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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A001835 a(n) = 4*a(n-1) - a(n-2), with a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481, 42037733184721, 156886956080403, 585510091136891
Offset: 0

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Author

Keywords

Comments

See A079935 for another version.
Number of ways of packing a 3 X 2*(n-1) rectangle with dominoes. - David Singmaster.
Equivalently, number of perfect matchings of the P_3 X P_{2(n-1)} lattice graph. - Emeric Deutsch, Dec 28 2004
The terms of this sequence are the positive square roots of the indices of the octagonal numbers (A046184) - Nicholas S. Horne (nairon(AT)loa.com), Dec 13 1999
Terms are the solutions to: 3*x^2 - 2 is a square. - Benoit Cloitre, Apr 07 2002
Gives solutions x > 0 of the equation floor(x*r*floor(x/r)) == floor(x/r*floor(x*r)) where r = 1 + sqrt(3). - Benoit Cloitre, Feb 19 2004
a(n) = L(n-1,4), where L is defined as in A108299; see also A001834 for L(n,-4). - Reinhard Zumkeller, Jun 01 2005
Values x + y, where (x, y) solves for x^2 - 3*y^2 = 1, i.e., a(n) = A001075(n) + A001353(n). - Lekraj Beedassy, Jul 21 2006
Number of 01-avoiding words of length n on alphabet {0,1,2,3} which do not end in 0. (E.g., for n = 2 we have 02, 03, 11, 12, 13, 21, 22, 23, 31, 32, 33.) - Tanya Khovanova, Jan 10 2007
sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571) + ... - Gary W. Adamson, Dec 18 2007
The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators = A001834, denominators = A001835. - Clark Kimberling, Aug 27 2008
From Gary W. Adamson, Jun 21 2009: (Start)
A001835 and A001353 = bisection of denominators of continued fraction [1, 2, 1, 2, 1, 2, ...]; i.e., bisection of A002530.
a(n) = determinant of an n*n tridiagonal matrix with 1's in the super- and subdiagonals and (3, 4, 4, 4, ...) as the main diagonal.
Also, the product of the eigenvalues of such matrices: a(n) = Product_{k=1..(n-1)/2)} (4 + 2*cos(2*k*Pi/n).
(End)
Let M = a triangle with the even-indexed Fibonacci numbers (1, 3, 8, 21, ...) in every column, and the leftmost column shifted up one row. a(n) starting (1, 3, 11, ...) = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n+1) is the number of compositions of n when there are 3 types of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2*n-2) X (2*n-2) tridiagonal matrix with sqrt(2)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Primes in the sequence are apparently those in A096147. - R. J. Mathar, May 09 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 4xy + y^2 + 2 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14xy + y^2 + 32 = 0. - Colin Barker, Feb 10 2014
The (1,1) element of A^n where A = (1, 1, 1; 1, 2, 1; 1, 1, 2). - David Neil McGrath, Jul 23 2014
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001075. - René Gy, Feb 25 2018
a(n+1) is the number of spanning trees of the graph T_n, where T_n is a 2 X n grid with an additional vertex v adjacent to (1,1) and (2,1). - Kevin Long, May 04 2018
a(n)/A001353(n) is the resistance of an n-ladder graph whose edges are replaced by one-ohm resistors. The resistance in ohms is measured at two nodes at one end of the ladder. It approaches sqrt(3) - 1 for n -> oo. See A342568, A357113, and A357115 for related information. - Hugo Pfoertner, Sep 17 2022
a(n) is the number of ways to tile a 1 X (n-1) strip with three types of tiles: small isosceles right triangles (with small side length 1), 1 X 1 squares formed by joining two of those right triangles along the hypotenuse, and large isosceles right triangles (with large side length 2) formed by joining two of those right triangles along a short leg. As an example, here is one of the a(6)=571 ways to tile a 1 X 5 strip with these kinds of tiles:
| / \ |\ /| |
|/_\|\/_||. - Greg Dresden and Arjun Datta, Jun 30 2023
From Klaus Purath, May 11 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 4xy + y^2 = -2 = A028872(-1). In general, the following applies to all sequences (t) satisfying t(i) = 4t(i-1) - t(i-2) with t(0) = 1 and two consecutive terms (x,y): x^2 - 4xy + y^2 = A028872(t(1)-2). This includes and interprets the Feb 04 2014 comments here and on A001075 by Colin Barker and the Dec 12 2012 comment on A001353 by Max Alekseyev. By analogy to this, for three consecutive terms (x,y,z) y^2 - xz = A028872(t(1)-2). This includes and interprets the Jul 10 2021 comment on A001353 by Bernd Mulansky.
If (t) is a sequence satisfying t(k) = 3t(k-1) + 3t(k-2) - t(k-3) or t(k) = 4t(k-1) - t(k-2) without regard to initial values and including this sequence itself, then a(n) = (t(k+2n+1) + t(k))/(t(k+n+1) + t(k+n)) always applies, as long as t(k+n+1) + t(k+n) != 0 for integer k and n >= 1. (End)
Binomial transform of 1, 0, 2, 4, 12, ... (A028860 without the initial -1) and reverse binomial transform of 1, 2, 6, 24, 108, ... (A094433 without the initial 1). - Klaus Purath, Sep 09 2024

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).
  • Leonhard Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.
  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics I, p. 292.

Crossrefs

Row 3 of array A099390.
Essentially the same as A079935.
First differences of A001353.
Partial sums of A052530.
Pairwise sums of A006253.
Bisection of A002530, A005246 and A048788.
First column of array A103997.
Cf. A001519, A003699, A082841, A101265, A125077, A001353, A001542, A096147 (subsequence of primes).

Programs

  • GAP
    a:=[1,1];; for n in [3..20] do a[n]:=4*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Haskell
    a001835 n = a001835_list !! n
    a001835_list =
       1 : 1 : zipWith (-) (map (4 *) $ tail a001835_list) a001835_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    [n le 2 select 1 else 4*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Sep 16 2016
    
  • Maple
    f:=n->((3+sqrt(3))^(2*n-1)+(3-sqrt(3))^(2*n-1))/6^n; [seq(simplify(expand(f(n))),n=0..20)]; # N. J. A. Sloane, Nov 10 2009
  • Mathematica
    CoefficientList[Series[(1-3x)/(1-4x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Jul 25 2011, after g.f. *)
    LinearRecurrence[{4,-1},{1,1},30] (* Harvey P. Dale, Jun 08 2013 *)
    Table[Round@Fibonacci[2n-1, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
    Table[(3*ChebyshevT[n, 2] - ChebyshevU[n, 2])/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)
  • PARI
    {a(n) = real( (2 + quadgen(12))^n * (1 - 1 / quadgen(12)) )} /* Michael Somos, Sep 19 2008 */
    
  • PARI
    {a(n) = subst( (polchebyshev(n) + polchebyshev(n-1)) / 3, x, 2)} /* Michael Somos, Sep 19 2008 */
    
  • Sage
    [lucas_number1(n,4,1)-lucas_number1(n-1,4,1) for n in range(25)] # Zerinvary Lajos, Apr 29 2009
    
