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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A005563 a(n) = n*(n+2) = (n+1)^2 - 1.

Original entry on oeis.org

0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600
Offset: 0

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Author

Keywords

Comments

Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).
Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - Len Smiley, Dec 08 2001
Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - Roberto E. Martinez II, Jan 07 2002
Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002
Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003
For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k iterations. - Gerald McGarvey, Aug 19 2004
a(n) mod 3 = 0 if and only if n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - Reinhard Zumkeller, Oct 16 2006
a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007
Nonnegative X values of solutions to the equation X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2). - Mohamed Bouhamida, Nov 06 2007
Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida, Nov 12 2007
From R. K. Guy, Feb 01 2008: (Start)
Toads and Frogs puzzle:
This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by
T T - F F
T - T F F
T F T - F
T F T F -
T F - F T
- F T F T
F - T F T
F F T - T
F F - T T
I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.
Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)
a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - Paul Curtz, Oct 28 2008
Row 3 of array A163280, n >= 1. - Omar E. Pol, Aug 08 2009
Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida, Sep 04 2009 [Comment edited by N. J. A. Sloane, Sep 24 2009]
Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - A.K. Devaraj, Sep 18 2009
For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). - Rick L. Shepherd, Sep 27 2009
For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - Gary W. Adamson, Jul 15 2010
Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g., a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - Gary W. Adamson, Jul 30 2010
a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - Peter Luschny, Jul 08 2011
a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - Aldo González Lorenzo, Oct 12 2011
Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - J. M. Bergot, May 03 2012
Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - Stanislav Sykora, May 26 2012
Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - Raphie Frank, Sep 28 2012
Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - Takumi Sato, Oct 10 2012
The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427... = A156035. - Kival Ngaokrajang, Sep 07 2013
The integers in the closed form solution of a(n) = 2*a(n-1) + a(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08 2001, are m and -m + 2 where m >= 3 is a positive integer. - Felix P. Muga II, Mar 18 2014
Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - Felix P. Muga II, Mar 18 2014
For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - Emeric Deutsch, Aug 07 2014
If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - Bruno Berselli, Sep 04 2014
For n >= 1, a(n) is the dimension of the simple Lie algebra A_n. - Wolfdieter Lang, Oct 21 2015
Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - David Covert, Jan 15 2016
For n > 0, a(n) mod (n+1) = a(n) / (n+1) = n. - Torlach Rush, Apr 04 2016
Conjecture: When using the Sieve of Eratosthenes and sieving (n+1..a(n)), with divisors (1..n) and n>0, there will be no more than a(n-1) composite numbers. - Fred Daniel Kline, Apr 08 2016
a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - Andres Cicuttin, Jun 02 2016
Also for n > 0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - R. J. Cano, Dec 21 2016
The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kival Ngaokrajang link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - Charles Kusniec, Jul 03 2017
Also the number of independent vertex sets in the n-barbell graph. - Eric W. Weisstein, Aug 16 2017
Interleaving of A000466 and A033996. - Bruce J. Nicholson, Nov 08 2019
a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020
From Muge Olucoglu, Jan 19 2021: (Start)
For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).
.
0 1 2 5 6 7
+---------------
*(0,1)| X X
(0,2)| X X
(1,5)| X X
*(2,6)| X X
*(5,7)| X X
(6,7)| X X
.
*: represents the elements that are covered. (End)
1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022
Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023
For n >= 1, also the number of minimum cyclic edge cuts in the (n+2)-trapezohedron graph. - Eric W. Weisstein, Nov 21 2024
For n >= 1, a(n) is the sum of the interior angles of a polygon with n+2 sides, in radians, multiplied by (n+2)/Pi. - Stuart E Anderson, Aug 06 2025

Examples

			G.f. = 3*x + 8*x^2 + 15*x^3 + 24*x^4 + 35*x^5 + 48*x^6 + 63*x^7 + 80*x^8 + ...
		

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see index under Toads and Frogs Puzzle.
  • Martin Gardner, Perplexing Puzzles and Tantalizing Teasers, p. 21 (for "The Dime and Penny Switcheroo").
  • R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.
  • Derek Holton, Math in School, 37 #1 (Jan 2008) 20-22.
  • Edouard Lucas, Récréations Mathématiques, Gauthier-Villars, Vol. 2 (1883) 141-143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

G.f.: x*(3-x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = A000290(n+1) - 1.
A002378(a(n)) = A002378(n)*A002378(n+1); e.g., A002378(15)=240=12*20. - Charlie Marion, Dec 29 2003
a(n) = A067725(n)/3. - Zerinvary Lajos, Mar 06 2007
a(n) = Sum_{k=1..n} A144396(k). - Zerinvary Lajos, May 11 2007
a(n) = A134582(n+1)/4. - Zerinvary Lajos, Feb 01 2008
A143053(a(n)) = A000290(n+1), for n > 0. - Reinhard Zumkeller, Jul 20 2008
a(n) = Real((n+1+i)^2). - Gerald Hillier, Oct 12 2008
A053186(a(n)) = 2*n. - Reinhard Zumkeller, May 20 2009
a(n) = (n! + (n+1)!)/(n-1)!, n > 0. - Gary Detlefs, Aug 10 2009
a(n) = floor(n^5/(n^3+1)) with offset 1 (a(1)=0). - Gary Detlefs, Feb 11 2010
a(n) = a(n-1) + 2*n + 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
Sum_{n>=1} 1/a(n) = 3/4. - Mohammad K. Azarian, Dec 29 2010
a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^(n-1)*(cos(x))^3), for n > 0. - Francesco Daddi, Aug 02 2011
a(n) = A002378(n) + floor(sqrt(A002378(n))); pronic number + its root. - Fred Daniel Kline, Sep 16 2011
a(n-1) = A008833(n) * A068310(n) for n > 1. - Reinhard Zumkeller, Nov 26 2011
G.f.: U(0) where U(k) = -1 + (k+1)^2/(1 - x/(x + (k+1)^2/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 19 2012
a(n) = 15*C(n+4,3)*C(n+4,5)/(C(n+4,2)*C(n+4,4)). - Gary Detlefs, Aug 05 2013
a(n) = (n+2)!/((n-1)! + n!), n > 0. - Ivan N. Ianakiev, Nov 11 2013
a(n) = 3*C(n+1,2) - C(n,2) for n >= 0. - Felix P. Muga II, Mar 11 2014
a(n) = (A016742(n+1) - 4)/4 for n >= 0. - Felix P. Muga II, Mar 11 2014
a(-2 - n) = a(n) for all n in Z. - Michael Somos, Aug 07 2014
A253607(a(n)) = 1. - Reinhard Zumkeller, Jan 05 2015
E.g.f.: x*(x + 3)*exp(x). - Ilya Gutkovskiy, Jun 03 2016
For n >= 1, a(n^2 + n - 2) = a(n-1) * a(n). - Miko Labalan, Oct 15 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/4. - Amiram Eldar, Nov 04 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 2.
Product_{n>=1} (1 - 1/a(n)) = -sqrt(2)*sin(sqrt(2)*Pi)/Pi. (End)
a(n) = A000290(n+2) - n*2. See Bounded Squares illustration. - Leo Tavares, Oct 05 2021
From Leo Tavares, Oct 10 2021: (Start)
a(n) = A008585(n) + 2*A000217(n-1). See Trapezoids illustration.
2*A005563 = A054000(n+1). See Trapagons illustration.
a(n) = 2*A000217(n) + n. (End)
a(n) = (n+2)!!/(n-2)!! for n > 1. - Jacob Szlachetka, Jan 02 2022

Extensions

Partially edited by Joerg Arndt, Mar 11 2010
More terms from N. J. A. Sloane, Aug 01 2010

A002057 Fourth convolution of Catalan numbers: a(n) = 4*binomial(2*n+3,n)/(n+4).

