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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A045925 a(n) = n*Fibonacci(n).

Original entry on oeis.org

0, 1, 2, 6, 12, 25, 48, 91, 168, 306, 550, 979, 1728, 3029, 5278, 9150, 15792, 27149, 46512, 79439, 135300, 229866, 389642, 659111, 1112832, 1875625, 3156218, 5303286, 8898708, 14912641, 24961200, 41734339, 69705888, 116311074, 193898158, 322961275, 537492672
Offset: 0

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Author

Keywords

Comments

Number of levels in all compositions of n+1 with only 1's and 2's.
Apart from first term: row sums of the triangle in A131410. - Reinhard Zumkeller, Oct 07 2012
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>3} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the third one. - Sergey Kitaev, Dec 08 2020

References

  • Jean Paul Van Bendegem, The Heterogeneity of Mathematical Research, a chapter in Perspectives on Interrogative Models of Inquiry, Volume 8 of the series Logic, Argumentation & Reasoning pp 73-94, Springer 2015. See Section 2.1.

Crossrefs

Partial sums: A014286. Cf. A000045.

Programs

  • Haskell
    a045925 n = a045925_list !! (n-1)
    a045925_list = zipWith (*) [0..] a000045_list
    -- Reinhard Zumkeller, Oct 01 2012
  • Magma
    [n*Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011
    
  • Maple
    a:= n-> n*(<<0|1>, <1|1>>^n)[1,2]:
    seq(a(n), n=0..37);  # Alois P. Heinz, May 07 2021
  • Mathematica
    Table[Fibonacci[n]*n, {n, 0, 33}] (* Zerinvary Lajos, Jul 09 2009 *)
    LinearRecurrence[{2, 1, -2, -1}, {0, 1, 2, 6}, 34] (* or *)
    CoefficientList[ Series[(x + x^3)/(-1 + x + x^2)^2, {x, 0, 35}], x] (* Robert G. Wilson v, Nov 14 2015 *)
  • PARI
    Lucas(n)=fibonacci(n-1)+fibonacci(n+1)
    a(n)=polcoeff(sum(m=1,n,eulerphi(m)*fibonacci(m)*x^m/(1-Lucas(m)*x^m+(-1)^m*x^(2*m)+x*O(x^n))),n) \\ Paul D. Hanna, Jan 12 2012
    
  • PARI
    a(n)=n*fibonacci(n) \\ Charles R Greathouse IV, Jan 12 2012
    
  • PARI
    concat(0, Vec(x*(1+x^2)/(1-x-x^2)^2 + O(x^100))) \\ Altug Alkan, Oct 28 2015
    

Formula

G.f.: x*(1+x^2)/(1-x-x^2)^2.
G.f.: Sum_{n>=1} phi(n)*Fibonacci(n)*x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) = Sum_{n>=1} n*Fibonacci(n)*x^n, where phi(n) = A000010(n) and Lucas(n) = A000204(n). - Paul D. Hanna, Jan 12 2012
a(n) = a(n-1) + a(n-2) + L(n-1). - Gary Detlefs, Dec 29 2012
a(n) = F(n+1) + Sum_{k=1..n-2} F(k)*L(n-k), F = A000045 and L = A000032. - Gary Detlefs, Dec 29 2012
a(n) = F(2*n)/Sum_{k=0..floor(n/2)} binomial(n-k,k)/(n-k). - Gary Detlefs, Jan 19 2013
a(n) = A014965(n) * A104714(n). - Michel Marcus, Oct 24 2013
a(n) = 3*A001629(n+1) - A001629(n+2) + A000045(n-1). - Ralf Stephan, Apr 26 2014
a(n) = 2*n*(F(n-2) + floor(F(n-3)/2)) + (n^3 mod 3*n), F = A000045. - Gary Detlefs, Jun 06 2014
E.g.f.: x*(exp(-x/phi)/phi + exp(x*phi)*phi)/sqrt(5), where phi = (1+sqrt(5))/2. - Vladimir Reshetnikov, Oct 28 2015
This is a divisibility sequence and is generated by x^4 - 2*x^3 - x^2 + 2*x + 1. - R. K. Guy, Nov 13 2015
a(n) = L'(n, 1), the first derivative of the n-th Lucas polynomial evaluated at 1. - Andrés Ventas, Nov 12 2021
Sum_{n>=0} a(n)/2^n = 10 (Euler, 1990). - Amiram Eldar, Jan 22 2022

Extensions

Incorrect formula removed by Gary Detlefs, Oct 27 2011

A023610 Convolution of Fibonacci numbers and {F(2), F(3), F(4), ...}.

Original entry on oeis.org

1, 3, 7, 15, 30, 58, 109, 201, 365, 655, 1164, 2052, 3593, 6255, 10835, 18687, 32106, 54974, 93845, 159765, 271321, 459743, 777432, 1312200, 2211025, 3719643, 6248479, 10482351, 17562870, 29391490, 49132669, 82048737, 136884293, 228160495, 379975140, 632293452
Offset: 0

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Author

Keywords

Comments

a(n-2) + 1 is the number of (3412,1243)-, (3412,2134)- and (3412,1324)-avoiding involutions in S_n, n>1. - Ralf Stephan, Jul 06 2003
The number of terms in all ordered partitions of (n+1) using only ones and twos. For example, a(3)=15 because there are 15 terms in 1+1+1+1;2+1+1;1+2+1;1+1+2;2+2 - Geoffrey Critzer, Apr 07 2008
a(n) is the number of n-matchings in the graph obtained by a zig-zag triangulation of a convex (2n+1)-gon. Example: a(2)=7 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have 7 2-matchings: {AB,CD},{AB,DE},{BC,AD},{BC,DE},{BC,EA},{CD,EA} and {DE,AC}. - Emeric Deutsch, Dec 25 2004
Partial sums of A029907. First differences of A002940. - Peter Bala, Oct 24 2007
Equals row sums of triangle A144154. - Gary W. Adamson, Sep 12 2008
Equals the number of 1's in Fibonacci Maximal notation for subsets of
(1, 2, 3, 5, 8, 13, ...) terms. For example (cf. A181630): 4, 5, and 6 are the 3 terms 101, 110, and 111 in Fibonacci Maximal. Total number of 1's for those terms = 7 = a(2). - Gary W. Adamson, Nov 02 2010
a(n) is half the number of strokes needed to draw all the domino tilings of a 2 X (n+2) rectangle. - Roberto Tauraso, Mar 15 2014
a(n) is the total number of 1's in all (n+1)-bit dual Zeckendorf representations of integers (A104326). For example, a(2) = 7 counts the 1's in 101, 110, 111. - Shenghui Yang, Feb 09 2025