  • Sage
    [(3*chebyshev_T(n,2) - chebyshev_U(n,2))/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019
    

Formula

G.f.: (1 - 3*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
a(1-n) = a(n).
a(n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n. - Dean Hickerson, Dec 01 2002
a(n) = (8 + a(n-1)*a(n-2))/a(n-3). - Michael Somos, Aug 01 2001
a(n+1) = Sum_{k=0..n} 2^k * binomial(n + k, n - k), n >= 0. - Len Smiley, Dec 09 2001
Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 10 2002
a(n) = 2*A061278(n-1) + 1 for n > 0. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n - i, i); then q(n, 2) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n+1) = Sum_{k=0..n} ((-1)^k)*((2*n+1)/(2*n + 1 - k))*binomial(2*n + 1 - k, k)*6^(n - k) (from standard T(n,x)/x, n >= 1, Chebyshev sum formula). The Smiley and Cloitre sum representation is that of the S(2*n, i*sqrt(2))*(-1)^n Chebyshev polynomial. - Wolfdieter Lang, Nov 29 2002
a(n) = S(n-1, 4) - S(n-2, 4) = T(2*n-1, sqrt(3/2))/sqrt(3/2) = S(2*(n-1), i*sqrt(2))*(-1)^(n - 1), with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x) = 0, S(-2, x) = -1, S(n, 4) = A001353(n+1), T(-1, x) = x.
a(n+1) = sqrt((A001834(n)^2 + 2)/3), n >= 0 (see Cloitre comment).
Sequence satisfies -2 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = (1/6)*(3*(2 - sqrt(3))^n + sqrt(3)*(2 - sqrt(3))^n + 3*(2 + sqrt(3))^n - sqrt(3)*(2 + sqrt(3))^n) (Mathematica's solution to the recurrence relation). - Sarah-Marie Belcastro, Jul 04 2009
If p[1] = 3, p[i] = 2, (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n+1) = det A. - Milan Janjic, Apr 29 2010
a(n) = (a(n-1)^2 + 2)/a(n-2). - Irene Sermon, Oct 28 2013
a(n) = A001353(n+1) - 3*A001353(n). - R. J. Mathar, Oct 30 2015
a(n) = a(n-1) + 2*A001353(n-1). - Kevin Long, May 04 2018
From Franck Maminirina Ramaharo, Nov 11 2018: (Start)
a(n) = (-1)^n*(A125905(n) + 3*A125905(n-1)), n > 0.
E.g.f.: exp^(2*x)*(3*cosh(sqrt(3)*x) - sqrt(3)*sinh(sqrt(3)*x))/3. (End)
From Peter Bala, Feb 12 2024: (Start)
For n in Z, a(n) = A001353(n) + A001353(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 2*A001353(j)*A001353(k). The case j = 1, k = 2 is given above. (End)

A181796 a(n) = number of divisors of n whose canonical prime factorizations contain no repeated positive exponents (cf. A130091).

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 2, 4, 3, 3, 2, 5, 2, 3, 3, 5, 2, 5, 2, 5, 3, 3, 2, 7, 3, 3, 4, 5, 2, 4, 2, 6, 3, 3, 3, 7, 2, 3, 3, 7, 2, 4, 2, 5, 5, 3, 2, 9, 3, 5, 3, 5, 2, 7, 3, 7, 3, 3, 2, 7, 2, 3, 5, 7, 3, 4, 2, 5, 3, 4, 2, 10, 2, 3, 5, 5, 3, 4, 2, 9, 5, 3, 2, 7, 3, 3, 3, 7, 2, 7, 3, 5, 3, 3, 3, 11, 2, 5, 5, 7, 2, 4, 2, 7, 4
Offset: 1

Views

Author

Matthew Vandermast, Nov 22 2010

Keywords

Comments

The canonical factorization of n into prime powers can be written as Product p(i)^e(i), for example. A host of equivalent notations can also be used (for another example, see Weisstein link). a(n) depends only on prime signature of n (cf. A025487).
a(n) >= A085082(n). (A085082(n) equals the number of members of A025487 that divide A046523(n), and each member of A025487 is divisible by at least one member of A130091 that divides no smaller member of A025487.) a(n) > A085082(n) iff n has in its canonical prime factorization at least two exponents greater than 1.
a(n) = number of such divisors of n that in their prime factorization all exponents are unique. - Antti Karttunen, May 27 2017
First differs from A335549 at a(90) = 7, A335549(90) = 8. First differs from A335516 at a(180) = 9, A335516(180) = 10. - Gus Wiseman, Jun 28 2020

Examples

			12 has a total of six divisors (1, 2, 3, 4, 6 and 12). Of those divisors, the number 1 has no prime factors, hence, no positive exponents at all (and no repeated positive exponents) in its canonical prime factorization. The lists of positive exponents for 2, 3, 4, 6 and 12 are (1), (1), (2), (1,1) and (2,1) respectively (cf. A124010). Of all six divisors, only the number 6 (2^1*3^1) has at least one positive exponent repeated (namely, 1). The other five do not; hence, a(12) = 5.
For n = 90 = 2 * 3^2 * 5, the divisors that satisfy the condition are: 1, 2, 3, 3^2, 5, 2 * 3^2, 3^2 * 5, altogether 7, (but for example 90 itself is not included), thus a(90) = 7.
		

Crossrefs

Diverges from A088873 at n=24 and from A085082 at n=36. a(36) = 7, while A085082(36) = 6.
Partitions with distinct multiplicities are A098859.
Sorted prime signature is A118914.
Unsorted prime signature is A124010.
a(n) is the number of divisors of n in A130091.
Factorizations with distinct multiplicities are A255231.
The largest of the counted divisors is A327498.
Factorizations using the counted divisors are A327523.

Programs

  • Mathematica
    Table[DivisorSum[n, 1 &, Length@ Union@ # == Length@ # &@ FactorInteger[#][[All, -1]] &], {n, 105}] (* Michael De Vlieger, May 28 2017 *)
  • PARI
    no_repeated_exponents(n) = { my(es = factor(n)[, 2]); if(length(Set(es)) == length(es),1,0); }
    A181796(n) = sumdiv(n,d,no_repeated_exponents(d)); \\ Antti Karttunen, May 27 2017
    
  • Python
    from sympy import factorint, divisors
    def ok(n):
        f=factorint(n)
        ex=[f[i] for i in f]
        for i in ex:
            if ex.count(i)>1: return 0
        return 1
    def a(n): return sum([1 for i in divisors(n) if ok(i)]) # Indranil Ghosh, May 27 2017

Formula

a(A000079(n)) = a(A002110(n)) = n+1.
a(A006939(n)) = A000110(n+1).
a(A181555(n)) = A002720(n).