Original entry on oeis.org

1, 4, 14, 48, 165, 572, 2002, 7072, 25194, 90440, 326876, 1188640, 4345965, 15967980, 58929450, 218349120, 811985790, 3029594040, 11338026180, 42550029600, 160094486370, 603784920024, 2282138106804, 8643460269248, 32798844771700, 124680849918352
Offset: 0

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Author

Keywords

Comments

a(n) is sum of the (flattened) list obtained by the iteration of: replace each integer k with the list 0,...,k+1 on the starting value 0. Length of this list is Catalan(n) or A000108. - Wouter Meeussen, Nov 11 2001
a(n-2) is the number of n-th generation vertices in the tree of sequences with unit increase labeled by 3 (cf. Zoran Sunic reference). - Benoit Cloitre, Oct 07 2003
Number of standard tableaux of shape (n+2,n-1). - Emeric Deutsch, May 30 2004
a(n) = CatalanNumber(n+3) - 2*CatalanNumber(n+2). Proof. From its definition as a convolution of Catalan numbers, a(n) counts lists of 4 Dyck paths of total size (semilength) = n. Connect the 4 paths by 3 upsteps (U) and append 3 downsteps (D). This is a reversible procedure. So a(n) is also the number of Dyck (n+3)-paths that end DDD (D for downstep). Let C(n) denote CatalanNumber(n) (A000108). Since C(n+3) is the total number of Dyck (n+3)-paths and C(n+2) is the number that end UD, we have (*) C(n+3) - C(n+2) is the number of Dyck (n+3)-paths that end DD. Also, (**) C(n+2) is the number of Dyck (n+3)-paths that end UDD (change the last D in a Dyck (n+2)-path to UDD). Subtracting (**) from (*) yields a(n) = C(n+3) - 2C(n+2) as claimed. - David Callan, Nov 21 2006
Convolution square of the Catalan sequence without one of the initial "1"'s: (1 + 4x + 14x^2 + 48x^3 + ...) = (1/x^2) * square(x + 2x^2 + 5x^3 + 14x^4 + ...)
a(n) is the number of binary trees with n+3 internal nodes in which both subtrees of the root are nonempty. Cf. A068875 [Sedgewick and Flajolet]. - Geoffrey Critzer, Jan 05 2013
With offset 4, a(n) is the number of permutations on {1,2,...,n} that are 123-avoiding, i.e., do not contain a three-term monotone subsequence, for which the first ascent is at positions (4,5); for example, there are 48 123-avoiding permutations on n=7 for which the first ascent is at spots (4,5). See Connolly link. There it is shown in general that the k-th Catalan Convolution is the number of 123-avoiding permutations for which the first ascent is at (k, k+1). (For n=k, the first ascent is defined to be at positions (k,k+1) if the permutation is the decreasing permutation with no ascents.) - Anant Godbole, Jan 17 2014
With offset 4, a(n) is the number of permutations on {1,2,...,n} that are 123-avoiding and for which the integer n is in the 4th spot; see Connolly link. - Anant Godbole, Jan 17 2014
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that have exactly one east step below the subdiagonal y = x-1. Details can be found in Section 3.1 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that bounce off the diagonal y = x to the right exactly once but do not bounce off y = x to the left. Details can be found in Section 4.2 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that horizontally cross the diagonal y = x exactly once but do not cross the diagonal vertically. Details can be found in Section 4.3 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
Apparently also Young tableaux of (non-partition) shape [n+1, 1, 1, n+1], see example file. - Joerg Arndt, Dec 30 2023

Examples

			From _Peter Bala_, Apr 14 2017: (Start)
This sequence appears on the main diagonal of a generalized Catalan triangle. Construct a lower triangular array (T(n,k)), n,k >= 0 by placing the sequence [0,0,0,1,1,1,1,...] in the first column and then filling in the remaining entries in the array using the rule T(n,k) = T(n,k-1) + T(n-1,k). The resulting array begins
  n\k| 0 1  2  3  4   5   6   7  ...
  ---+-------------------------------
   0 | 0
   1 | 0 0
   2 | 0 0  0
   3 | 1 1  1  1
   4 | 1 2  3  4  4
   5 | 1 3  6 10 14  14
   6 | 1 4 10 20 34  48  48
   7 | 1 5 15 35 69 117 165 165
   ...
(see Tedford 2011; this is essentially the array C_4(n,k) in the notation of Lee and Oh). Compare with A279004. (End)
		

References

  • Pierre de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 11, coefficients of P_4(z).
  • C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., Vol. 14 (1922), pp. 55-62, 122-138 and 143-146.
  • Robert Sedgewick and Phillipe Flajolet, Analysis of Algorithms, Addison Wesley, 1996, page 225.
  • 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

T(n, n+4) for n=0, 1, 2, ..., array T as in A047072. Also a diagonal of A059365 and of A009766.
Cf. A001003.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A145596 (row sums), A279004.

Programs

  • GAP
    List([0..25],n->4*Binomial(2*n+3,n)/(n+4)); # Muniru A Asiru, Mar 05 2018
    
  • Magma
    [4*Binomial(2*n+3,n)/(n+4): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016
    
  • Maple
    a := n -> 32*4^n*GAMMA(5/2+n)*(1+n)/(sqrt(Pi)*GAMMA(5+n)):
    seq(a(n),n=0..23); # Peter Luschny, Dec 14 2015
    A002057List := proc(m) local A, P, n; A := [1]; P := [1,1,1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-1]]);
    A := [op(A), P[-1]] od; A end: A002057List(27); # Peter Luschny, Mar 26 2022
  • Mathematica
    Table[Plus@@Flatten[Nest[ #/.a_Integer:> Range[0, a+1]&, {0}, n]], {n, 0, 10}]
    Table[4 Binomial[2n+3,n]/(n+4),{n,0,30}] (* or *) CoefficientList[ Series[ (1-Sqrt[1-4 x]+2 x (-2+Sqrt[1-4 x]+x))/(2 x^4),{x,0,30}],x] (* Harvey P. Dale, May 05 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n+=2; 2*binomial(2*n, n-2) / n)}; /* Michael Somos, Jul 31 2005 */
    