Crossrefs

Cf. A000045 (Fibonacci numbers).
Column 1 of triangle A063967.

Programs

  • Haskell
    a023610 n = a023610_list !! n
    a023610_list = f [1] $ drop 3 a000045_list where
       f us (v:vs) = (sum $ zipWith (*) us $ tail a000045_list) : f (v:us) vs
    -- Reinhard Zumkeller, Jan 18 2014
    
  • Mathematica
    Table[Sum[Binomial[n - i, i]*(n - i), {i, 0, n}], {n, 1, 33}] (* Geoffrey Critzer, May 04 2009 *)
  • PARI
    a(n)=(n+2)*fibonacci(n+4)/5+(n-1)*fibonacci(n+2)/5 \\ Charles R Greathouse IV, Jun 11 2015
  • Sage
    def A023610():
        a, b, c, d = 1, 3, 7, 15
        while True:
            yield a
            a, b, c, d = b, c, d, 2*(d-b)+c-a
    a = A023610(); [next(a) for i in range(33)]  # Peter Luschny, Nov 20 2013
    

Formula

O.g.f.: (x+1)/(1-x-x^2)^2. - Len Smiley, Dec 11 2001
a(n) = (1/5)*((n+2)*F(n+4) + (n-1)*F(n+2)), with F(n)=A000045(n). - Ralf Stephan, Jul 06 2003
a(n) = Sum_{k=0..n+1} (n-k+1)*binomial(n-k+1, k). - Paul Barry, Nov 05 2005
Recurrence: a(n+2) = a(n+1) + a(n) + Fib(n+4), n >= 0. For n >= 2, a(n-2) = (-1)^n*((-2n+3)*Fib(-n) - (-n)*Fib(-n-1))/5 = (-1)^n*A010049(-n), the second-order Fibonacci numbers of negative index, where Fib(-n) = (-1)^(n+1)*Fib(n). - Peter Bala, Oct 24 2007
a(n) = (n+1)*F(n+2) - A001629(n+1) where F(n) is the n-th Fibonacci number. - Geoffrey Critzer, Apr 07 2008
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4), n >= 4. - L. Edson Jeffery, Mar 29 2013
a(n+1) = A004798(n) + A000045(n+2) for n >= 0. - John Molokach, Jul 04 2013
a(n) = A001629(n+1) + A001629(n+2). - Philippe Deléham, Oct 30 2013
E.g.f.: exp(x/2)*(5*(5 + 7*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(11 + 15*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023

A036355 Fibonacci-Pascal triangle read by rows.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 5, 5, 3, 5, 10, 14, 10, 5, 8, 20, 32, 32, 20, 8, 13, 38, 71, 84, 71, 38, 13, 21, 71, 149, 207, 207, 149, 71, 21, 34, 130, 304, 478, 556, 478, 304, 130, 34, 55, 235, 604, 1060, 1390, 1390, 1060, 604, 235, 55, 89, 420, 1177, 2272, 3310, 3736, 3310, 2272, 1177, 420, 89
Offset: 0

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Author

Floor van Lamoen, Dec 28 1998

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n-k,k) using steps (1,0),(2,0),(0,1),(0,2). - Joerg Arndt, Jun 30 2011, corrected by Greg Dresden, Aug 25 2020
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			Triangle begins
   1;
   1,   1;
   2,   2,   2;
   3,   5,   5,    3;
   5,  10,  14,   10,    5;
   8,  20,  32,   32,   20,    8;
  13,  38,  71,   84,   71,   38,   13;
  21,  71, 149,  207,  207,  149,   71,  21;
  34, 130, 304,  478,  556,  478,  304, 130,  34;
  55, 235, 604, 1060, 1390, 1390, 1060, 604, 235, 55;
with indices
  T(0,0);
  T(1,0),  T(1,1);
  T(2,0),  T(2,1),  T(2,2);
  T(3,0),  T(3,1),  T(3,2),  T(3,3);
  T(4,0),  T(4,1),  T(4,2),  T(4,3),  T(4,4);
For example, T(4,2) = 14 and there are 14 lattice paths from (0,0) to (4-2,2) = (2,2) using steps (1,0),(2,0),(0,1),(0,2). - _Greg Dresden_, Aug 25 2020
		

Crossrefs

Row sums form sequence A002605. T(n, 0) forms the Fibonacci sequence (A000045). T(n, 1) forms sequence A001629.
Derived sequences: A036681, A036682, A036683, A036684, A036692 (central terms).
Some other Fibonacci-Pascal triangles: A027926, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074.