A063007 T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 12, 30, 20, 1, 20, 90, 140, 70, 1, 30, 210, 560, 630, 252, 1, 42, 420, 1680, 3150, 2772, 924, 1, 56, 756, 4200, 11550, 16632, 12012, 3432, 1, 72, 1260, 9240, 34650, 72072, 84084, 51480, 12870, 1, 90, 1980, 18480, 90090, 252252, 420420, 411840, 218790, 48620
Offset: 0

Views

Author

Henry Bottomley, Jul 02 2001

Keywords

Comments

T(n,k) is the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's Cluster algebra of finite type B_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of triangle A008459 (squares of binomial coefficients). For example, x^2+6*x+6 = y^2+4*y+1. - Paul Boddington, Mar 07 2003
T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k N=(0,1) steps. E.g. T(2,0)=1 because we have DD; T(2,1) = 6 because we have NED, NDE, EDN, END, DEN and DNE; T(2,2)=6 because we have NNEE, NENE, NEEN, EENN, ENEN and ENNE. - Emeric Deutsch, Apr 20 2004
Another version of [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 2, 0; 1, 6, 6, 0; 1, 12, 30, 20, 0; ..., where DELTA is the operator defined in A084938. - Philippe Deléham Apr 15 2005
Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with increasing powers of x.
From Peter Bala, Oct 28 2008: (Start)
Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p.60]. See A008459 for the corresponding h-vectors for associahedra of type B_n and A001263 and A033282 respectively for the h-vectors and f-vectors for associahedra of type A_n.
An alternative description of this triangle in terms of f-vectors is as follows. Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i,j <= n+1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the f-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A008459 is the corresponding array of h-vectors for these type A_n polytopes. See A127674 (without the signs) for the array of f-vectors for type C_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes.
The S-transform on the ring of polynomials is the linear transformation of polynomials that is defined on the basis monomials x^k by S(x^k) = binomial(x,k) = x(x-1)...(x-k+1)/k!. Let P_n(x) denote the S-transform of the n-th row polynomial of this array. In the notation of [Hetyei] these are the Stirling polynomials of the type B associahedra. The first few values are P_1(x) = 2*x + 1, P_2(x) = 3*x^2 + 3*x + 1 and P_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials P_n(-x) satisfy a Riemann hypothesis. See A142995 for further details. The sequence of values P_n(k) for k = 0,1,2,3, ... produces the n-th row of A108625. (End)
This is the row reversed version of triangle A104684. - Wolfdieter Lang, Sep 12 2016
T(n, k) is also the number of (n-k)-dimensional faces of a convex n-dimensional Lipschitz polytope of real functions f defined on the set X = {1, 2, ..., n+1} which satisfy the condition f(n+1) = 0 (see Gordon and Petrov). - Stefano Spezia, Sep 25 2021
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial ((x+1)*(x+2)*(x+3)*...*(x+n) / n!)^2 in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022
Chapoton's observation above is correct: the precise expansion is ((x+1)*(x+2)*(x+3)*...*(x+n)/ n!)^2 = Sum_{k = 0..n} (-1)^k*T(n,n-k)*binomial(x+2*n-k, 2*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives ((x+1)*(x+2)*(x+3)/3!)^2 = 20*binomial(x+6,6) - 30*binomial(x+5,5) + 12*binomial(x+4,4) - binomial(x+3,3). - Peter Bala, Jun 24 2023

Examples

			The triangle T(n, k) starts:
  n\k 0  1    2     3     4      5      6      7      8     9
  0:  1
  1:  1  2
  2:  1  6    6
  3:  1 12   30    20
  4:  1 20   90   140    70
  5:  1 30  210   560   630    252
  6:  1 42  420  1680  3150   2772    924
  7:  1 56  756  4200 11550  16632  12012   3432
  8:  1 72 1260  9240 34650  72072  84084  51480  12870
  9:  1 90 1980 18480 90090 252252 420420 411840 218790 48620
... reformatted by _Wolfdieter Lang_, Sep 12 2016
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92, in Ser's book) is
   1;
  -1/2,  1/2;
   1/3, -1/2,    1/6;
  -1/4,  9/20,  -1/4,   1/20;
   1/5, -2/5,    2/7,  -1/10,  1/70;
  -1/6,  5/14, -25/84,  5/36, -1/28,  1/252;
   1/7, -9/28,  25/84, -1/6,   9/154, -1/84, 1/924;
   ... (End)
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 366.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, Table I, p. 92.
  • D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: from Neolithic Scots to John McKay, CRM Proc. Lecture Notes 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349-366.

Crossrefs

See A331430 for an essentially identical triangle, except with signed entries.
Columns include A000012, A002378, A033487 on the left and A000984, A002457, A002544 on the right.
Main diagonal is A006480.
Row sums are A001850. Alternating row sums are A033999.
Cf. A033282 (f-vectors type A associahedra), A108625, A080721 (f-vectors type D associahedra).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Haskell
    a063007 n k = a063007_tabl !! n !! k
    a063007_row n = a063007_tabl !! n
    a063007_tabl = zipWith (zipWith (*)) a007318_tabl a046899_tabl
    -- Reinhard Zumkeller, Nov 18 2014
    
  • Magma
    /* As triangle: */ [[Binomial(n,k)*Binomial(n+k,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 03 2015
  • Maple
    p := (n,x) -> orthopoly[P](n,1+2*x): seq(seq(coeff(p(n,x),x,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Binomial[n, k]Binomial[n + k, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Dec 24 2011 *)
    Table[CoefficientList[Hypergeometric2F1[-n, n + 1, 1, -x], x], {n, 0, 9}] // Flatten
    (* Peter Luschny, Mar 09 2018 *)
  • PARI
    {T(n, k) = local(t); if( n<0, 0, t = (x + x^2)^n; for( k=1, n, t=t'); polcoeff(t, k) / n!)} /* Michael Somos, Dec 19 2002 */
    
  • PARI
    {T(n, k) = binomial(n, k) * binomial(n+k, k)} /* Michael Somos, Sep 22 2013 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n+k)! / (k!^2 * (n-k)!))} /* Michael Somos, Sep 22 2013 */
    