  • PARI
    x='x+O('x^100); Vec((1-(1-4*x)^(1/2)+2*x*(-2+(1-4*x)^(1/2)+x))/(2*x^4)) \\ Altug Alkan, Dec 14 2015
    
  • SageMath
    [2*(n+1)*catalan_number(n+2)/(n+4) for n in (0..30)] # G. C. Greubel, May 27 2022

Formula

a(n) = A033184(n+4, 4) = 4*binomial(2*n+3, n)/(n+4) = 2*(n+1)*A000108(n+2)/(n+4).
G.f.: c(x)^4 with c(x) g.f. of A000108 (Catalan).
Row sums of A145596. Column 4 of A033184. By specializing the identities for the row polynomials given in A145596 we obtain the results a(n) = Sum_{k = 0..n} (-1)^k*binomial(n+1,k+1)*a(k)*4^(n-k) and a(n) = Sum_{k = 0..floor(n/2)} binomial(n+1,2*k+1) * Catalan(k+1) * 2^(n-2*k). From the latter identity we can derive the congruences a(2n+1) == 0 (mod 4) and a(2n) == Catalan(n+1) (mod 4). It follows that a(n) is odd if and only if n = (2^m - 4) for some m >= 2. - Peter Bala, Oct 14 2008
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=3, a(n-3) = (-1)^(n-3) * coeff(charpoly(A,x), x^3). - Milan Janjic, Jul 08 2010
G.f.: (1-sqrt(1-4*x) + 2*x*(-2+sqrt(1-4*x) + x))/(2*x^4). - Harvey P. Dale, May 05 2011
a(n+1) = A214292(2*n+4,n). - Reinhard Zumkeller, Jul 12 2012
D-finite with recurrence: (n+4)a(n) = 8*(2*n-1)*a(n-3) - 20*(n+1)*a(n-2) + 4*(2*n+5)*a(n-1). - Fung Lam, Jan 29 2014
D-finite with recurrence: (n+4)*a(n) - 2*(3*n+7)*a(n-1) + 4*(2*n+1)*a(n-2) = 0. - R. J. Mathar, Jun 03 2014
Asymptotics: a(n) ~ 4^(n+3)/sqrt(4*Pi*n^3). - Fung Lam, Mar 31 2014
a(n) = 32*4^n*Gamma(5/2+n)*(1+n)/(sqrt(Pi)*Gamma(5+n)). - Peter Luschny, Dec 14 2015
a(n) = C(n+1) - 2*C(n) where C is Catalan number A000108. Yuchun Ji, Oct 18 2017 [Note: Offset is off by 2]
E.g.f.: d/dx ( 2*exp(2*x)*BesselI(2,2*x)/x ). - Ilya Gutkovskiy, Nov 01 2017
From Bradley Klee, Mar 05 2018: (Start)
With F(x) = 16/(1+sqrt(1-4*x))^4 g.f. of A002057, xi(x) = F(x/4)*(x/4)^2, K(16*x) = 2F1(1/2,1/2;1;16*x) g.f. of A002894, q(x) g.f. of A005797, and q'(x) g.f. of A274344:
K(x) = (1+sqrt(xi(x)))*K(xi(x)).
2*K(1-x) = (1+sqrt(xi(x)))*K(1-xi(x)).
q(x) = sqrt(q(xi(16*x)/16)) = q'(xi(16*x)/16)/sqrt(xi(16*x)/16). (End)
From Amiram Eldar, Jan 02 2022: (Start)
Sum_{n>=0} 1/a(n) = 5/4 + Pi/(18*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 183*log(phi)/(25*sqrt(5)) - 77/100, where phi is the golden ratio (A001622). (End)
a(n) = Integral_{x=0..4} x^n*W(x) dx where W(x) = -x^(3/2)*(1 - x/2)*sqrt(4 - x)/Pi, defined on the open interval (0,4). - Karol A. Penson, Nov 13 2022

A003517 Number of permutations of [n+1] with exactly 1 increasing subsequence of length 3.

Original entry on oeis.org

1, 6, 27, 110, 429, 1638, 6188, 23256, 87210, 326876, 1225785, 4601610, 17298645, 65132550, 245642760, 927983760, 3511574910, 13309856820, 50528160150, 192113383644, 731508653106, 2789279908316, 10649977831752, 40715807302800, 155851062397940, 597261490737912
Offset: 2

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Comments

a(n-4) = number of n-th generation vertices in the tree of sequences with unit increase labeled by 5 (cf. Zoran Sunic reference). - Benoit Cloitre, Oct 07 2003
Number of standard tableaux of shape (n+3,n-2). - Emeric Deutsch, May 30 2004
a(n) = A214292(2*n,n-3) for n > 2. - Reinhard Zumkeller, Jul 12 2012
a(n) is the number of North-East paths from (0,0) to (n,n) that cross the diagonal y = x horizontally exactly once. By symmetry, it is also the number of North-East paths from (0,0) to (n,n) that cross the diagonal y = x vertically exactly once. Details can be found in Section 3.3 in Pan and Remmel's link. - Ran Pan, Feb 02 2016
a(n) is the number of permutations pi of [n+3] such that s(pi)=321456...(n+3), where s denotes West's stack-sorting map. - Colin Defant, Jan 14 2019
a(n) is also the number of permutations of [n+1] avoiding the pattern 321. For permutations avoiding any of the other permutations of [3] (that is, any of 132, 213, 231, or 312) see A002054. - N. J. A. Sloane, Nov 26 2022

Examples

			a(3)=6 because the only permutations of 1234 with exactly 1 increasing subsequence of length 3 are 1423, 4123, 1342, 2314, 2341, 3124.
		

References

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

Crossrefs

T(n, n+6) for n=0, 1, 2, ..., array T as in A047072.
See also A002054.
First differences are in A026017.
A diagonal of any of the essentially equivalent arrays: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.

Programs

  • Maple
    A003517List := proc(m) local A, P, n; A := [1]; P := [1,1,1,1,1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-1]]);
    A := [op(A), P[-1]] od; A end: A003517List(25); # Peter Luschny, Mar 26 2022
  • Mathematica
    f[x_] = (Sqrt[1 - 4 x] - 1)^6/(64 x^4); CoefficientList[Series[f[x], {x, 0, 25}], x][[3 ;; 26]] (* Jean-François Alcover, Jul 13 2011, after g.f. *)
    Table[6 Binomial[2n+1,n-2]/(n+4),{n,2,30}] (* Harvey P. Dale, Feb 27 2012 *)
  • PARI
    a(n)=6*binomial(2*n+1,n-2)/(n+4) \\ Charles R Greathouse IV, May 18 2015
    
  • PARI
    x='x+O('x^50); Vec(x^2*((1-(1-4*x)^(1/2))/(2*x))^6) \\ Altug Alkan, Nov 01 2015