Programs

  • Haskell
    a036355 n k = a036355_tabl !! n !! k
    a036355_row n = a036355_tabl !! n
    a036355_tabl = [1] : f [1] [1,1] where
       f us vs = vs : f vs (zipWith (+)
                           (zipWith (+) ([0,0] ++ us) (us ++ [0,0]))
                           (zipWith (+) ([0] ++ vs) (vs ++ [0])))
    -- Reinhard Zumkeller, Apr 23 2013
  • Mathematica
    nmax = 11; t[n_, m_] := t[n, m] = tp[n-1, m-1] + tp[n-2, m-2] + tp[n-1, m] + tp[n-2, m]; tp[n_, m_] /; 0 <= m <= n && n >= 0 := t[n, m]; tp[n_, m_] = 0; t[0, 0] = 1; Flatten[ Table[t[n, m], {n, 0, nmax}, {m, 0, n}]] (* Jean-François Alcover, Nov 09 2011, after formula *)
  • PARI
    /* same as in A092566 but use */
    steps=[[1,0], [2,0], [0,1], [0,2]];
    /* Joerg Arndt, Jun 30 2011 */
    

Formula

T(n, m) = T'(n-1, m-1)+T'(n-2, m-2)+T'(n-1, m)+T'(n-2, m), where T'(n, m) = T(n, m) if 0<=m<=n and n >= 0 and T'(n, m)=0 otherwise. Initial term T(0, 0)=1.
G.f.: 1/(1-(1+y)*x-(1+y^2)*x^2). - Vladeta Jovovic, Oct 11 2003

A001871 Expansion of 1/(1 - 3*x + x^2)^2.

Original entry on oeis.org

1, 6, 25, 90, 300, 954, 2939, 8850, 26195, 76500, 221016, 632916, 1799125, 5082270, 14279725, 39935214, 111228804, 308681550, 853904015, 2355364650, 6480104231, 17786356776, 48715278000, 133167004200, 363372003625, 989900286774
Offset: 0

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Author

Keywords

Comments

Convolution of A001906(n), n >= 1 (even-indexed Fibonacci numbers) with itself.
A001787 and this sequence arise in counting ordered trees of height at most k where only the rightmost branch at the root actually achieves this height and the count is by the number of edges, with k = 3 for A001787 and k = 4 for this sequence.
Gives the number of 3412-avoiding permutations containing exactly one subsequence of type 321. - Dan Daly (ddaly(AT)du.edu), Apr 24 2008

References

  • 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

Partial sums of A001870 (one half of odd-indexed A001629(n), n >= 2, Fibonacci convolution).

Programs

  • Magma
    I:=[1, 6, 25, 90]; [n le 4 select I[n] else 6*Self(n-1)-11*Self(n-2)+6*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 10 2012
    
  • Maple
    f:= gfun:-rectoproc({a(n)=6*a(n-1)-11*a(n-2)+6*a(n-3)-a(n-4),
    a(0)=1,a(1)=6,a(2)=25,a(3)=90},a(n),remember):
    map(f, [$0..50]); # Robert Israel, May 05 2017
    # alternative
    A001871 := proc(n)
        option remember ;
        if n <= 3 then
            op(n+1,[1,6,25,90]) ;
        else
            6*procname(n-1)-11*procname(n-2)+6*procname(n-3)-procname(n-4) ;
        end if;
    end proc:
    seq(A001871(n),n=0..10) ; # R. J. Mathar, Dec 16 2024
  • Mathematica
    CoefficientList[Series[1/(1-3x+x^2)^2,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)
  • PARI
    a(n)=((4*n+2)*fibonacci(2*n)+(7*n+5)*fibonacci(2*n+1))/5
    
  • PARI
    Vec(1/(1-3*x+x^2)^2 + O(x^100)) \\ Altug Alkan, Oct 31 2015

Formula

a(n) = (2*(2*n+1)*F(2*(n+1))+3*(n+1)*F(2*n+1))/5 with F(n) = A000045 (Fibonacci numbers).
a(n) = -a(-4-n) = ((4*n+2)*F(2*n) + (7*n+5)*F(2*n+1))/5 with F(n) = A000045 (Fibonacci numbers).
a(n) = (2*a(n-1) + (n+1)*F(2n+4))/3, where F(n) = A000045 (Fibonacci numbers). - Emeric Deutsch, Oct 08 2002
G.f.: 1/(1 - 3*x + x^2)^2. - Simon Plouffe in his 1992 dissertation
a(n) = (Sum_{k=0..n} S(k, 3)*S(n-k, 3)), where S(n, x) = U(n, x/2) is the n-th Chebyshev polynomial of the 2nd kind, A049310. - Paul Barry, Nov 14 2003
a(n) = Sum_{k=1..n+1} F(2k)*F(2(n-k+2)), where F(k) is the k-th Fibonacci number. - Dan Daly (ddaly(AT)du.edu), Apr 24 2008
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Vincenzo Librandi, Mar 14 2011
a(n) = 2*A001870(n) - A238846(n). - Philippe Deléham, Mar 06 2014
a(n) ~ (7 + 3*sqrt(5))*n*cos(n*arccos(3/2))/5. - Stefano Spezia, Mar 29 2022
From Peter Bala, Nov 05 2024: (Start)
a(n) = Sum_{k = 0..n} (n + 2*k + 1)*binomial(n+k, 2*k).
a(n) = (n+1) * hypergeom([-n, n+1, (n+3)/2], [1/2, (n+1)/2], -1/4).
Second-order recurrence: n*a(n) = 3*(n + 1)*a(n-1) - (n + 2)*a(n-2) with a(0) = 1 and a(1) = 6. (End)
E.g.f.: exp(3*x/2)*(5*(5 + 18*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(9 + 40*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Extensions

Additional comments from Wolfdieter Lang, Apr 07 2000

A010049 Second-order Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 3, 5, 10, 18, 33, 59, 105, 185, 324, 564, 977, 1685, 2895, 4957, 8462, 14406, 24465, 41455, 70101, 118321, 199368, 335400, 563425, 945193, 1583643, 2650229, 4430290, 7398330, 12342849, 20573219, 34262337, 57013865, 94800780, 157517532, 261545777
Offset: 0

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Author

Keywords

Comments

Number of parts in all compositions of n+1 with no 1's. E.g. a(5)=10 because in the compositions of 6 with no part equal to 1, namely 6,4+2,3+3,2+4,2+2+2, the total number of parts is 10. - Emeric Deutsch, Dec 10 2003

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 83.
  • Cornelius Gerrit Lekkerkerker, Voorstelling van natuurlijke getallen door een som van getallen van Fibonacci, Simon Stevin, Vol. 29 (1952), pp. 190-195.