Formula

T(n, k) = (n+k)!/(k!^2*(n-k)!) = T(n-1, k)*(n+k)/(n-k) = T(n, k-1)*(n+k)*(n-k+1)/k^2 = T(n-1, k-1)*(n+k)*(n+k-1)/k^2.
binomial(x, n)^2 = Sum_{k>=0} T(n,k) * binomial(x, n+k). - Michael Somos, May 11 2012
T(n, k) = A109983(n, k+n). - Michael Somos, Sep 22 2013
G.f.: G(t, z) = 1/sqrt(1-2*z-4*t*z+z^2). Row generating polynomials = P_n(1+2z), i.e., T(n, k) = [z^k] P_n(1+2*z), where P_n are the Legendre polynomials. - Emeric Deutsch, Apr 20 2004
Sum_{k>=0} T(n, k)*A000172(k) = Sum_{k>=0} T(n, k)^2 = A005259(n). - Philippe Deléham, Jun 08 2005
1 + z*d/dz(log(G(t,z))) = 1 + (1 + 2*t)*z + (1 + 8*t + 8*t^2)*z^2 + ... is the o.g.f. for a signed version of A127674. - Peter Bala, Sep 02 2015
If R(n,t) denotes the n-th row polynomial then x^3 * exp( Sum_{n >= 1} R(n,t)*x^n/n ) = x^3 + (1 + 2*t)*x^4 + (1 + 5*t + 5*t^2)*x^5 + (1 + 9*t + 21*t^2 + 14*t^3)*x^6 + ... is an o.g.f for A033282. - Peter Bala, Oct 19 2015
P(n,x) := 1/(1 + x)*Integral_{t = 0..x} R(n,t) dt are (modulo differences of offset) the row polynomials of A033282. - Peter Bala, Jun 23 2016
From Peter Bala, Mar 09 2018: (Start)
R(n,x) = Sum_{k = 0..n} binomial(2*k,k)*binomial(n+k,n-k)*x^k.
R(n,x) = Sum_{k = 0..n} binomial(n,k)^2*x^k*(1 + x)^(n-k).
n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x).
R(n,x) = (-1)^n*R(n,-1 - x).
R(n,x) = 1/n! * (d/dx)^n ((x^2 + x)^n). (End)
The row polynomials are R(n,x) = hypergeom([-n, n + 1], [1], -x). - Peter Luschny, Mar 09 2018
T(n,k) = C(n+1,k)*A009766(n,k). - Bob Selcoe, Jan 18 2020 (Connects this triangle with the Catalan triangle. - N. J. A. Sloane, Jan 18 2020)
If we let A(n,k) = (-1)^(n+k)*(2*k+1)*(n*(n-1)*...*(n-(k-1)))/((n+1)*...*(n+(k+1))) for n >= 0 and k = 0..n, and we consider both T(n,k) and A(n,k) as infinite lower triangular arrays, then they are inverses of one another. (Empty products are by definition 1.) See the example below. The rational numbers |A(n,k)| appear in Table II on p. 92 in Ser's (1933) book. - Petros Hadjicostas, Jul 11 2020
From Peter Bala, Nov 28 2021: (Start)
Row polynomial R(n,x) = Sum_{k >= n} binomial(k,n)^2 * x^(k-n)/(1+x)^(k+1) for x > -1/2.
R(n,x) = 1/(1 + x)^(n+1) * hypergeom([n+1, n+1], [1], x/(1 + x)).
R(n,x) = (1 + x)^n * hypergeom([-n, -n], [1], x/(1 + x)).
R(n,x) = hypergeom([(n+1)/2, -n/2], [1], -4*x*(1 + x)).
If we set R(-1,x) = 1, we can run the recurrence n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x) backwards to give R(-n,x) = R(n-1,x).
R(n,x) = [t^n] ( (1 + t)*(1 + x*(1 + t)) )^n. (End)
n*T(n,k) = (2*n-1)*T(n-1,k) + (4*n-2)*T(n-1,k-1) - (n-1)*T(n-2,k). - Fabián Pereyra, Jun 30 2022
From Peter Bala, Oct 07 2024: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n, k) * x^k o (1 + x)^(n-k), where o denotes the black diamond product of power series as defined by Dukes and White (see Bala, Section 4.4, exercise 3).
Denote this triangle by T. Then T * transpose(T) = A143007, the square array of crystal ball sequences for the A_n X A_n lattices.
Let S denote the triangle ((-1)^(n+k)*T(n, k))n,k >= 0, a signed version of this triangle. Then S^(-1) * T = A007318, Pascal's triangle; it appears that T * S^(-1) = A110098.
T = A007318 * A115951. (End)

A021009 Triangle of coefficients of Laguerre polynomials n!*L_n(x) (rising powers of x).

Original entry on oeis.org

1, 1, -1, 2, -4, 1, 6, -18, 9, -1, 24, -96, 72, -16, 1, 120, -600, 600, -200, 25, -1, 720, -4320, 5400, -2400, 450, -36, 1, 5040, -35280, 52920, -29400, 7350, -882, 49, -1, 40320, -322560, 564480, -376320, 117600, -18816, 1568, -64, 1, 362880, -3265920
Offset: 0

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Keywords

Comments

In absolute values, this sequence also gives the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals (j-1)^2 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
A partial permutation on a set X is a bijection between two subsets of X. |T(n,n-k)| equals the numbers of partial permutations of an n-set having domain cardinality equal to k. Let E denote the operator D*x*D, where D is the derivative operator d/dx. Then E^n = Sum_{k = 0..n} |T(n,k)|*x^k*D^(n+k). - Peter Bala, Oct 28 2008
The unsigned triangle is the generalized Riordan array (exp(x), x) with respect to the sequence n!^2 as defined by Wang and Wang (the generalized Riordan array (exp(x), x) with respect to the sequence n! is Pascal's triangle A007318, and with respect to the sequence n!*(n+1)! is A105278). - Peter Bala, Aug 15 2013
The unsigned triangle appears on page 83 of Ser (1933). - N. J. A. Sloane, Jan 16 2020

Examples

			The triangle a(n,m) starts:
n\m   0       1      2       3      4      5    6  7  8
0:    1
1:    1      -1
2:    2      -4      1
3:    6     -18      9      -1
4:   24     -96     72     -16      1
5:  120    -600    600    -200     25     -1
6:  720   -4320   5400   -2400    450    -36    1
7: 5040  -35280  52920  -29400   7350   -882   49  -1
8:40320 -322560 564480 -376320 117600 -18816 1568 -64 1
...
From _Wolfdieter Lang_, Jan 31 2013 (Start)
Recurrence (usual one): a(4,1) = 7*(-18) - 6 - 3^2*(-4) = -96.
Recurrence (simplified version): a(4,1) = 5*(-18) - 6 = -96.
Recurrence (Sage program): |a(4,1)| = 6 + 3*18 + 4*9 = 96. (End)
Embedded recurrence (Maple program): a(4,1) = -4!*(1 + 3) = -96.
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 799.
  • G. Rota, Finite Operator Calculus, Academic Press, New York, 1975.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 83.

Crossrefs

Row sums give A009940, alternating row sums are A002720.
Column sequences (unsigned): A000142, A001563, A001809-A001812 for m=0..5.
Central terms: A295383.
For generators and generalizations see A132440.

Programs

  • Magma
    /* As triangle: */ [[((-1)^k)*Factorial(n)*Binomial(n, k)/Factorial(k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 18 2020
  • Maple
    A021009 := proc(n,k) local S; S := proc(n,k) option remember; `if`(k = 0, 1, `if`( k > n, 0, S(n-1,k-1)/k + S(n-1,k))) end: (-1)^k*n!*S(n,k) end: seq(seq(A021009(n,k), k=0..n), n=0..8); # Peter Luschny, Jun 21 2017
    # Alternative for the unsigned case (function RiordanSquare defined in A321620):
    RiordanSquare(add(x^m, m=0..10), 10, true); # Peter Luschny, Dec 06 2018
  • Mathematica
    Flatten[ Table[ CoefficientList[ n!*LaguerreL[n, x], x], {n, 0, 9}]] (* Jean-François Alcover, Dec 13 2011 *)
  • PARI
    p(n) = denominator(bestapprPade(Ser(vector(2*n, k, (k-1)!))));
    concat(1, concat(vector(9, n, Vec(-p(n)))))  \\ Gheorghe Coserea, Dec 01 2016
    