Formula

a(n) = 6*binomial(2*n+1, n-2)/(n+4).
G.f.: x^2*C(x)^6, where C(x) is g.f. for the Catalan numbers (A000108). - Emeric Deutsch, May 30 2004
E.g.f.: exp(2*x)*(Bessel_I(2,2*x) - Bessel_I(4,2*x)). - Paul Barry, Jun 04 2007
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n >= 5, a(n-3) = (-1)^(n-5)*coeff(charpoly(A,x),x^5). - Milan Janjic, Jul 08 2010
a(n) = Sum_{i>=1, j>=1, k>=1, i+j+k=n+1} Catalan(i)*Catalan(j)*Catalan(k). T. D. Noe, Dec 22 2010
D-finite with recurrence -(n+4)*(n-2)*a(n) + 2*n*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
From Amiram Eldar, Jan 02 2022: (Start)
Sum_{n>=2} 1/a(n) = 7/2 - 34*Pi/(27*sqrt(3)).
Sum_{n>=2} (-1)^n/a(n) = 828*log(phi)/(25*sqrt(5)) - 2819/450, where phi is the golden ratio (A001622). (End)
a(n) ~ 3*4^(n+1)/(n^(3/2)*sqrt(Pi)). - Stefano Spezia, Apr 17 2024
a(n) = A000108(n+3) - 4*A000108(n+2) + 3*A000108(n+1). - Taras Goy, Jul 15 2024
a(n) = 6*(2*n+1)!*(n-1)!/((2*n-4)!*(n+4)!)*A000108(n-2). - Taras Goy, Dec 21 2024

A005572 Number of walks on cubic lattice starting and finishing on the xy plane and never going below it.

Original entry on oeis.org

1, 4, 17, 76, 354, 1704, 8421, 42508, 218318, 1137400, 5996938, 31940792, 171605956, 928931280, 5061593709, 27739833228, 152809506582, 845646470616, 4699126915422, 26209721959656, 146681521121244, 823429928805936
Offset: 0

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Comments

Also number of paths from (0,0) to (n,0) in an n X n grid using only Northeast, East and Southeast steps and the East steps come in four colors. - Emeric Deutsch, Nov 03 2002
Number of skew Dyck paths of semilength n+1 with the left steps coming in two colors. - David Scambler, Jun 21 2013
Number of 2-colored Schroeder paths from (0,0) to (2n+2,0) with no level steps H=(2,0) at an even level. There are two ways to color an H-step at an odd level. Example: a(1)=4 because we have UUDD, UHD (2 choices) and UDUD. - José Luis Ramírez Ramírez, Apr 27 2015

Examples

			a(3) = 76 = sum of top row terms of M^3; i.e., (37 + 29 + 9 + 1).
		

References

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

Crossrefs

Binomial transform of A002212. Sequence shifted right twice is A025228.

Programs

  • Maple
    a := n -> simplify(2^n*hypergeom([3/2, -n], [3], -2)):
    seq(a(n), n=0..21); # Peter Luschny, Feb 03 2015
    a := n -> simplify(GegenbauerC(n, -n-1, -2))/(n+1):
    seq(a(n), n=0..21); # Peter Luschny, May 09 2016
  • Mathematica
    RecurrenceTable[{a[0]==1,a[1]==4,a[n]==((2n+1)a[n-1]-3(n-1)a[n-2]) 4/(n+2)}, a[n],{n,30}] (* Harvey P. Dale, Oct 04 2011 *)
    a[n_]:=If[n==0,1,Coefficient[(1+4x+x^2)^(n+1),x^n]/(n+1)]
    Table[a[n],{n,0,40}] (* Emanuele Munarini, Apr 06 2012 *)
  • Maxima
    a(n):=coeff(expand((1+4*x+x^2)^(n+1)),x^n)/(n+1); makelist(a(n),n,0,12); /* Emanuele Munarini, Apr 06 2012 */
    
  • PARI
    a(n)=polcoeff((1-4*x-sqrt(1-8*x+12*x^2+x^3*O(x^n)))/2,n+2)
    
  • PARI
    { A005572(n) = sum(k=0,n\2, binomial(n,2*k) * binomial(2*k,k) * 4^(n-2*k) / (k+1) ) } /* Max Alekseyev, Feb 02 2015 */
    
  • PARI
    {a(n)=sum(k=0,n, binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1) )}
    for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Feb 02 2015
    
  • Sage
    def A005572(n):
        A108198 = lambda n,k: (-1)^k*catalan_number(k+1)*rising_factorial(-n,k)/factorial(k)
        return sum(A108198(n,k)*2^(n-k) for k in (0..n))
    [A005572(n) for n in range(22)] # Peter Luschny, Feb 05 2015

Formula

Generating function A(x) satisfies 1 + (xA)^2 = A - 4xA.
a(0) = 1 and, for n > 0, a(n) = 4a(n-1) + Sum_{i=1..n-1} a(i-1)*a(n-i-1). - John W. Layman, Jan 07 2000
G.f.: (1 - 4*x - sqrt(1 - 8*x + 12*x^2))/(2*x^2).
D-finite with recurrence: a(n) = ((2*n+1)*a(n-1) - 3*(n-1)*a(n-2))*4/(n+2), n > 0.
a(m+n) = Sum_{k>=0} A052179(m, k)*A052179(n, k) = A052179(m+n, 0). - Philippe Deléham, Sep 15 2005
a(n) = 4*a(n-1) + A052177(n-1) = A052179(n, 0) = 6*A005573(n)-A005573(n-1) = Sum_{j=0..floor(n/2)} 4^(n-2*j)*C(n, 2*j)*C(2*j, j)/(j+1). - Henry Bottomley, Aug 23 2001
a(n) = Sum_{k=0..n} A097610(n,k)*4^k. - Philippe Deléham, Dec 03 2009
Let A(x) be the g.f., then B(x) = 1 + x*A(x) = 1 + 1*x + 4*x^2 + 17*x^3 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-2*x) (continued fraction); more generally B(x) = C(x/(1-2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
From Gary W. Adamson, Jul 21 2011: (Start)
a(n) = sum of top row terms of M^n, M = an infinite square production matrix as follows:
3, 1, 0, 0, ...
1, 3, 1, 0, ...
1, 1, 3, 1, ...
1, 1, 1, 3, ...
... (End)
a(n) ~ 3*6^(n+1/2)/(n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Oct 05 2012
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k) * binomial(2k,k) * 4^(n-2k) / (k+1). - Max Alekseyev, Feb 02 2015
From Paul D. Hanna, Feb 02 2015: (Start)
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * binomial(2*k+2, k)/(k+1).
a(n) = Sum_{k=0..n} binomial(n,k) * 2^(n-k) * A000108(k+1).
a(n) = [x^n] (1 + 4*x + x^2)^(n+1) / (n+1).
G.f.: (1/x) * Series_Reversion( x/(1 + 4*x + x^2) ). (End)
a(n) = 2^n*hypergeom([3/2, -n], [3], -2). - Peter Luschny, Feb 03 2015
a(n) = 4^n*hypergeom([-n/2, (1-n)/2], [2], 1/4). - Robert Israel, Feb 04 2015
a(n) = Sum_{k=0..n} A108198(n,k)*2^(n-k). - Peter Luschny, Feb 05 2015
a(n) = 2*(12^(n/2))*(n!/(n+2)!)*GegenbauerC(n, 3/2,2/sqrt(3)), where GegenbauerC are Gegenbauer polynomials in Maple notation. This is a consequence of Robert Israel's formula. - Karol A. Penson, Feb 20 2015
a(n) = (2^(n+1)*3^((n+1)/2)*P(n+1,1,2/sqrt(3)))/((n+1)*(n+2)) where P(n,u,x) are the associated Legendre polynomials of the first kind. - Peter Luschny, Feb 24 2015
a(n) = -6^(n+1)*sqrt(3)*Integral{t=0..Pi}(cos(t)*(2+cos(t))^(-n-2))/(Pi*(n+2)). - Peter Luschny, Feb 24 2015
From Karol A. Penson and Wojciech Mlotkowski, Mar 16 2015: (Start)
Integral representation as the n-th moment of a positive function defined on a segment x=[2, 6]. This function is the Wigner's semicircle distribution shifted to the right by 4. This representation is unique. In Maple notation,
a(n) = int(x^n*sqrt(4-(x-4)^2)/(2*Pi), x=2..6),
a(n) = 2*6^n*Pochhammer(3/2, n)*hypergeom([-n, 3/2], [-n-1/2], 1/3)/(n+2)!
(End)
a(n) = GegenbauerC(n, -n-1, -2)/(n+1). - Peter Luschny, May 09 2016
E.g.f.: exp(4*x) * BesselI(1,2*x) / x. - Ilya Gutkovskiy, Jun 01 2020
From Peter Bala, Aug 18 2021: (Start)
G.f. A(x) = 1/(1 - 2*x)*c(x/(1 - 2*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. Cf. A129400.
Conjecture: a(n) is even except for n of the form 2*(2^k - 1). [added Feb 03: the conjecture follows from the formula a(n) = Sum_{k = 0..n} 2^(n-k)*binomial(n, k)*Catalan(k+1) given above.] (End)
From Peter Bala, Feb 03 2024: (Start)
G.f.: 1/(1 - 2*x) * c(x/(1 - 2*x))^2 = 1/(1 - 6*x) * c(-x/(1 - 6*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
a(n) = 6^n * Sum_{k = 0..n} (-6)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 6^n * hypergeom([-n, 3/2], [3], 2/3). (End)