Crossrefs

Programs

  • GAP
    a:=List([0..40],n->Sum([0..n-1],k->(k+1)*Binomial(n-k-1,k)));; Print(a); # Muniru A Asiru, Dec 31 2018
    
  • Haskell
    a010049 n = a010049_list !! n
    a010049_list = uncurry c $ splitAt 1 a000045_list where
       c us (v:vs) = (sum $ zipWith (*) us (1 : reverse us)) : c (v:us) vs
    -- Reinhard Zumkeller, Nov 01 2013
    
  • Magma
    [((2*n+3)*Fibonacci(n)-n*Fibonacci(n-1))/5: n in [0..40]]; // Vincenzo Librandi, Dec 31 2018
    
  • Maple
    with(combinat): A010049 := proc(n) options remember; if n <= 1 then n else A010049(n-1)+A010049(n-2)+fibonacci(n-2); fi; end;
  • Mathematica
    CoefficientList[Series[(z - z^2)/(z^2 + z - 1)^2, {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
    CoefficientList[Series[x (1 - x) / (1 - x - x^2)^2, {x, 0, 60}], x] (* Vincenzo Librandi, Jun 11 2013 *)
    LinearRecurrence[{2, 1, -2, -1}, {0, 1, 1, 3}, 38] (* Amiram Eldar, Jan 11 2020 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-2,1,2]^n*[0;1;1;3])[1,1] \\ Charles R Greathouse IV, Jul 20 2016
    
  • Sage
    def A010049():
        a, b, c, d = 0, 1, 1, 3
        while True:
            yield a
            a, b, c, d = b, c, d, 2*(d-b)+c-a
    a = A010049(); [next(a) for i in range(38)]  # Peter Luschny, Nov 20 2013
    
  • SageMath
    def A010049(n): return (1/5)*(n*lucas_number2(n-1, 1, -1) + 3*fibonacci(n))
    [A010049(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022

Formula

First differences of A001629.
From Wolfdieter Lang, May 03 2000: (Start)
a(n) = ((2*n+3)*F(n) - n*F(n-1))/5, F(n)=A000045(n) (Fibonacci numbers) (Turban reference eq.(2.12)).
G.f.: x*(1-x)/(1-x-x^2)^2. (Turban reference eq.(2.10)). (End)
Recurrence: a(0)=0, a(1)=1, a(2)=1, a(n+2) = a(n+1) + a(n) + F(n). - Benoit Cloitre, Sep 02 2002
Set A(n) = a(n+1) + a(n-1), B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + Lucas(n) and B(n+2) = B(n+1) + B(n) + Fibonacci(n). The polynomials F_2(n,-x) = Sum_{k=0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials F(n,-x) defined in A094440. For a similar conjecture for polynomials involving the second-order Lucas numbers see A134410. - Peter Bala, Oct 24 2007
a(n) = -A001629(n+2) + 2*A001629(n+1) + A000045(n+1). - R. J. Mathar, Nov 16 2007
Starting (1, 1, 3, 5, 10, ...), = row sums of triangle A135830. - Gary W. Adamson, Nov 30 2007
a(n) = F(n) + Sum_{k=0..n-1} F(k)*F(n-1-k), where F = A000045. - Reinhard Zumkeller, Nov 01 2013
a(n) = Sum_{k=0..n-1} (k+1)*binomial(n-k-1, k). - Peter Luschny, Nov 20 2013
a(n) = Sum_{i=0..n-1} Sum_{j=0..i} F(j-1)*F(i-j-1), where F = A000045. - Carlos A. Rico A., Jul 14 2016
a(n) = Sum_{k = F(n+1)..F(n+2)-1} A007895(k), where F(n) is the n-th Fibonacci number (Lekkerkerker, 1952). - Amiram Eldar, Jan 11 2020
a(n) = (1/5)*(n*A000032(n-1) + 3*A000045(n)). - G. C. Greubel, Apr 06 2022
E.g.f.: 2*exp(x/2)*(5*x*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Dec 04 2023

Extensions

More terms from Emeric Deutsch, Dec 10 2003

A127670 Discriminants of Chebyshev S-polynomials A049310.

Original entry on oeis.org

1, 4, 32, 400, 6912, 153664, 4194304, 136048896, 5120000000, 219503494144, 10567230160896, 564668382613504, 33174037869887488, 2125764000000000000, 147573952589676412928, 11034809241396899282944, 884295678882933431599104, 75613185918270483380568064
Offset: 1

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Author

Wolfdieter Lang, Jan 23 2007

Keywords

Comments

a(n-1) is the number of fixed n-cell polycubes that are proper in n-1 dimensions (Barequet et al., 2010).
From Rigoberto Florez, Sep 02 2018: (Start)
a(n-1) is the discriminant of the Morgan-Voyce Fibonacci-type polynomial B(n).
Morgan-Voyce Fibonacci-type polynomials are defined as B(0) = 0, B(1) = 1 and B(n) = (x+2)*B(n-1) - B(n-2) for n > 1.
The absolute value of the discriminant of Fibonacci polynomial F(n) is a(n-1).
Fibonacci polynomials are defined as F(0) = 0, F(1) = 1 and F(n) = x*F(n-1) + F(n-2) for n > 1. (End)
The first 6 values are the dimensions of the polynomial ring in 3n variables xi, yi, zi for 1 <= i <= n modulo the ideal generated by x1^a y1^b z1^c + ... + xn^a yn^b zn^c for 0 < a+b+c <= n (see Fact 2.8.1 in Haiman's paper). - Mike Zabrocki, Dec 31 2019