  • PARI
    {T(n, k) = if( n<0, 0, n! * polcoeff( sum(i=0, n, binomial(n, n-i) * (-x)^i / i!), k))}; /* Michael Somos, Dec 01 2016 */
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n)); \\ Michel Marcus, Feb 06 2021
    
  • Sage
    def A021009_triangle(dim): # computes unsigned T(n,k).
        M = matrix(ZZ,dim,dim)
        for n in (0..dim-1): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+(2*k+1)*M[n-1,k]+(k+1)^2*M[n-1,k+1]
        return M
    A021009_triangle(9) # Peter Luschny, Sep 19 2012
    

Formula

a(n, m) = ((-1)^m)*n!*binomial(n, m)/m! = ((-1)^m)*((n!/m!)^2)/(n-m)! if n >= m, otherwise 0.
E.g.f. for m-th column: (-x/(1-x))^m /((1-x)*m!), m >= 0.
Representation (of unsigned a(n, m)) as special values of Gauss hypergeometric function 2F1, in Maple notation: n!*(-1)^m*hypergeom([ -m, n+1 ], [ 1 ], 1)/m!. - Karol A. Penson, Oct 02 2003
Sum_{m>=0} (-1)^m*a(n, m) = A002720(n). - Philippe Deléham, Mar 10 2004
E.g.f.: (1/(1-x))*exp(x*y/(x-1)). - Vladeta Jovovic, Apr 07 2005
Sum_{n>=0, m>=0} a(n, m)*(x^n/n!^2)*y^m = exp(x)*BesselJ(0, 2*sqrt(x*y)). - Vladeta Jovovic, Apr 07 2005
Matrix square yields the identity matrix: L^2 = I. - Paul D. Hanna, Nov 22 2008
From Tom Copeland, Oct 20 2012: (Start)
Symbolically, with D=d/dx and LN(n,x)=n!L_n(x), define :Dx:^j = D^j x^j, :xD:^j = x^j D^j, and LN(.,x)^j = LN(j,x) = row polynomials of A021009.
Then some useful relations are
1) (:Dx:)^n = LN(n,-:xD:) [Rodriguez formula]
2) (xDx)^n = x^n D^n x^n = x^n LN(n,-:xD:) [See Al-Salam ref./A132440]
3) (DxD)^n = D^n x^n D^n = LN(n,-:xD:) D^n [See ref. in A132440]
4) umbral composition LN(n,LN(.,x))= x^n [See Rota ref.]
5) umbral comp. LN(n,-:Dx:) = LN(n,-LN(.,-:xD:)) = 2^n LN(n,-:xD:/2)= n! * (n-th row e.g.f.(x) of A038207 with x replaced by :xD:).
An example for 2) is the operator (xDx)^2 = (xDx)(xDx) = xD(x^2 + x^3D)= 2x^2 + 4x^3 D + x^4 D^2 = x^2 (2 + 4x D + x^2 D^2) = x^2 (2 + 4 :xD: + :xD:^2) = x^2 LN(2,-:xD:) = x^2 2! L_2(-:xD:).
An example of the umbral composition in 5) is given in A038207.
The op. xDx is related to the Euler/binomial transformation for power series/o.g.f.s. through exp(t*xDx) f(x) = f[x/(1-t*x)]/(1-t*x) and to the special Moebius/linear fractional/projective transformation z exp(-t*zDz)(1/z)f(z) = f(z/(1+t*z)).
For a general discussion of umbral calculus see the Gessel link. (End)
From Wolfdieter Lang, Jan 31 2013: (Start)
Standard recurrence derived from the three term recurrence of the orthogonal polynomials system {n!*L(n,x)}: L(n,x) = (2*n - 1 - x)*L(n-1,x) - (n-1)^2*L(n-2,x), n>=1, L(-1,x) = 0, L(0,x) = 1.
a(n,m) = (2*n-1)*a(n-1,m) - a(n-1,m-1) - (n-1)^2*a(n-2,m),
n >=1, with a(n,-1) = 0, a(0,0) = 1, a(n,m) = 0 if n < m. (compare this with Peter Luschny's program for the unsigned case |a(n,m)| = (-1)^m*a(n,m)).
Simplified recurrence (using column recurrence from explicit form for a(n,m) given above):
a(n,m) = (n+m)*a(n-1,m) - a(n-1,m-1), n >= 1, a(0,0) = 1, a(n,-1) = 0, a(n,m) = 0 if n < m. (End)
|T(n,k)| = [x^k] (-1)^n*U(-n,1,-x), where U(a,b,x) is Kummer's hypergeometric U function. - Peter Luschny, Apr 11 2015
T(n,k) = (-1)^k*n!*S(n,k) where S(n,k) is recursively defined by: "if k = 0 then 1 else if k > n then 0 else S(n-1,k-1)/k + S(n-1,k)". - Peter Luschny, Jun 21 2017
The unsigned case is the exponential Riordan square (see A321620) of the factorial numbers. - Peter Luschny, Dec 06 2018
Omitting the diagonal and signs, this array is generated by the commutator [D^n,x^n] = D^n x^n - x^n D^n = Sum_{i=0..n-1} ((n!/i!)^2/(n-i)!) x^i D^i on p. 9 of both papers by Belov-Kanel and Kontsevich. - Tom Copeland, Jan 23 2020

Extensions

Name changed and table given by Wolfdieter Lang, Nov 28 2011

A052852 Expansion of e.g.f.: (x/(1-x))*exp(x/(1-x)).

Original entry on oeis.org

0, 1, 4, 21, 136, 1045, 9276, 93289, 1047376, 12975561, 175721140, 2581284541, 40864292184, 693347907421, 12548540320876, 241253367679185, 4909234733857696, 105394372192969489, 2380337795595885156, 56410454014314490981, 1399496554158060983080
Offset: 0

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Author

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

Keywords

Comments

A simple grammar.
Number of {121,212}-avoiding n-ary words of length n. - Ralf Stephan, Apr 20 2004
The infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...))) converges to the rational number A052852(n)/A000262(n) when n is a positive integer. - David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008
a(n) is the total number of components summed over all nilpotent partial permutations of [n]. - Geoffrey Critzer, Feb 19 2022

Crossrefs

Row sums of unsigned triangle A062139 (generalized a=2 Laguerre).

