) + 3*F(2n-6) + 4*F(2n-8) + ...
F(2n+1) = 1 + 1*F(2n-1) + 2*F(2n-3) + 3*F(2n-5) + 4*F(2n-7) + ...
Convolutions with 1, 3, 6, 10, ..., n*(n+1)/2:
1*F(2n) + 3*F(2n-2) + 6*F(2n-4) + 10*F(2n-6) + ... = F(2n+3) - 1.
1*F(2n-1) + 3*F(2n-3) + 6*F(2n-5) + 10*F(2n-7) + ... = F(2n+2) - n - 1.
G.f.: 1/( 1 - G(0)*(1 + x)*x), where G(k) = 1 + x/(1 - x*(k+2)/(x*(k+2) + (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + (1-x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
a(n) = H(2*n, 1, 1/2) for n > 0 where H(n, a, b) = hypergeom([a - n/2, b - n/2], [1 - n], -4). - Peter Luschny, Sep 03 2019
INVERT transform of the identity function. - Alois P. Heinz, Feb 09 2021

Extensions

More terms from Ray Chandler, Nov 06 2003
Further terms from Edward Scheinerman (ers(AT)jhu.edu), Feb 28 2007

A033890 a(n) = Fibonacci(4*n + 2).

Original entry on oeis.org

1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673
Offset: 0

Views

Author

Keywords

Comments

(x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - Floor van Lamoen, Dec 10 2001
This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Indices of square numbers which are also 12-gonal. - Sture Sjöstedt, Jun 01 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - Kival Ngaokrajang, Nov 03 2013
The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - Bernard Schott, Mar 19 2021

Crossrefs

Programs

  • Magma
    [Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011
  • Maple
    A033890 := proc(n)
        option remember;
        if n <= 1 then
            op(n+1,[1,8]);
        else
            7*procname(n-1)-procname(n-2) ;
        end if;
    end proc: # R. J. Mathar, Apr 30 2017
  • Mathematica
    Table[Fibonacci[4n + 2], {n, 0, 14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
    LinearRecurrence[{7, -1}, {1, 8}, 50] (* G. C. Greubel, Jul 13 2017 *)
    a[n_] := (GoldenRatio^(2 (1 + 2 n)) - GoldenRatio^(-2 (1 + 2 n)))/Sqrt[5]
    Table[a[n] // FullSimplify, {n, 0, 21}] (* Gerry Martens, Aug 20 2025 *)
  • PARI
    a(n)=fibonacci(4*n+2);
    

Formula

G.f.: (1+x)/(1-7*x+x^2).
a(n) = 7*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=8.
a(n) = S(n,7) + S(n-1,7) = S(2*n,sqrt(9) = 3), where S(n,x) = U(n,x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), S(n,3) = A001906(n+1).
a(n) = ((7+3*sqrt(5))^n - (7-3*sqrt(5))^n + 2*((7+3*sqrt(5))^(n-1) - ((7-3*sqrt(5))^(n-1)))) / (3*(2^n)*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -9). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 for L(n,+7). - Reinhard Zumkeller, Jun 01 2005
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),7/2) + f(a(n-2),7/2). - Marcos Carreira, Dec 27 2006
a(n+1) = 8*a(n) - 8*a(n-1) + a(n-2); a(1)=1, a(2)=8, a(3)=55. - Sture Sjöstedt, May 27 2009
a(n) = A167816(4*n+2). - Reinhard Zumkeller, Nov 13 2009
a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n) = A000045(A016825(n)). - Michel Marcus, Mar 22 2015
a(n) = A001906(2*n+1). - R. J. Mathar, Apr 30 2017
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + 3*sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Apr 14 2025
From Peter Bala, Jun 08 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/9 [telescoping series: 3/(a(n) - 1/a(n)) = 1/Fibonacci(4*n+4) + 1/Fibonacci(4*n)].
Product_{n >= 1} (a(n) + 3)/(a(n) - 3) = 5/2 [telescoping product:
(a(n) + 3)/(a(n) - 3) = b(n)/b(n-1), where b(n) = (Lucas(4*n+4) - 3)/(Lucas(4*n+4) + 3)].
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(9/5) [telescoping product:
(a(n) + 1)/(a(n) - 1) = c(n)/c(n-1) for n >= 1, where c(n) = Fibonacci(2*n+2)/Lucas(2*n+2)]. (End)
From Gerry Martens, Aug 20 2025: (Start)
a(n) = ((3 + sqrt(5))^(1 + 2*n) - (3 - sqrt(5))^(1 + 2*n)) / (2^(1 + 2*n)*sqrt(5)).
a(n) = Sum_{k=0..2*n} binomial(2*n + k + 1, 2*k + 1). (End)

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

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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.