Examples

			n=3: The zeros are [sqrt(2),0,-sqrt(2)]. The Vn(xn[1],...,xn[n]) matrix is [[1,1,1],[sqrt(2),0,-sqrt(2)],[2,0,2]]. The squared determinant is 32 = a(3). - _Wolfdieter Lang_, Aug 07 2011
		

References

  • Gill Barequet, Solomon W. Golomb, and David A. Klarner, Polyominoes. (This is a revision, by G. Barequet, of the chapter of the same title originally written by the late D. A. Klarner for the first edition, and revised by the late S. W. Golomb for the second edition.) Preprint, 2016, http://www.csun.edu/~ctoth/Handbook/chap14.pdf.
  • G. Barequet and M. Shalah, Automatic Proofs for Formulae Enumerating Proper Polycubes, 31st International Symposium on Computational Geometry (SoCG'15). Editors: Lars Arge and János Pach; pp. 19-22, 2015.
  • Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990; p. 219 for T and U polynomials.

Crossrefs

Cf. A007701 (T-polynomials), A086804 (U-polynomials), A171860 and A191092 (fixed n-cell polycubes proper in n-2 and n-3 dimensions, resp.).
A317403 is essentially the same sequence.
Diagonal 1 of A195739.

Programs

  • Magma
    [((n+1)^n/(n+1)^2)*2^n: n in [1..20]]; // Vincenzo Librandi, Jun 23 2014
  • Mathematica
    Table[((n + 1)^n)/(n + 1)^2 2^n, {n, 1, 30}] (* Vincenzo Librandi, Jun 23 2014 *)

Formula

a(n) = ((n+1)^(n-2))*2^n, n >= 1.
a(n) = (Det(Vn(xn[1],...,xn[n])))^2 with the determinant of the Vandermonde matrix Vn with elements (Vn)i,j:= xn[i]^j, i=1..n, j=0..n-1 and xn[i]:=2*cos(Pi*i/(n+1)), i=1..n, are the zeros of S(n,x):=U(n,x/2).
a(n) = ((-1)^(n*(n-1)/2))*Product_{j=1..n} ((d/dx)S(n,x)|_{x=xn[j]}), n >= 1, with the zeros xn[j], j=1..n, given above.
a(n) = A007830(n-2)*A000079(n), n >= 2. - Omar E. Pol, Aug 27 2011
E.g.f.: -LambertW(-2*x)*(2+LambertW(-2*x))/(4*x). - Vaclav Kotesovec, Jun 22 2014

Extensions

Slightly edited by Gill Barequet, May 24 2011

A099920 a(n) = (n+1)*F(n), F(n) = Fibonacci numbers A000045.

Original entry on oeis.org

0, 2, 3, 8, 15, 30, 56, 104, 189, 340, 605, 1068, 1872, 3262, 5655, 9760, 16779, 28746, 49096, 83620, 142065, 240812, 407353, 687768, 1159200, 1950650, 3277611, 5499704, 9216519, 15426870, 25793240, 43080608, 71884197, 119835652
Offset: 0

Views

Author

Paul Barry and Ralf Stephan, Oct 15 2004

Keywords

Comments

A Fibonacci-Lucas convolution.
The number of edges in the Lucas cube L_(n+1) [Klavzar]. - R. J. Mathar, Nov 05 2008
Sums of rows of the triangle in A108037. - Reinhard Zumkeller, Oct 07 2012
a(n-1) is the total binary weight of all bimultus bitstrings of length n. A bitstring is bimultus if each of its 1's possess at least one neighboring 1 and each of its 0's possess at least one neighboring 0. - Steven Finch, May 26 2020

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 35.

Crossrefs

Equals A010049(n) + A001629(n+1).

Programs

  • Haskell
    a099920 n = a099920_list !! n
    a099920_list = zipWith (*) [1..] a000045_list
    -- Reinhard Zumkeller, Oct 07 2012
    
  • Magma
    [(n+1)*Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011
    
  • Mathematica
    Table[(n + 1) Fibonacci[n], {n, 0, 40}] (* Harvey P. Dale, Jan 18 2012 *)
    LinearRecurrence[{2, 1, -2, -1}, {0, 2, 3, 8}, 40] (* Harvey P. Dale, Jan 18 2012 *)
    CoefficientList[Series[(2 - x) x/(-1 + x + x^2)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Jul 28 2023 *)
  • PARI
    a(n)=(n+1)*fibonacci(n) \\ Charles R Greathouse IV, Jun 11 2015

Formula

G.f.: x*(2-x)/(1-x-x^2)^2;
a(n) = Sum_{k=0..n} F(n-k)*(L(k-1) + 0^k).
a(n) = Sum_{k=0..n+1} F(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4); a(0)=0, a(1)=2, a(2)=3, a(3)=8. - Harvey P. Dale, Jan 18 2012
a(n) = a(n-1) + a(n-2) + A000032(n-1) (Lucas numbers). - Bob Selcoe, Aug 19 2015
a(n) = 2*A001629(n+1) - A001629(n). - R. J. Mathar, Feb 04 2022

Extensions

Entry revised by N. J. A. Sloane, Jan 23 2006. The offset changed, so some of the formulas may now be slightly off.

A001870 Expansion of (1-x)/(1 - 3*x + x^2)^2.