Programs

  • Magma
    [n eq 0 select 0 else Factorial(n)*Evaluate(LaguerrePolynomial(n-1, 0), -1): n in [0..30]]; // G. C. Greubel, Feb 23 2021
  • Maple
    spec := [S,{B=Set(C),C=Sequence(Z,1 <= card),S=Prod(B,C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
    a := n -> ifelse(n = 0, 0, n!*hypergeom([-n+1], [1], -1)): seq(simplify(a(n)), n = 0..18);  # Peter Luschny, Dec 30 2024
  • Mathematica
    Table[n!*SeriesCoefficient[(x/(1-x))*E^(x/(1-x)),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 09 2012 *)
    Table[If[n==0, 0, n!*LaguerreL[n-1, 0, -1]], {n, 0, 30}] (* G. C. Greubel, Feb 23 2021 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(serlaplace((x/(1-x))*exp(x/(1-x))))) \\ G. C. Greubel, May 15 2018
    
  • Sage
    [0 if n==0 else factorial(n)*gen_laguerre(n-1, 0, -1) for n in (0..30)] # G. C. Greubel, Feb 23 2021
    

Formula

D-finite with recurrence: a(1)=1, a(0)=0, (n^2+2*n)*a(n)+(-4-2*n)*a(n+1)+ a(n+2)=0.
a(n) = Sum_{m=0..n} n!*binomial(n+2, n-m)/m!. - Wolfdieter Lang, Jun 19 2001
a(n) = n*A002720(n-1). [Riordan] - Vladeta Jovovic, Mar 18 2005
Related to an n-dimensional series: for n>=1, a(n) = (n!/e)*Sum_{k_n>=k_{n-1}>=...>=k_1>=0} 1/(k_n)!. - Benoit Cloitre, Sep 30 2006
E.g.f.: (x/(1-x))*exp((x/(1-x))) = (x/(1-x))*G(0); G(k)=1+x/((2*k+1)*(1-x)-x*(1-x)*(2*k+1)/(x+(1-x)*(2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
a(n) = D^n(x*exp(x)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A000262 and A005493. - Peter Bala, Nov 25 2011
a(n) ~ exp(2*sqrt(n)-n-1/2)*n^(n+1/4)/sqrt(2). - Vaclav Kotesovec, Oct 09 2012
a(n) = n!*hypergeom([-n+1], [1], -1) for n>=1. - Peter Luschny, Oct 18 2014 [Simplified by Natalia L. Skirrow, 30 December 2024]
a(n) = Sum_{k=0..n} L(n,k)*k; L(n,k) the unsigned Lah numbers. - Peter Luschny, Oct 18 2014
a(n) = n!*LaguerreL(n-1, 0, -1) for n>=1. - Peter Luschny, Apr 08 2015, simplified Dec 30 2024
The series reversion of the e.g.f. equals W(x)/(1 + W(x)) = x - 2^2*x^2/2! + 3^3*x^3/3! - 4^4*x^4/4! + ..., essentially the e.g.f. for a signed version of A000312, where W(x) is Lambert's W-function (see A000169). - Peter Bala, Jun 14 2016
Equals column A059114(n, 2) for n >= 1. - G. C. Greubel, Feb 23 2021
a(n) = Sum_{k=1..n} k * A271703(n,k). - Geoffrey Critzer, Feb 19 2022

A004253 a(n) = 5*a(n-1) - a(n-2), with a(1)=1, a(2)=4.

Original entry on oeis.org

1, 4, 19, 91, 436, 2089, 10009, 47956, 229771, 1100899, 5274724, 25272721, 121088881, 580171684, 2779769539, 13318676011, 63813610516, 305749376569, 1464933272329, 7018916985076, 33629651653051, 161129341280179, 772017054747844, 3698955932459041, 17722762607547361
Offset: 1

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Author

Keywords

Comments

Number of domino tilings in K_3 X P_2n (or in S_4 X P_2n).
Number of perfect matchings in graph C_{3} X P_{2n}.
Number of perfect matchings in S_4 X P_2n.
In general, Sum_{k=0..n} binomial(2*n-k, k)*j^(n-k) = (-1)^n * U(2*n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,5), where L is defined as in A108299; see also A030221 for L(n,-5). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4} which do not end in 0 (e.g., at n=2, we have 02, 03, 04, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44). - Tanya Khovanova, Jan 10 2007
(sqrt(21)+5)/2 = 4.7912878... = exp(arccosh(5/2)) = 4 + 3/4 + 3/(4*19) + 3/(19*91) + 3/(91*436) + ... - Gary W. Adamson, Dec 18 2007
a(n+1) is the number of compositions of n when there are 4 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(3)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Right-shifted Binomial Transform of the left-shifted A030195. - R. J. Mathar, Oct 15 2012
Values of x (or y) in the solutions to x^2 - 5xy + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
From Wolfdieter Lang, Oct 15 2020: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 5*x*y = -3 (see the preceding comment) are given by [x(n) = S(n, 5) - S(n-1, 5), y(n) = x(n-1)], for n =-oo..+oo, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form has discriminant D = +21. There is only this family representing -3 properly with x and y positive, and there are no improper solutions.
See the formula for a(n) = x(n-1), for n >= 1, in terms of S-polynomials below.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
From Wolfdieter Lang, Feb 08 2021: (Start)
All proper and improper solutions of the generalized Pell equation X^2 - 21*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n+1) from the preceding comment by X(n) = x(n) + x(n-1) = S(n-1, 5) - S(n-2, 5) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 5), for all integer numbers n. For positive integers X(n) = A003501(n) and Y(n) = A004254(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper families of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer n. This follows from the mentioned paper by Robert K. Moniot. (End)
Equivalent definition: a(n) = ceiling(a(n-1)^2 / a(n-2)), with a(1)=1, a(2)=4, a(3)=19. The problem for USA Olympiad (see Andreescu and Gelca reference) asks to prove that a(n)-1 is always a multiple of 3. - Bernard Schott, Apr 13 2022

References

  • Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 311, pp. 104 and 466-467 (proposed for the USA Mathematical Olympiad by G. Heuer).
  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
  • F. A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A003501, A004254, A030221, A049310, A004254 (partial sums), A290902 (first differences).
Row 5 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • GAP
    a:=[1,4];; for n in [3..30] do a[n]:=5*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Oct 23 2019
  • Magma
    [ n eq 1 select 1 else n eq 2 select 4 else 5*Self(n-1)-Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=5*a[n-1]-a[n-2] od: seq(a[n], n=1..22); # Zerinvary Lajos, Jul 26 2006
  • Mathematica
    LinearRecurrence[{5, -1}, {1, 4}, 22] (* Jean-François Alcover, Sep 27 2017 *)
  • PARI
    Vec((1-x)/(1-5*x+x^2)+O(x^30)) \\ Charles R Greathouse IV, Jul 01 2013
    
  • Sage
    [lucas_number1(n,5,1)-lucas_number1(n-1,5,1) for n in range(1, 23)] # Zerinvary Lajos, Nov 10 2009
    