Extensions

Additional comments from Michael Somos, Jun 10 2000

A003518 a(n) = 8*binomial(2*n+1,n-3)/(n+5).

Original entry on oeis.org

1, 8, 44, 208, 910, 3808, 15504, 62016, 245157, 961400, 3749460, 14567280, 56448210, 218349120, 843621600, 3257112960, 12570420330, 48507033744, 187187399448, 722477682080, 2789279908316, 10772391370048, 41620603020640, 160878516023680, 622147386185325
Offset: 3

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Comments

a(n-6) is the number of n-th generation nodes in the tree of sequences with unit increase labeled by 7 (cf. Zoran Sunic reference). - Benoit Cloitre, Oct 07 2003
Number of standard tableaux of shape (n+4,n-3). - Emeric Deutsch, May 30 2004

Examples

			G.f. = x^3 + 8*x^4 + 44*x^5 + 208*x^6 + 910*x^7 + 3808*x^8 + 15504*x^9 + ...
		

References

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

Crossrefs

Cf. A002057.
First differences are in A026018.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.

Programs

  • Magma
    [8*Binomial(2*n+1,n-3)/(n+5): n in [3..30]]; // Vincenzo Librandi, Jan 23 2017
  • Mathematica
    Table[8 Binomial[2 n + 1, n - 3]/(n + 5), {n, 3, 25}] (* Michael De Vlieger, Oct 26 2016 *)
    CoefficientList[Series[((1 - Sqrt[1 - 4 x])/(2 x))^8, {x, 0, 30}], x] (* Vincenzo Librandi, Jan 23 2017 *)
  • PARI
    {a(n) = if( n<3, 0, 8 * binomial(2*n + 1, n-3) / (n + 5))}; /* Michael Somos, Mar 14 2011 */
    
  • PARI
    my(x='x+O('x^50)); Vec(x^3*((1-(1-4*x)^(1/2))/(2*x))^8) \\ Altug Alkan, Nov 01 2015
    

Formula

G.f.: x^3*C(x)^8, where C(x)=(1-sqrt(1-4*x))/(2*x) is g.f. for the Catalan numbers (A000108). - Emeric Deutsch, May 30 2004
The convolution of A002057 with itself. - Gerald McGarvey, Nov 08 2007
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=7, a(n-4)=(-1)^(n-7)*coeff(charpoly(A,x),x^7). - Milan Janjic, Jul 08 2010
a(n) = A214292(2*n,n-4) for n > 3. - Reinhard Zumkeller, Jul 12 2012
Integral representation as the n-th moment of the signed weight function W(x) on (0,4), i.e.: a(n+3) = Integral_{x=0..4} x^n*W(x) dx, n >= 0, with W(x) = (1/2)*x^(7/2)*(x-2)*(x^2-4*x+2)*sqrt(4-x)/Pi. - Karol A. Penson, Oct 26 2016
From Ilya Gutkovskiy, Jan 22 2017: (Start)
E.g.f.: 4*BesselI(4,2*x)*exp(2*x)/x.
a(n) ~ 4^(n+2)/(sqrt(Pi)*n^(3/2)). (End)
D-finite with recurrence: -(n+5)*(n-3)*a(n) +2*n*(2*n+1)*a(n-1)=0. - R. J. Mathar, Feb 20 2020
From Amiram Eldar, Jan 02 2022: (Start)
Sum_{n>=3} 1/a(n) = 43*Pi/(36*sqrt(3)) - 81/80.
Sum_{n>=3} (-1)^(n+1)/a(n) = 6213*log(phi)/(50*sqrt(5)) - 10339/400, where phi is the golden ratio (A001622). (End)

Extensions

More terms from Jon E. Schoenfield, May 06 2010

A003519 a(n) = 10*C(2n+1, n-4)/(n+6).

Original entry on oeis.org

1, 10, 65, 350, 1700, 7752, 33915, 144210, 600875, 2466750, 10015005, 40320150, 161280600, 641886000, 2544619500, 10056336264, 39645171810, 155989499540, 612815891050, 2404551645100, 9425842448792, 36921502679600, 144539291740025, 565588532895750, 2212449261033375
Offset: 4

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Comments

Number of standard tableaux of shape (n+5,n-4). - Emeric Deutsch, May 30 2004
a(n) is the number of North-East paths from (0,0) to (n,n) that cross the diagonal y = x horizontally exactly twice. By symmetry, it is also the number of North-East paths from (0,0) to (n,n) that cross the diagonal y = x vertically exactly twice. Details can be found in Section 3.3 in Pan and Remmel's link. - Ran Pan, Feb 02 2016

References

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

Crossrefs

A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.