Original entry on oeis.org

1, 5, 19, 65, 210, 654, 1985, 5911, 17345, 50305, 144516, 411900, 1166209, 3283145, 9197455, 25655489, 71293590, 197452746, 545222465, 1501460635, 4124739581, 11306252545, 30928921224, 84451726200, 230204999425
Offset: 0

Views

Author

Keywords

Comments

a(n) = ((n+1)*F(2*n+3)+(2*n+3)*F(2*(n+1)))/5 with F(n)=A000045(n) (Fibonacci numbers). One half of odd-indexed A001629(n), n >= 2 (Fibonacci convolution).
Convolution of F(2n+1) (A001519) and F(2n+2) (A001906(n+1)). - Graeme McRae, Jun 07 2006
Number of reentrant corners along the lower contours of all directed column-convex polyominoes of area n+3 (a reentrant corner along the lower contour is a vertical step that is followed by a horizontal step). a(n) = Sum_{k=0..ceiling((n+1)/2)} k*A121466(n+3,k). - Emeric Deutsch, Aug 02 2006
From Wolfdieter Lang, Jan 02 2012: (Start)
a(n) = A024458(2*n), n >= 1 (bisection, even arguments).
a(n) is also the odd part of the bisection of the half-convolution of the sequence A000045(n+1), n >= 0, with itself. See a comment on A201204 for the definition of the half-convolution of a sequence with itself. There one also finds the rule for the o.g.f. which in this case is Chato(x)/2 with the o.g.f. Chato(x) = 2*(1-x)/(1-3*x+x^2)^2 of A001629(2*n+3), n >= 0.
(End)

References

  • 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

a(n) = A060921(n+1, 1)/2.
Partial sums of A030267. First differences of A001871.
Cf. A121466.
Cf. A023610.

Programs

  • GAP
    F:=Fibonacci;; List([0..30], n-> ((n+1)*F(2*n+3)+(2*n+3)*F(2*(n+1)))/5); # G. C. Greubel, Jul 15 2019
  • Haskell
    a001870 n = a001870_list !! n
    a001870_list = uncurry c $ splitAt 1 $ tail a000045_list where
       c us vs'@(v:vs) = (sum $ zipWith (*) us vs') : c (v:us) vs
    -- Reinhard Zumkeller, Oct 31 2013
    
  • Magma
    I:=[1, 5, 19, 65]; [n le 4 select I[n] else 6*Self(n-1) -11*Self(n-2)+6*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 10 2012
    
  • Maple
    A001870:=-(-1+z)/(z**2-3*z+1)**2; # Simon Plouffe in his 1992 dissertation.
  • Mathematica
    CoefficientList[Series[(1-x)/(1-3*x+x^2)^2,{x,0,40}],x] (* Vincenzo Librandi, Jun 10 2012 *)
    LinearRecurrence[{6,-11,6,-1},{1,5,19,65},30] (* Harvey P. Dale, Aug 17 2013 *)
    With[{F=Fibonacci}, Table[((n+1)*F[2*n+3]+(2*n+3)*F[2*n+2])/5, {n,0,30}]] (* G. C. Greubel, Jul 15 2019 *)
  • PARI
    Vec((1-x)/(1-3*x+x^2)^2+O(x^30)) \\ Charles R Greathouse IV, Sep 23 2012
    
  • Sage
    f=fibonacci; [((n+1)*f(2*n+3)+(2*n+3)*f(2*n+2))/5 for n in (0..30)] # G. C. Greubel, Jul 15 2019
    

Formula

a(n) = Sum_{k=1..n+1} k*binomial(n+k+1, 2k). - Emeric Deutsch, Jun 11 2003
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 10 2012
a(n) = (A238846(n) + A001871(n))/2. - Philippe Deléham, Mar 06 2014
a(n) = ((2*n-1)*Fibonacci(2*n) - n*Fibonacci(2*n-1))/5 [Czabarka et al.]. - N. J. A. Sloane, Sep 18 2018
E.g.f.: exp(3*x/2)*(5*(5 + 11*x)*cosh(sqrt(5)*x/2) + sqrt(5)*(13 + 25*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Extensions

More terms from Christian G. Bower

A030267 Compose the natural numbers with themselves, A(x) = B(B(x)) where B(x) = x/(1-x)^2 is the generating function for natural numbers.

Original entry on oeis.org

1, 4, 14, 46, 145, 444, 1331, 3926, 11434, 32960, 94211, 267384, 754309, 2116936, 5914310, 16458034, 45638101, 126159156, 347769719, 956238170, 2623278946, 7181512964, 19622668679, 53522804976, 145753273225, 396323283724, 1076167858046, 2918447861686
Offset: 1

Views

Author

Keywords

Comments

Sum of pyramid weights of all nondecreasing Dyck paths of semilength n. (A pyramid in a Dyck word (path) is a factor of the form U^h D^h, where U=(1,1), D=(1,-1) and h is the height of the pyramid. A pyramid in a Dyck word w is maximal if, as a factor in w, it is not immediately preceded by a u and immediately followed by a d. The pyramid weight of a Dyck path (word) is the sum of the heights of its maximal pyramids.) Example: a(4) = 46. Indeed, there are 14 Dyck paths of semilength 4. One of them, namely UUDUDDUD is not nondecreasing because the valleys are at heights 1 and 0. The other 13, with the maximal pyramids shown between parentheses, are: (UD)(UD)(UD)(UD), (UD)(UD)(UUDD), (UD)(UUDD)(UD), (UD)U(UD)(UD)D, (UD)(UUUDDD), (UUDD)(UD)(UD), (UUDD)(UUDD), (UUUDDD)(UD), U(UD)(UD)(UD)D, U(UD)(UUDD)D, U(UUDD)(UD)D, UU(UD)(UD)DD and (UUUUDDDD). The pyramid weights of these paths are 4, 4, 4, 3, 4, 4, 4, 4, 3, 3, 3, 2, and 4, respectively. Their sum is 46. a(n) = Sum_{k = 1..n} k*A121462(n, k). - Emeric Deutsch, Jul 31 2006
Number of 1s in all compositions of n, where compositions are understood with two different kinds of 1s, say 1 and 1' (n >= 1). Example: a(2) = 4 because the compositions of 2 are 11, 11', 1'1, 1'1', 2, having a total of 2 + 1 + 1 + 0 + 0 = 4 1s. Also number of k's in all compositions of n + k (k = 2, 3, ...). - Emeric Deutsch, Jul 21 2008
From Petros Hadjicostas, Jun 24 2019: (Start)
If c = (c(m): m >= 1) is the input sequence and b_k = (b_k(n): n >= 1) is the output sequence under the AIK[k] = INVERT[k] transform (see Bower's web link below), then the bivariate g.f. of the list of sequences (b_k: k >= 1) = ((b_k(n): n >= 1): k >= 1) is Sum_{n, k >= 1} b_k(n)*x^n*y^k = y*C(x)/(1 - y*C(x)), where C(x) = Sum_{m >= 1} c(m)*x^m is the g.f. of the input sequence.
Here, b_k(n) is the number of all (linear) compositions of n with k parts where a part of size m is colored with one of c(m) colors. Thus, Sum_{k = 1..n} k*b_k(n) is the total number of parts in all compositions of n.
If we differentiate the bivariate g.f. function above, i.e., Sum_{n, k >= 1} b_k(n)*x^n*y^k, with respect to y and set y = 1, we get the g.f. of the sequence (Sum_{k = 1..n} k*b_k(n): n >= 1). It is C(x)/(1 - C(x))^2.
When c(m) = m for all m >= 1, we have m-color compositions of n that were first studied by Agarwal (2000). The cyclic version of these m-color compositions were studied by Gibson (2017) and Gibson et al. (2018).
When c(m) = m for each m >= 1, we have C(x) = x/(1 - x)^2, and so C(x)/(1 - C(x))^2 = x * (1 - x)^2/(1 - 3*x + x^2)^2, which is the g.f. of the current sequence.
Hence, a(n) is the total number of parts in all m-color compositions of n (in the sense of Agarwal (2000)).
(End)
Series reversal gives A153294 starting from index 1, with alternating signs: 1, -4, 18, -86, 427, -2180, ... - Vladimir Reshetnikov, Aug 03 2019