Crossrefs

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

A078812 Triangle read by rows: T(n, k) = binomial(n+k-1, 2*k-1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 10, 6, 1, 5, 20, 21, 8, 1, 6, 35, 56, 36, 10, 1, 7, 56, 126, 120, 55, 12, 1, 8, 84, 252, 330, 220, 78, 14, 1, 9, 120, 462, 792, 715, 364, 105, 16, 1, 10, 165, 792, 1716, 2002, 1365, 560, 136, 18, 1, 11, 220, 1287, 3432, 5005, 4368, 2380, 816, 171, 20, 1
Offset: 0

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Author

Michael Somos, Dec 05 2002

Keywords

Comments

Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.
Apart from signs, identical to A053122.
Coefficient array for Morgan-Voyce polynomial B(n,x); see A085478 for references. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of compositions of n having k parts when there are q kinds of part q (q=1,2,...). Example: T(4,2) = 10 because we have (1,3),(1,3'),(1,3"), (3,1),(3',1),(3",1),(2,2),(2,2'),(2',2) and (2',2'). - Emeric Deutsch, Apr 09 2005
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
This sequence is jointly generated with A085478 as a triangular array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = u(n-1,x) + x*v(n-1)x and v(n,x) = u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Concerning Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle T(n,k), 0 <= k <= n, read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 27 2012
From Wolfdieter Lang, Aug 30 2012: (Start)
With offset [0,0] the triangle with entries R(n,k) = T(n+1,k+1):= binomial(n+k+1, 2*k+1), n >= k >= 0, and zero otherwise, becomes the Riordan lower triangular convolution matrix R = (G(x)/x, G(x)) with G(x):=x/(1-x)^2 (o.g.f. of A000027). This means that the o.g.f. of column number k of R is (G(x)^(k+1))/x. This matrix R is the inverse of the signed Riordan lower triangular matrix A039598, called in a comment there S.
The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598.
For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)
From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
For 1 <= k <= n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01. - Milan Janjic, Dec 20 2016
The infinite sum (Sum_{i >= 0} (T(s+i,1+i) / 2^(s+2*i)) * zeta(s+1+2*i)) = 1 allows any zeta(s+1) to be expressed as a sum of rational multiples of zeta(s+1+2*i) having higher arguments. For example, zeta(3) can be expressed as a sum involving zeta(5), zeta(7), etc. The summation for each s >= 1 uses the s-th diagonal of the triangle. - Robert B Fowler, Feb 23 2022
The convolution triangle of the nonnegative integers. - Peter Luschny, Oct 07 2022

Examples

			Triangle begins, 1 <= k <= n:
                          1
                        2   1
                      3   4   1
                    4  10   6   1
                  5  20  21   8   1
                6  35  56  36  10   1
              7  56 126 120  55  12   1
            8  84 252 330 220  78  14   1
From _Peter Bala_, Feb 11 2025: (Start)
The array factorizes as an infinite product of lower triangular arrays:
  / 1               \    / 1              \ / 1              \ / 1             \
  | 2    1           |   | 2   1          | | 0  1           | | 0  1          |
  | 3    4   1       | = | 3   2   1      | | 0  2   1       | | 0  0  1       | ...
  | 4   10   6   1   |   | 4   3   2  1   | | 0  3   2  1    | | 0  0  2  1    |
  | 5   20  21   8  1|   | 5   4   3  2  1| | 0  4   3  2  1 | | 0  0  3  2  1 |
  |...               |   |...             | |...             | |...            |
Cf. A092276. (End)
		

Crossrefs

This triangle is formed from odd-numbered rows of triangle A011973 read in reverse order.
Row sums give A001906. With signs: A053122.
The column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for k=1..6, resp. For k=7..24 they are A010966..(+2)..A011000 and for k=25..50 they are A017713..(+2)..A017763.