Formula

G.f.: x*(1 - x) / (1 - 5*x + x^2). Simon Plouffe in his 1992 dissertation.[offset 0]
For n>1, a(n) = A005386(n) + A005386(n-1). - Floor van Lamoen, Dec 13 2006
a(n) ~ (1/2 + 1/14*sqrt(21))*(1/2*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002[offset 0]
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then q(n, 3)=a(n). - Benoit Cloitre, Nov 10 2002 [offset 0]
For n>0, a(n)*a(n+3) = 15 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*3^k. - Paul Barry, Jul 26 2004[offset 0]
a(n) = (-1)^n*U(2n, i*sqrt(3)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005[offset 0]
[a(n), A004254(n)] = the 2 X 2 matrix [1,3; 1,4]^n * [1,0]. - Gary W. Adamson, Mar 19 2008
a(n) = ((sqrt(21)-3)*((5+sqrt(21))/2)^n + (sqrt(21)+3)*((5-sqrt(21))/2)^n)/2/sqrt(21). - Seiichi Kirikami, Sep 06 2011
a(n) = S(n-1, 5) - S(n-2, 5) = (-1)^n*S(2*n, i*sqrt(3)), n >= 1, with the Chebyshev S polynomials (A049310), and S(n-1, 5) = A004254(n), for n >= 0. See a Paul Barry formula (offset corrected). - Wolfdieter Lang, Oct 15 2020
From Peter Bala, Feb 10 2024: (Start)
a(n) = a(1-n).
a(n) = A004254(n) + A004254(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 3*A004254(j)*A004254(k). The case j = 1, k = 2 is given above.
a(n)^2 + a(n+1)^2 - 5*a(n)*a(n+1) = - 3.
More generally, a(n)^2 + a(n+k)^2 - (A004254(k+1) - A004254(k-1))*a(n)*a(n+k) = -3*A004254(k)^2. (End)
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/3 (telescoping series: for n >= 2, 3/(a(n) - 1/a(n)) = 1/A004254(n-1) - 1/A004254(n)). - Peter Bala, May 21 2025
E.g.f.: exp(5*x/2)*(7*cosh(sqrt(21)*x/2) - sqrt(21)*sinh(sqrt(21)*x/2))/7 - 1. - Stefano Spezia, Jul 02 2025

Extensions

Additional comments from James Sellers and N. J. A. Sloane, May 03 2002
More terms from Ray Chandler, Nov 17 2003

A049685 a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).

Original entry on oeis.org

1, 6, 41, 281, 1926, 13201, 90481, 620166, 4250681, 29134601, 199691526, 1368706081, 9381251041, 64300051206, 440719107401, 3020733700601, 20704416796806, 141910183877041, 972666870342481, 6666757908520326, 45694638489299801, 313195711516578281
Offset: 0

Views

Author

Keywords

Comments

In general, Sum_{k=0..n} binomial(2*n-k,k)j^(n-k) = (-1)^n*U(2n, I*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,7), where L is defined as in A108299; see also A033890 for L(n,-7). - Reinhard Zumkeller, Jun 01 2005
Take 7 numbers consisting of 5 ones together with any two successive terms from this sequence. This set has the property that the sum of their squares is 7 times their product. (R. K. Guy, Oct 12 2005.) See also A111216.
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
From Wolfdieter Lang, Feb 09 2021: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 7*x*y = -5 are given by [x(n) = S(n, 7) - S(n-1, 7), y(n) = x(n-1)], for all integer numbers n, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-n, x) = -S(n-2, x), for n >= 2. x(n) = a(n), for n >= 0.
This indefinite binary quadratic form has discriminant D = +45. There is only this family representing -5 properly with x and y positive, and there are no improper solutions.
All proper and improper solutions of the generalized Pell equation X^2 - 45*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n) from the preceding comment, by X(n) = x(n) + x(n-1) = S(n-1, 7) - S(n-2, 7) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 7), for all integer numbers n. For positive integers X(n) = A056854(n) and Y(n) = A004187(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper family of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer numbers n.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

Examples

			a(3) = L(4*3 + 2)/3 = 843/3 = 281. - _Indranil Ghosh_, Feb 06 2017
		

Crossrefs

Row 7 of array A094954. First differences of A004187.
Cf. similar sequences listed in A238379.

Programs

  • Magma
    [Lucas(4*n+2)/3: n in [0..30]]; // G. C. Greubel, Dec 17 2017
  • Mathematica
    Table[LucasL[4*n+2]/3, {n,0,50}] (* or *) LinearRecurrence[{7,-1}, {1,6}, 50] (* G. C. Greubel, Dec 17 2017 *)
  • PARI
    a(n)=(fibonacci(4*n+1)+fibonacci(4*n+3))/3 \\ Charles R Greathouse IV, Jun 16 2014
    
  • Sage
    [lucas_number1(n,7,1)-lucas_number1(n-1,7,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
    

Formula

Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i); then q(n, 5)=a(n); a(n) = 7a(n-1) - a(n-2). - Benoit Cloitre, Nov 10 2002
From Ralf Stephan, May 29 2004: (Start)
a(n+2) = 7a(n+1) - a(n).
G.f.: (1-x)/(1-7x+x^2).
a(n)*a(n+3) = 35 + a(n+1)*a(n+2). (End)
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*5^k. - Paul Barry, Aug 30 2004
If another "1" is inserted at the beginning of the sequence, then A002310, A002320 and A049685 begin with 1, 2; 1, 3; and 1, 1; respectively and satisfy a(n+1) = (a(n)^2+5)/a(n-1). - Graeme McRae, Jan 30 2005
a(n) = (-1)^n*U(2n, i*sqrt(5)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
[a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = S(n, 7) - S(n-1, 7) with Chebyshev S polynomials S(n-1, 7) = A004187(n), for n >= 0. - Wolfdieter Lang, Feb 09 2021
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/3. - Stefano Spezia, Apr 14 2025
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/9) * sqrt(1 - T(2*n+1, -7/2)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 6, 0, 41, 0, 281, 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 = -9, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/5 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A290903(n-1) - 1/A290903(n).) (End)

A277373 a(n) = Sum_{k=0..n} binomial(n,n-k)*n^(n-k)*n!/(n-k)!.

Original entry on oeis.org

1, 2, 14, 168, 2840, 61870, 1649232, 51988748, 1891712384, 78031713690, 3598075308800, 183396819358192, 10239159335648256, 621414669926828102, 40733145577028065280, 2867932866586451980500, 215859025837098699948032, 17295664826665032427023922, 1469838791737283957748596736
Offset: 0

Views

Author

Peter Luschny, Oct 12 2016

Keywords

Comments

Limit_{n -> infinity} (LaguerreL(n,-n)/BesselI(0,2*n))^(1/n) = exp(-2 + 1/phi) * phi^2 = 0.657347578792874..., where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 12 2016
For m > 0, n!*LaguerreL(n, -m*n) ~ sqrt(1/2 + (m+2)/(2*sqrt(m*(m+4)))) * (2+m+sqrt(m*(m+4)))^n * exp(n*(sqrt(m*(m+4))-m-2)/2) * n^n / 2^n. - Vaclav Kotesovec, Oct 14 2016
For m > 4, (-1)^n * n! * LaguerreL(n, m*n) ~ sqrt(1/2 + (m-2)/(2*sqrt(m*(m-4)))) * exp((m - 2 - sqrt(m*(m-4)))*n/2) * ((m - 2 + sqrt(m*(m-4)))/2)^n * n^n. - Vaclav Kotesovec, Feb 20 2020

Crossrefs

Cf. A002720 (n!L(n,-1)), A087912 (n!L(n,-2)), A277382 (n!L(n,-3)), A277372 (n!L(n,-n)-n^n), A277423 (n!L(n,n)), A144084 (polynomials).
Cf. A277391 (n!L(n,-2*n)), A277392 (n!L(n,-3*n)), A277418 (n!L(n,-4*n)), A277419 (n!L(n,-5*n)), A277420 (n!L(n,-6*n)), A277421 (n!L(n,-7*n)), A277422 (n!L(n,-8*n)).
Main diagonal of A289192.