Examples

			From _Petros Hadjicostas_, Jun 24 2019: (Start)
Recall that with m-color compositions, a part of size m may be colored with one of m colors.
We have a(1) = 1 because we only have one colored composition, namely 1_1, that has only 1 part.
We have a(2) = 4 because we have the following colored compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4.
We have a(3) = 14 because we have the following colored compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 2_1 + 1_1, 2_2 + 1_1, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 2 + 2 + 3 = 14.
We have a(14) = 46 because we have the following colored compositions of n = 4:
(i) 4_1, 4_2, 4_3, 4_4; with a total of 4 parts.
(ii) 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 3_1 + 1_1, 3_2 + 1_1, 3_3 + 1_1, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_1, 2_2 + 2_2; with a total of 2 x 10 = 20 parts.
(iii) 1_1 + 1_1 + 2_1, 1_1 + 1_1 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 2_1 + 1_1 + 1_1, 2_2 + 1_1 + 1_1; with a total of 3 x 6 = 18 parts.
(iv) 1_1 + 1_1 + 1_1 + 1_1; with a total of 4 parts.
Hence, a(4) = 4 + 20 + 18 + 4 = 46.
(End)
		

References

  • R. P. Grimaldi, Compositions and the alternate Fibonacci numbers, Congressus Numerantium, 186, 2007, 81-96.

Crossrefs

Partial sums of A038731. First differences of A001870.
Cf. A001629 (right-shifted inverse Binomial Transform), A023610 (inverse Binomial Transform of left-shifted sequence), A030279, A045623, A088305, A121462, A153294, A279282, A307415, A308723.

Programs

  • Maple
    with(combinat): L[0]:=2: L[1]:=1: for n from 2 to 60 do L[n]:=L[n-1] +L[n-2] end do: seq(2*fibonacci(2*n)*1/5+(1/5)*n*L[2*n],n=1..30); # Emeric Deutsch, Jul 21 2008
  • Mathematica
    Table[Sum[k Binomial[n+k-1,2k-1],{k,n}],{n,30}] (* or *) LinearRecurrence[ {6,-11,6,-1},{1,4,14,46},30] (* Harvey P. Dale, Aug 01 2011 *)
  • PARI
    a(n)=(2*n*fibonacci(2*n+1)+(2-n)*fibonacci(2*n))/5

Formula

a(n) = -a(-n) = (2n * F(2n+1) + (2 - n) * F(2n))/5 with F(n) = A000045(n) (Fibonacci numbers).
G.f.: x * (1 - x)^2/(1 - 3*x + x^2)^2.
a(n) = Sum_{k = 1..n} k*C(n + k - 1, 2*k - 1).
a(n) = (2/5)*F(2*n) + (1/5)*n*L(2*n), where F(k) are the Fibonacci numbers (F(0)=0, F(1)=1) and L(k) are the Lucas numbers (L(0) = 2, L(1) = 1). - Emeric Deutsch, Jul 21 2008
a(0) = 1, a(1) = 4, a(2) = 14, a(3) = 46, a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4). - Harvey P. Dale, Aug 01 2011
a(n) = ((3 - sqrt(5))^n*(5*n - 2*sqrt(5)) + (3 + sqrt(5))^n*(5*n + 2*sqrt(5)))/ (25*2^n). - Peter Luschny, Mar 07 2022
E.g.f.: exp(3*x/2)*(15*x*cosh(sqrt(5)*x/2) + sqrt(5)*(4 + 5*x)*sinh(sqrt(5)*x/2))/25. - Stefano Spezia, Mar 04 2025

Extensions

Name clarified using a comment of the author by Peter Luschny, Aug 03 2019

A058071 A Fibonacci triangle: triangle T(n,k) = Fibonacci(k+1)*Fibonacci(n-k+1), for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 2, 2, 3, 5, 3, 4, 3, 5, 8, 5, 6, 6, 5, 8, 13, 8, 10, 9, 10, 8, 13, 21, 13, 16, 15, 15, 16, 13, 21, 34, 21, 26, 24, 25, 24, 26, 21, 34, 55, 34, 42, 39, 40, 40, 39, 42, 34, 55, 89, 55, 68, 63, 65, 64, 65, 63, 68, 55, 89, 144, 89, 110, 102, 105, 104, 104, 105, 102, 110, 89, 144
Offset: 0