Programs

  • Magma
    [10*Binomial(2*n+1, n-4)/(n+6): n in [4..35]]; // Vincenzo Librandi, Feb 03 2016
  • Maple
    seq(10*binomial(2*n+1,n-4)/(n+6), n=4..50); # Robert Israel, Feb 02 2016
  • Mathematica
    Table[10 Binomial[2 n + 1, n - 4]/(n + 6), {n, 4, 28}] (* Michael De Vlieger, Feb 03 2016 *)
  • PARI
    a(n) = 10*binomial(2*n+1, n-4)/(n+6); \\ Michel Marcus, Feb 02 2016
    

Formula

G.f.: x^4*C(x)^10, where C(x)=[1-sqrt(1-4x)]/(2x) is g.f. for the Catalan numbers (A000108). - Emeric Deutsch, May 30 2004
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=9, a(n-5)=(-1)^(n-9)*coeff(charpoly(A,x),x^9). [Milan Janjic, Jul 08 2010]
a(n) = A214292(2*n,n-5) for n > 4. - Reinhard Zumkeller, Jul 12 2012
From Robert Israel, Feb 02 2016: (Start)
D-finite with recurrence a(n+1) = 2*(n+1)*(2n+3)/((n+7)*(n-3)) * a(n).
a(n) ~ 20 * 4^n/sqrt(Pi*n^3). (End)
E.g.f.: 5*BesselI(5,2*x)*exp(2*x)/x. - Ilya Gutkovskiy, Jan 23 2017
From Amiram Eldar, Jan 02 2022: (Start)
Sum_{n>=4} 1/a(n) = 34*Pi/(45*sqrt(3)) - 44/175.
Sum_{n>=4} (-1)^n/a(n) = 53004*log(phi)/(125*sqrt(5)) - 79048/875, where phi is the golden ratio (A001622). (End)

A005568 Product of two successive Catalan numbers C(n)*C(n+1).

Original entry on oeis.org

1, 2, 10, 70, 588, 5544, 56628, 613470, 6952660, 81662152, 987369656, 12228193432, 154532114800, 1986841476000, 25928281261800, 342787130211150, 4583937702039300, 61923368957373000, 844113292629453000, 11600528392993339800, 160599522947154548400, 2238236829690383152800
Offset: 0

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Comments

Also equal to the number of standard tableaux of 2n cells with height less than or equal to 4. A005817(2n) - Mike Zabrocki, Feb 22 2007
Also equal to Sum binomial(2n,2i)*C(i)*C(n-i) = (4/Pi^2) Integral_{y=0..Pi} Integral_{x=0..Pi} (2*cos(x)+2*cos(y))^(2n)*sin^2(x)*sin^2(y) dx dy, since this counts walks of 2n steps in the nonnegative quadrant of an integer lattice that return to the origin (cf. R. K. Guy link below). - Andrew V. Sutherland, Nov 29 2007
Also, number of walks within N^2 (the first quadrant of Z^2) starting at (0,0), ending on the vertical axis and consisting of 2 n steps taken from {(-1, 0), (-1, 1), (1, -1), (1, 0)}. - Manuel Kauers, Nov 18 2008 - Manuel Kauers, Nov 18 2008
Also, number of walks within N^2 (the first quadrant of Z^2) starting and ending at (0,0) and consisting of 2 n steps taken from {(-1, 0), (0, -1), (0, 1), (1, 0)}. - Manuel Kauers, Nov 18 2008
a(2n-2) is also the sum of the numbers of standard Young tableaux of size 2n of (2,2) rectangular hook shapes (k+2,k+2,2^{n-2-k}, 0 <= k <= n-2. - Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 10 2010
Also, number of tree-rooted planar maps with n edges. - Noam Zeilberger, Aug 18 2017

References

  • M. Lothaire, Applied Combinatorics on Words, Cambridge, 2005. See Prop. 9.1.9, p. 452. [From N. J. A. Sloane, Apr 03 2012]
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..21],n->Binomial(2*n,n)*Binomial(2*(n+1),n+1)/((n+1)*(n+2))); # Muniru A Asiru, Dec 13 2018
    
  • Magma
    [Catalan(n)*Catalan(n+1): n in [0..21]]; // Vincenzo Librandi, Feb 06 2020
  • Maple
    A000108:=n->binomial(2*n,n)/(n+1):
    seq(A000108(n)*A000108(n+1),n=0..21); # Emeric Deutsch, Mar 05 2007
  • Mathematica
    f[n_] := CatalanNumber[n] CatalanNumber[n + 1] (* Or *) (4n + 2) Binomial[2 n, n]^2/(n^3 + 4n^2 +5n + 2) (* Or *) (2 n)! (2 + 2 n)!/(n! ((1 + n)!)^2 (2 + n)!); Array[f, 22, 0] (* Robert G. Wilson v *)
    Times@@@Partition[CatalanNumber[Range[0,30]],2,1] (* Harvey P. Dale, Jul 23 2012 *)
  • PARI
    (alias(C,binomial));a(n)=(C(2*n,n)-C(2*n,n-1))*(C(2*n+2,n+1)-C(2*n+2,n)) /* Michael Somos, Jun 22 2005 */
    
  • Sage
    [catalan_number(i)*catalan_number(i+1) for i in range(0,22)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = binomial(2*n,n)*binomial(2*n+2,n+1)/((n+1)(n+2)).
a(n) = 2*(2*n+1)*binomial(2*n,n)^2/((n+2)(n+1)^2).
D-finite with recurrence (n+2)*(n+1)*a(n) = 4*(2*n-1)*(2*n+1)*a(n-1). - Corrected R. J. Mathar, Feb 05 2020
G.f. in Maple notation: (1/2)/x+1/768/(x^2*Pi)*((32-512*x)*EllipticK(4*x^(1/2))+(-32-512*x)*EllipticE(4*x^(1/2))). - Karol A. Penson, Oct 24 2003
G.f.: 3F2( (1, 1/2, 3/2); (2, 3))(16*x) = (1 - 2F1((-1/2, 1/2); (2))( 16*x))/(2*x). - Olivier Gérard Feb 16 2011
G.f.: (1/(6*x))*(3+(16*x-1)*(2*hypergeom([1/2, 1/2],[1],16*x) + (16*x+1)*hypergeom([3/2, 3/2],[2],16*x))). - Mark van Hoeij, Nov 02 2009
G.f.: (1-hypergeom([-1/2,1/2],[2],16*x))/(2*x). - Mark van Hoeij, Aug 14 2014
E.g.f.: (1/3)*(8*x^2*BesselI(0, 2*x)^2 - 4*BesselI(0, 2*x)*BesselI(1, 2*x)*x - BesselI(1, 2*x)^2 - 8*BesselI(1, 2*x)^2*x^2)/x. - Vladeta Jovovic, Dec 29 2003
E.g.f. Sum_{n>=0} a(n)*x^(2n)/(2n)! = BesselI(1, 2x)^2/x^2. - Michael Somos, Jun 22 2005
From Paul D. Hanna, Nov 26 2009: (Start)
G.f.: A(x) = [(1/x)*Series_Reversion(x/F(x)^2)]^(1/2) where F(x) = g.f. of A004304, where A004304(n) is the number of nonseparable planar tree-rooted maps with n edges.
G.f.: A(x) = F(x*A(x)^2) where A(x/F(x)^2) = F(x) where F(x) = g.f. of A004304.
G.f.: A(x) = G(x*A(x)) where A(x/G(x)) = G(x) where G(x) = g.f. of A168450.
G.f.: A(x) = (1/x)*Series_Reversion(x/G(x)) where G(x) = g.f. of A168450.
Self-convolution yields A168452.
(End)
Representation of a(n) as the n-th power moment of a positive function on the segment [0,16]; in Mathematica notation, a(n) = NIntegrate[x^n*(8 ((1+x/16)*EllipticE[1-x/16]-1/8*x*EllipticK[1-x/16]))/(3*(Pi^2)*Sqrt[x]),{x,0,16}]. This solution of the Hausdorff power moment problem is unique. - Karol A. Penson, Oct 05 2011
G.f. y=A(x) satisfies: 0 = x^2*(16*x-1)*y''' + 6*x*(16*x-1)*y'' + 6*(18*x-1)*y' + 12*y. - Gheorghe Coserea, Jun 14 2018
Sum_{n>=0} a(n)/4^(2*n+1) = 2 - 16/(3*Pi). - Amiram Eldar, Apr 02 2022