Programs

  • Magma
    [(&+[Binomial(n, n-k)*Binomial(n, k)*n^(n-k)*Factorial(k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, May 16 2018
  • Maple
    A277373 := n -> n!*LaguerreL(n, -n): seq(simplify(A277373(n)), n=0..18);
    # second Maple program:
    a:= n-> n! * add(binomial(n, i)*n^i/i!, i=0..n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Jun 27 2017
  • Mathematica
    Table[n!*LaguerreL[n, -n], {n, 0, 30}] (* G. C. Greubel, May 16 2018 *)
  • PARI
    a(n) = sum(k=0,n, binomial(n,n-k)*n^(n-k)*n!/(n-k)!) \\ Charles R Greathouse IV, Feb 07 2017
    
  • PARI
    a(n) = n!*pollaguerre(n, 0, -n); \\ Michel Marcus, Feb 05 2021
    
  • Sage
    @cached_function
    def L(n, x):
        if n == 0: return 1
        if n == 1: return 1 - x
        return (L(n-1,x) * (2*n-1-x) - L(n-2,x)*(n-1))/n
    A277373 = lambda n: factorial(n)*L(n, -n)
    print([A277373(n) for n in (0..20)])
    

Formula

a(n) = p(n,n) where p(n,x) = Sum_{k=0..n} binomial(n,n-k)*x^(n-k)*n!/(n-k)!. The coefficients of these polynomials are in A144084 (sorted by falling powers).
a(n) = n!*LaguerreL(n, -n).
a(n) = (-1)^n*KummerU(-n, 1, -n).
a(n) = n^n*hypergeom([-n, -n], [], 1/n) for n>=1.
a(n) ~ n^n * phi^(2*n+1) * exp(n/phi-n) / 5^(1/4), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 12 2016
a(n) = n! * [x^n] exp(n*x/(1-x))/(1-x). - Alois P. Heinz, Jun 28 2017
a(n) = n!^2 * [x^n] exp(x) * BesselI(0,2*sqrt(n*x)). - Ilya Gutkovskiy, Jun 19 2022

A023998 Number of block permutations on an n-set which are uniform, i.e., corresponding blocks have same size.

Original entry on oeis.org

1, 1, 3, 16, 131, 1496, 22482, 426833, 9934563, 277006192, 9085194458, 345322038293, 15024619744202, 740552967629021, 40984758230303149, 2527342803112928081, 172490568947825135203, 12952575262915522547136, 1064521056888312620947794, 95305764621957309071404877
Offset: 0

Views

Author

Des FitzGerald (D.FitzGerald(AT)utas.edu.au)

Keywords

Comments

Number of games of simple patience with n cards. Take a shuffled deck of n cards labeled 1..n; as each card is dealt it is placed either on a higher-numbered card or starts a new pile to the right. Cards are not moved once they are placed. Suggested by reading Aldous and Diaconis. - N. J. A. Sloane, Dec 19 1999
Number of set partitions of [2n] such that within each block the numbers of odd and even elements are equal. a(2) = 3: 1234, 12|34, 14|23; a(3) = 16: 123456, 1234|56, 1236|45, 1245|36, 1256|34, 12|3456, 12|34|56, 12|36|45, 1346|25, 1456|23, 14|2356, 14|23|56, 16|2345, 16|23|45, 14|25|36, 16|25|34. - Alois P. Heinz, Jul 14 2016

Examples

			For n=3 there are 25 block permutations, of which 9 of the form ({1} maps to {1,2}; {2,3} maps to {3}), are not uniform. Hence a(3) = 25 - 9 = 16.
Alternatively, for n=3 the 6 permutations of 3 cards produce 16 games, as follows: 123 -> {1,2,3}; 132 -> {1,32}, {1,3,2}; 213 -> {21,3}, {2,1,3}; 231 -> {21,3}, {2,31}, {2,3,1}; 312 -> {31,2}, {32,1}, {3,1,2}; 321 -> {321}, {32,1}, {31,2}, {3,21}, {3,2,1}.
G.f.: A(x) = 1 + x + 3*x^2/2!^2 + 16*x^3/3!^2 + 131*x^4/4!^2 + 1496*x^5/5!^2 + ...
log(A(x)) = x + x^2/2!^2 + x^3/3!^2 + x^4/4!^2 + x^5/5!^2 + ...
		

Crossrefs

Cf. A132813.
Column k=2 of A275043.
Main diagonal of A321296 and of A322670.

Programs

  • Haskell
    a023998 n = a023998_list !! n
    a023998_list = 1 : f 2 [1] a132813_tabl where
       f x ys (zs:zss) = y : f (x + 1) (ys ++ [y]) zss where
                         y = sum $ zipWith (*) ys zs
    -- Reinhard Zumkeller, Apr 04 2014
  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
          add(b(n-i)*binomial(n-1, i-1)/i!, i=1..n))
        end:
    a:= n-> b(n)*n!:
    seq(a(n), n=0..25);  # Alois P. Heinz, May 11 2016
  • Mathematica
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] Binomial[n-1, k] a[k], {k, 0, n-1}];
    Array[a, 25, 0] (* Jean-François Alcover, Jul 28 2016 *)
    nmax = 20; CoefficientList[Series[E^(-1 + BesselI[0, 2*Sqrt[x]]), {x, 0, nmax}], x]*Range[0, nmax]!^2 (* Vaclav Kotesovec, Jun 09 2019 *)
  • PARI
    a(n)=if(n==0,1,sum(k=0,n-1,binomial(n,k)*binomial(n-1,k)*a(k))) \\ Paul D. Hanna, Aug 15 2007
    
  • PARI
    {a(n)=n!^2*polcoeff(exp(sum(m=1, n, x^m/m!^2)+x*O(x^n)), n)} /* Paul D. Hanna */
    
  • PARI
    N=66; x='x+O('x^N); /* that many terms */
    Vec(serlaplace(serlaplace(exp(sum(n=1, N, x^n/n!^2))))) /* show terms */
    /* Joerg Arndt, Jul 12 2011 */
    
  • PARI
    v=vector(N); v[1]=1;
    for (n=1,N-1, v[n+1]=sum(k=0,n-1, binomial(n,k)*binomial(n-1,k)*v[k+1]) );
    v /* show terms */
    /* Joerg Arndt, Jul 12 2011 */
    

Formula

a(n) = Sum_{k=0..n-1} C(n,k)*C(n-1,k)*a(k) for n>0 with a(0)=1. - Paul D. Hanna, Aug 15 2007
G.f.: Sum_{n>=0} a(n)*x^n/n!^2 = exp( Sum_{n>=1} x^n/n!^2 ). [Paul D. Hanna, Jan 04 2011; merged from duplicate entry A179119]
Row sums of A061691.
Generating function: Let J(z) = Sum_{n>=0} z^n/n!^2. Then exp(J(z)-1) = Sum_{n>=0} a(n)*z^n/n!^2 = 1 + z + 3*z^2/2!^2 + 16*z^3/3!^2 + .... - Peter Bala, Jul 11 2011

Extensions

More terms from Vladeta Jovovic, Sep 03 2002

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

Views

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
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