Views

Author

N. J. A. Sloane, Nov 24 2000

Keywords

Comments

Or, multiplication table of the positive Fibonacci numbers read by antidiagonals.
Or, triangle of products of nonzero Fibonacci numbers.
Or, a two-dimensional square Fibonacci array read by antidiagonals, with offset 1: T(1,1) = T(1,2) = T(2,1) = T(2,2) = 1; thereafter T(m,n) = max {T(m,n-2) + T(m,n-1), T(m-2,n) + T(m-1,n), T(m-2,n-2) + T(m-1,n-1)}. If "max" is changed to "min" we get A283845. - N. J. A. Sloane, Mar 31 2017
Row sums are A001629 (Fibonacci numbers convolved with themselves.). The main diagonal and first subdiagonal are Fibonacci numbers, for other entries T(n,k) = T(n-1,k) + T(n-2,k). The central numbers form A006498. - Gerald McGarvey, Jun 02 2005
Alternating row sums = (1, 0, 3, 0, 8, ...), given by Fibonacci(2n) if n even, else zero.
Row n = edge-counting vector for the Fibonacci cube F(n+1) embedded in the natural way in the hypercube Q(n+1). - Emanuele Munarini, Apr 01 2008
The augmentation of A058071 is the triangle A193595. To fit the definition of augmented triangle at A103091, it is helpful to represent A058071 using p(n,k)=F(k+1)*F(n+1-k) for 0<=k<=n. - Clark Kimberling, Jul 31 2011
T(n,k) = number of appearances of a(k) in p(n) in the n-th convergent p(n)/q(n) of the formal infinite continued fraction [a(0), a(1), ...]; e.g., p(3) = a(0)*a(1)*a(2)*a(3) + a(0)*a(1) + a(0)*a(3) + a(2)*a(3) + 1. Also, T(n,k) = number of appearances of a(k+1) in q(n+1); e.g., q(3) = a(1)*a(2)*a(3) + a(1) + a(3). - Clark Kimberling, Dec 21 2015
Each row is a palindrome, and the central term of row 2n is the square of the F(n+1), where F = A000045 (Fibonacci numbers). - Clark Kimberling, Dec 21 2015
Also called Hosoya's triangle, after the Japanese chemist Haruo Hosoya (b. 1936). - Amiram Eldar, Jun 10 2021

Examples

			Triangle begins as:
   1;
   1,  1;
   2,  1,  2;
   3,  2,  2,  3;
   5,  3,  4,  3,  5;
   8,  5,  6,  6,  5,  8;
  13,  8, 10,  9, 10,  8, 13;
  21, 13, 16, 15, 15, 16, 13, 21;
  34, 21, 26, 24, 25, 24, 26, 21, 34;
  ...
As a square array:
   1,  1,  2,  3,  5,  8, 13, 21, ...
   1,  1,  2,  3,  5,  8, 13, 21, ...
   2,  2,  4,  6, 10, 16, 26, ...
   3,  3,  6,  9, 15, 24, ...
   5,  5, 10, 15, 25, ...
   8,  8, 16, 24, ...
  13, 13, 26, ...
  21, 21, ...
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Chap. 15, Hosoya's Triangle, Wiley, New York, 2001.

Crossrefs

Programs

  • Haskell
    a058071 n k = a058071_tabl !! n !! k
    a058071_row n = a058071_tabl !! n
    a058071_tabl = map (\fs -> zipWith (*) fs $ reverse fs) a104763_tabl
    -- Reinhard Zumkeller, Aug 15 2013
    
  • Magma
    [Fibonacci(k+1)*Fibonacci(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 06 2022
    
  • Mathematica
    row[n_] := Table[Fibonacci[k]*Fibonacci[n-k+1], {k, 1, n}]; Table[row[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Dec 16 2013 *)
  • PARI
    T(n,k)=fibonacci(k)*fibonacci(n+2-k) \\ Charles R Greathouse IV, Feb 07 2017
    
  • SageMath
    flatten([[fibonacci(k+1)*fibonacci(n-k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 06 2022

Formula

Row n: F(1)*F(n), F(2)*F(n-1), ..., F(n)*F(1).
G.f.: T(x,y) = 1/((1-x-x^2)(1-xy-x^2y^2)). Recurrence: T(n+4,k+2) = T(n+3,k+2) + T(n+3,k+1) + T(n+2,k+2) - T(n+2,k+1) + T(n+2,k) - T(n+1,k+1) - T(n+1,k) - T(n,k). - Emanuele Munarini, Apr 01 2008
T(n,k) = A104763(n+1,k+1) * A104763(n+1,n+1-k). - Reinhard Zumkeller, Aug 15 2013
Column k is the (generalized) Fibonacci sequence having first two terms F(k+1), F(k+1). - Clark Kimberling, Dec 21 2015
From G. C. Greubel, Apr 06 2022: (Start)
T(n,k) = Fibonacci(k+1)*Fibonacci(n-k+1).
Sum_{k=0..n} T(n, k) = A001629(n+2).
Sum_{k=0..floor(n/2)} T(n, k) = A024458(n+1).
Sum_{k=1..n-1} T(n, k) = A004798(n-1), n >= 2.
Sum_{k=0..floor(n/2)} T(n-k, k) = A250111(n+2).
T(n, 0) = A000045(n+1).
T(2*n, n) = A007598(n+1).
T(2*n+1, n) = A001654(n+1).
T(n, n-k) = T(n, k). (End)

Extensions

More terms from James Sellers, Nov 27 2000
Edited by N. J. A. Sloane, Sep 15 2008 at the suggestion of R. J. Mathar
Name edited by G. C. Greubel, Apr 06 2022
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