Extensions

More terms from Emeric Deutsch, Feb 20 2004
More terms from Manuel Kauers, Nov 18 2008
Two hypergeometric g.f.s, van Hoeij's formula checked and formula field edited by Olivier Gérard, Feb 16 2011

A005586 a(n) = n*(n+4)*(n+5)/6.

Original entry on oeis.org

0, 5, 14, 28, 48, 75, 110, 154, 208, 273, 350, 440, 544, 663, 798, 950, 1120, 1309, 1518, 1748, 2000, 2275, 2574, 2898, 3248, 3625, 4030, 4464, 4928, 5423, 5950, 6510, 7104, 7733, 8398, 9100, 9840, 10619, 11438, 12298, 13200, 14145, 15134, 16168, 17248
Offset: 0

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Author

Keywords

Comments

Number of walks on square lattice.
Number of standard tableaux of shape (n+2,3) (n >= 1). - Emeric Deutsch, May 20 2004
Number of left factors of Dyck paths from (0,0) to (n+5,n-1). E.g. a(1)=5 because we have UDUDUD, UDUUDD, UUDDUD, UUDUDD and UUUDDD, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Jan 25 2005
Column 4 of Catalan triangle A009766. - Zerinvary Lajos, Nov 25 2006
Sum of first n triangular numbers minus next triangular number. - Vladimir Joseph Stephan Orlovsky, Oct 13 2009
Number of packed increasing tableaux of shape 3 X (n+1) with alphabet [n+4]. - Oliver Pechenik, Jan 03 2022

Examples

			G.f. = 5*x + 14*x^2 + 28*x^3 + 48*x^4 + 75*x^5 + 110*x^6 + 154*x^7 + ...
		

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. 796.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n)=A053121(n+5,n-1).

Programs

  • Magma
    [n*(n+4)*(n+5)/6: n in [0..50]]; // Vincenzo Librandi, Jun 09 2013
  • Maple
    [seq(binomial(n,3 )-binomial(n,1),n=4..48)]; # Zerinvary Lajos, Nov 25 2006
    a:=n->sum ((j-3)*j/2,j=0..n): seq(a(n),n=4..48); # Zerinvary Lajos, Dec 17 2006
    A005586:=z*(5-6*z+2*z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
    seq(sum(binomial(n,m), m=1..3)-n^2,n=5..49); # Zerinvary Lajos, Jun 19 2008
  • Mathematica
    Clear[lst,n,a,f]; f[n_]:=n*(n+1)/2; a=0;lst={};Do[a+=f[n];AppendTo[lst,a-f[n+1]],{n,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Oct 13 2009 *)
    CoefficientList[Series[x (5 - 6 x + 2 x^2) / (1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
    Table[(n(n+4)(n+5))/6,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,5,14,28},50] (* Harvey P. Dale, Jul 14 2018 *)
  • PARI
    {a(n) = n * (n+4) * (n+5) / 6}; /* Michael Somos, Apr 13 2007 */
    

Formula

G.f.: x * (5 - 6*x + 2*x^2) / (1 - x)^4.
E.g.f.: (5*x + 2*x^2 + x^3/6) * exp(x). - Michael Somos, Apr 13 2007
Let t(n) = n*(n+1)/2, te(n) = (n+1)*(n+2)*(n+3)/6. Then a(n-4) = -2*t(n) + te(n-1), e.g., a(2) = -2*t(6) + te(5) = -2*21 + 56 = 14, where te(n) are the tetrahedral numbers A000292 and t(n) are the triangular numbers A000217. - Jon Perry, Jul 23 2003
a(n) = C(5+n, 3)-C(5+n, 2). - Zerinvary Lajos, Jan 09 2006
a(n) = C(n,3) - C(n,1), n>=4. - Zerinvary Lajos, Nov 25 2006
a(n) = - A005581(-4-n) for all n in Z. - Michael Somos, Apr 13 2007
a(n) = A214292(n+4,2). - Reinhard Zumkeller, Jul 12 2012
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 77/200.
Sum_{n>=1} (-1)^(n+1)/a(n) = 363/200 - 12*log(2)/5. (End)
a(n) = A005581(n+2)-2. - R. J. Mathar, Nov 22 2024

Extensions

M3842=A005555 in the 1995 EIS was the same sequence as this.
More terms from Zerinvary Lajos, Jan 09 2006

A005566 Number of walks of length n on square lattice, starting at origin, staying in first quadrant.

Original entry on oeis.org

1, 2, 6, 18, 60, 200, 700, 2450, 8820, 31752, 116424, 426888, 1585584, 5889312, 22084920, 82818450, 312869700, 1181952200, 4491418360, 17067389768, 65166397296, 248817153312, 953799087696, 3656229836168, 14062422446800, 54086240180000, 208618354980000
Offset: 0

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Keywords

Comments

a(n) is the number of involutions of length 2n which are invariant under the reverse-complement map and have no decreasing subsequences of length 5. - Eric S. Egge, Oct 21 2008

Examples

			G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 60*x^4 + 200*x^5 + 700*x^6 + 2450*x^7 + ... - _Michael Somos_, Oct 17 2019
		

References

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

Crossrefs

a(2*n) = A000894(n), a(2*n+1) = 2*A060150(n+1).