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k+1, 2*k+1) ))); # G. C. Greubel, Aug 01 2019
  • Haskell
    a078812 n k = a078812_tabl !! n !! k
    a078812_row n = a078812_tabl !! n
    a078812_tabl = [1] : [2, 1] : f [1] [2, 1] where
       f us vs = ws : f vs ws where
         ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
                          (us ++ [0, 0])
    -- Reinhard Zumkeller, Dec 16 2013
    
  • Magma
    /* As triangle */ [[Binomial(n+k-1, 2*k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jun 01 2018
    
  • Maple
    for n from 1 to 11 do seq(binomial(n+k-1,2*k-1),k=1..n) od; # yields sequence in triangular form; Emeric Deutsch, Apr 09 2005
    # Uses function PMatrix from A357368. Adds a row and column above and to the left.
    PMatrix(10, n -> n); # Peter Luschny, Oct 07 2022
  • Mathematica
    (* First program *)
    u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
    u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
    v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%] (* A085478 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%] (* A078812 *) (* Clark Kimberling, Feb 25 2012 *)
    (* Second program *)
    Table[Binomial[n+k+1, 2*k+1], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 01 2019 *)
  • Maxima
    T(n,m):=sum(binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)*binomial(n+1,m+k+1),k,0,n-m); /* Vladimir Kruchinin, Apr 13 2016 */
    
  • PARI
    {T(n, k) = if( n<0, 0, binomial(n+k-1, 2*k-1))};
    
  • PARI
    {T(n, k) = polcoeff( polcoeff( x*y / (1 - (2 + y) * x + x^2) + x * O(x^n), n), k)};
    
  • Sage
    @cached_function
    def T(k,n):
        if k==n: return 1
        if k==0: return 0
        return sum(i*T(k-1,n-i) for i in (1..n-k+1))
    A078812 = lambda n,k: T(k,n)
    [[A078812(n,k) for k in (1..n)] for n in (1..8)] # Peter Luschny, Mar 12 2016
    
  • Sage
    [[binomial(n+k+1, 2*k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019
    

Formula

G.f.: x*y / (1 - (2 + y)*x + x^2). To get row n, expand this in powers of x then expand the coefficient of x^n in increasing powers of y.
From Philippe Deléham, Feb 16 2004: (Start)
If indexing begins at 0 we have
T(n,k) = (n+k+1)!/((n-k)!*(2k+1))!.
T(n,k) = Sum_{j>=0} T(n-1-j, k-1)*(j+1) with T(n, 0) = n+1, T(n, k) = 0 if n < k.
T(n,k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j) with T(n,k) = 0 if k < 0, T(0, 0)=1 and T(0, k) = 0 for k > 0.
G.f. for the column k: Sum_{n>=0} T(n, k)*x^n = (x^k)/(1-x)^(2k+2).
Row sums: Sum_{k>=0} T(n, k) = A001906(n+1). (End)
Antidiagonal sums are A000079(n) = Sum_{k=0..floor(n/2)} binomial(n+k+1, n-k). - Paul Barry, Jun 21 2004
Riordan array (1/(1-x)^2, x/(1-x)^2). - Paul Barry, Oct 22 2006
T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n, T(n,k) = T(n-1,k-1) + 2*T(n-1,k) - T(n-2,k). - Philippe Deléham, Jan 26 2010
For another version see A128908. - Philippe Deléham, Mar 27 2012
T(n,m) = Sum_{k=0..n-m} (binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)* binomial(n+1,m+k+1)). - Vladimir Kruchinin, Apr 13 2016
T(n, k) = T(n-1, k) + (T(n-1, k-1) + T(n-2, k-1) + T(n-3, k-1) + ...) for k >= 2 with T(n, 1) = n. - Peter Bala, Feb 11 2025
From Peter Bala, May 04 2025: (Start)
With the column offset starting at 0, the n-th row polynomial B(n, x) = 1/sqrt(x + 4) * Chebyshev_U(2*n+1, (1/2)*sqrt(x + 4)) = (-1)^n * Chebyshev_U(n, -(1/2)*(x + 2)).
B(n, x) / Product_{k = 1..2*n} (1 + 1/B(k, x)) = b(n, x), the n-th row polynomial of A085478. (End)

Extensions

Edited by N. J. A. Sloane, Apr 28 2008
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