Programs

  • Magma
    [Binomial(n, Floor(n/2))*Binomial(n+1, Floor((n+1)/2)): n in [0..30]]; // Vincenzo Librandi, Feb 18 2015
  • Mathematica
    f[n_] := Binomial[n, Floor[n/2]] Binomial[n + 1, Floor[(n + 1)/2]]; Array[f, 25, 0] (* Robert G. Wilson v *)

Formula

a(n) = binomial(n, floor(n/2))*binomial(n+1, floor((n+1)/2)).
E.g.f.: (BesselI(0, 2*x)+BesselI(1, 2*x))^2. - Vladeta Jovovic, Apr 28 2003
EXPCONV of A001405 with itself, i.e., a(n) = sum_{k=0}^n binomial(n,k)*A001405(k)*A001405(n-k). - Max Alekseyev, May 18 2006
G.f.: (16*x^2-1)*hypergeom([3/2, 3/2],[2],16*x^2) + (1/(2x)+2)*hypergeom([1/2, 1/2],[1],16*x^2) - 1/(2x). - Mark van Hoeij, Oct 13 2009
G.f.: (hypergeom([1/2,1/2],[1],16*x^2) - 1)/(2*x) + hypergeom([1/2,3/2],[2],16*x^2). - Mark van Hoeij, Aug 14 2014
a(n) = A241530(n)*2*floor(n/2)/(floor(n/2)+1). - Peter Luschny, Apr 25 2014
D-finite with recurrence (n+2)*(n+1)*a(n) +4*(-2*n-1)*a(n-1) -16*n*(n-1)*a(n-2)=0. - R. J. Mathar, Mar 07 2015
0 = a(n)*(+16*a(n+2) -6*a(n+3)) +a(n+1)*(-2*a(n+2) +a(n+3)) if n >= 0. - Michael Somos, Oct 17 2019
a(n) = binomial(floor(n + 1/2), floor(n/2)) * binomial(ceiling(n + 1/2), ceiling(n/2)). - Peter Luschny, Dec 14 2024

Extensions

Additional comments from David W. Wilson, May 05 2001
a(25)-a(26) from Vincenzo Librandi, Feb 18 2015

A005558 a(n) is the number of n-step walks on square lattice such that 0 <= y <= x at each step.

Original entry on oeis.org

1, 1, 3, 6, 20, 50, 175, 490, 1764, 5292, 19404, 60984, 226512, 736164, 2760615, 9202050, 34763300, 118195220, 449141836, 1551580888, 5924217936, 20734762776, 79483257308, 281248448936, 1081724803600, 3863302870000, 14901311070000, 53644719852000
Offset: 0

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Keywords

Comments

Number of n-step walks that start at the origin, constrained to stay in the first octant (0 <= y <= x). (Conjectured) - Benjamin Phillabaum, Mar 11 2011, corrected by Robert Israel, Oct 07 2015
For n >= 1, a(n-1) is the number of Dyck Paths with semilength n having floor((n+2)/2) U's in odd numbered positions. Example: (U is in odd numbered position and u is in even numbered position) Dyck path with n=5, floor ((5+2)/2)=3: UuddUuUddd. - Roger Ford, May 27 2017
The ratio of the number of n-step walks on the octant with an equal number of North steps and South steps to the total number of n-step walks on the octant is A005817(n)/a(n). For the reduced ratio, if n is divisible by 4 or n-1 is divisible by 4 the ratio is 1:floor(n/4)+1 and for all other values of n the ratio is 2:floor(n/2)+2. Example n = 4: A005817(4) = 10; EEEE, EEEW, EEWE, EWEE, EWEW, EEWW, ENSE, ENES, ENSW, EENS; a(4) = 20; 10:20 reduces to 1:2. - Roger Ford, Nov 04 2019

References

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

Crossrefs

See A138350 for a signed version.
Bisections are A000891 and A000888/2.
Cf. A000108, A005817. Column y=0 of A052174.

Programs

  • Magma
    [Binomial(n+1, Ceiling(n/2))*Binomial(n, Floor(n/2)) - Binomial(n+1, Ceiling((n-1)/2))*Binomial(n, Floor((n-1)/2)): n in [0..30]]; // Vincenzo Librandi, Sep 30 2015
    
  • Maple
    A:= proc(n,x,y) option remember;
        local j, xpyp, xp,yp, res;
        xpyp:= [[x-1,y],[x+1,y],[x,y-1],[x,y+1]];
        res:= 0;
        for j from 1 to 4 do
          xp:= xpyp[j,1];
          yp:= xpyp[j,2];
          if xp < 0 or xp > yp or xp + yp > n then next fi;
          res:= res + procname(n-1,xp,yp)
        od;
    return res
    end proc:
    A(0,0,0) := 1:
    seq(add(add(A(n,x,y), y = x .. n - x), x = 0 .. floor(n/2)), n = 0 .. 50); # Robert Israel, Oct 07 2015
  • Mathematica
    a[n_] := 1/2*Binomial[2*Floor[n/2]+1, Floor[n/2]+1]*CatalanNumber[1/2*(n+Mod[n, 2])]*(Mod[n, 2]+2); Table[a[n]//Abs, {n, 0, 27}] (* Jean-François Alcover, Mar 13 2014 *)
  • PARI
    a(n)=binomial(n+1,ceil(n/2))*binomial(n,floor(n/2)) - binomial(n+1,ceil((n-1)/2))*binomial(n,floor((n-1)/2))
    
  • Python
    from sympy import ceiling as c, binomial
    def a(n):
        return binomial(n + 1, c(n/2))*binomial(n, n//2) - binomial(n + 1, c((n - 1)/2))*binomial(n, (n - 1)//2)
    print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 02 2017

Formula

a(n) = C(n+1, ceiling(n/2))*C(n, floor(n/2)) - C(n+1, ceiling((n-1)/2))*C(n, floor((n-1)/2)). - Paul D. Hanna, Apr 16 2004
G.f.: (1/(4x^2))*((16*x^2-1)*(hypergeom([1/2, 1/2],[1],16*x^2)+2*x*(4*x-1)*hypergeom([3/2, 3/2],[2],16*x^2))-2*x+1). - Mark van Hoeij, Oct 13 2009
E.g.f (conjectured): BesselI(1,2*x)*(BesselI(0,2*x)+BesselI(1,2*x))/x. - Benjamin Phillabaum, Feb 25 2011
Conjecture: (2*n+1)*(n+3)*(n+2)*a(n) - 4*(2*n^2+4*n+3)*a(n-1) - 16*n*(2*n+3)*(n-1)*a(n-2) = 0. - R. J. Mathar, Apr 02 2017
Conjecture: (n+3)*(n+2)*a(n) - 4*(n^2+3*n+1)*a(n-1) + 16*(-n^2+n+1)*a(n-2) + 64*(n-1)*(n-2)*a(n-3) = 0. - R. J. Mathar, Apr 02 2017
a(n) = Sum_{k=0..floor(n/2)} n!/(k!*k!*(floor(n/2)-k)!*(floor((n+1)/2)-k)!*(k+1)) (conjectured). - Roger Ford, Aug 04 2017
a(n) = A000108(floor((n+1)/2))*A000108(floor(n/2))*(2*(floor(n/2))+1). - Roger Ford, Nov 15 2019
a(n) = Product_{k=3..n} (4*floor((k-1)/2) + 2) / (floor((k+2)/2)). - Roger Ford, Apr 29 2024
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