cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A002457 a(n) = (2n+1)!/n!^2.

Original entry on oeis.org

1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780, 3879876, 16224936, 67603900, 280816200, 1163381400, 4808643120, 19835652870, 81676217700, 335780006100, 1378465288200, 5651707681620, 23145088600920, 94684453367400, 386971244197200, 1580132580471900
Offset: 0

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Author

Keywords

Comments

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer, Apr 13 2001
Hankel transform is (-1)^n*A014480(n). - Paul Barry, Apr 26 2009
Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 2009
Convolution of A000302 and A000984. - Philippe Deléham, May 18 2009
1/a(n) is the integral of (x(1-x))^n on interval [0,1]. Apparently John Wallis computed these integrals for n=0,1,2,3,.... A004731, shifted left by one, gives numerators/denominators of related integrals (1-x^2)^n on interval [0,1]. - Marc van Leeuwen, Apr 14 2010
Extend the triangular peaks of Dyck paths of semilength n down to the baseline forming (possibly) larger and overlapping triangles. a(n) = sum of areas of these triangles. Also a(n) = triangular(n) * Catalan(n). - David Scambler, Nov 25 2010
Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n of B equals a(n-1). - T. D. Noe, May 01 2011
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+1. - David Scambler, Apr 29 2013
Denominator of central elements of Leibniz's Harmonic Triangle A003506.
Central terms of triangle A116666. - Reinhard Zumkeller, Nov 02 2013
Number of distinct strings of length 2n+1 using n letters A, n letters B, and 1 letter C. - Hans Havermann, May 06 2014
Number of edges in the Hasse diagram of the poset of partitions in the n X n box ordered by containment (from Havermann's comment above, C represents the square added in the edge). - William J. Keith, Aug 18 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r then V(n, 1/2^n) = V(n-1, 1/2^n) / a((n-1)/2) for all odd n. - Peter Luschny, Oct 12 2015
a(n) is the result of processing the n+1 row of Pascal's triangle A007318 with the method of A067056. Example: Let n=3. Given the 4th row of Pascal's triangle 1,4,6,4,1, we get 1*(4+6+4+1) + (1+4)*(6+4+1) + (1+4+6)*(4+1) + (1+4+6+4)*1 = 15+55+55+15 = 140 = a(3). - J. M. Bergot, May 26 2017
a(n) is the number of (n+1) X 2 Young tableaux with a two horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021] and A000984 for one horizontal wall. - Michael Wallner, Jan 31 2022
a(n) is the number of facets of the symmetric edge polytope of the cycle graph on 2n+1 vertices. - Mariel Supina, May 12 2022
Diagonal of the rational function 1 / (1 - x - y)^2. - Ilya Gutkovskiy, Apr 23 2025

Examples

			G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
  • C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
  • M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
  • 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).
  • J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

Crossrefs

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
A diagonal of A331430.
The rightmost diagonal of the triangle A331431.

Programs

Formula

G.f.: (1-4x)^(-3/2) = 1F0(3/2;;4x).
a(n-1) = binomial(2*n, n)*n/2 = binomial(2*n-1, n)*n.
a(n-1) = 4^(n-1)*Sum_{i=0..n-1} binomial(n-1+i, i)*(n-i)/2^(n-1+i).
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 21 2001
(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.
Sum_{i=0..n} i * binomial(n, i)^2 = n*binomial(2*n, n)/2. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003
E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Sep 22 2003
a(n) = Sum_{i+j+k=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k). - Benoit Cloitre, Nov 09 2003
a(n) = (2*n+1)*A000984(n) = A005408(n)*A000984(n). - Zerinvary Lajos, Dec 12 2010
a(n-1) = Sum_{k=0..n} A039599(n,k)*A000217(k), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson, Sep 02 2007
Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2). - Jaume Oliver Lafont, Mar 07 2009
a(n) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009
a(n) = A000217(n) * A000108(n). - David Scambler, Nov 25 2010
a(n) = f(n, n-3) where f is given in A034261.
a(n) = A005430(n+1)/2 = A002011(n)/4.
a(n) = binomial(2n+2, 2) * binomial(2n, n) / binomial(n+1, 1), a(n) = binomial(n+1, 1) * binomial(2n+2, n+1) / binomial(2, 1) = binomial(2n+2, n+1) * (n+1)/2. - Rui Duarte, Oct 08 2011
G.f.: (G(0) - 1)/(4*x) where G(k) = 1 + 2*x*((2*k + 3)*G(k+1) - 1)/(k + 1). - Sergei N. Gladkovskii, Dec 03 2011 [Edited by Michael Somos, Dec 06 2013]
G.f.: 1 - 6*x/(G(0)+6*x) where G(k) = 1 + (4*x+1)*k - 6*x - (k+1)*(4*k-2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: Q(0), where Q(k) = 1 + 4*(2*k + 1)*x*(2*k + 2 + Q(k+1))/(k+1). - Sergei N. Gladkovskii, May 10 2013 [Edited by Michael Somos, Dec 06 2013]
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+3)/(4*x*(2*k+3) + 2*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
a(n) = 2^(4n)/Sum_{k=0..n} (-1)^k*C(2n+1,n-k)/(2k+1). - Mircea Merca, Nov 12 2013
a(n) = (2*n)!*[x^(2*n)] HeunC(0,0,-2,-1/4,7/4,4*x^2) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunC is the Heun confluent function. - Peter Luschny, Nov 22 2013
0 = a(n) * (16*a(n+1) - 2*a(n+2)) + a(n+1) * (a(n+2) - 6*a(n+1)) for all n in Z. - Michael Somos, Dec 06 2013
a(n) = 4^n*binomial(n+1/2, 1/2). - Peter Luschny, Apr 24 2014
a(n) = 4^n*hypergeom([-2*n,-2*n-1,1/2],[-2*n-2,1],2)*(n+1)*(2*n+1). - Peter Luschny, Sep 22 2014
a(n) = 4^n*hypergeom([-n,-1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = 2*4^n*Gamma(3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
Sum_{n >= 0} 2^(n+1)/a(n) = Pi, related to Newton/Euler's Pi convergence transformation series. - Tony Foster III, Jul 28 2016. See the Weisstein Pi link, eq. (23). - Wolfdieter Lang, Aug 26 2016
Boas-Buck recurrence: a(n) = (6/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, and a(0) = 1. Proof from a(n) = A046521(n+1,1). See comment in A046521. - Wolfdieter Lang, Aug 10 2017
a(n) = (1/3)*Sum_{i = 0..n+1} C(n+1,i)*C(n+1,2*n+1-i)*C(3*n+2-i,n+1) = (1/3)*Sum_{i = 0..2*n+1} (-1)^(i+1)*C(2*n+1,i)*C(n+i+1,i)^2. - Peter Bala, Feb 07 2018
a(n) = (2*n+1)*binomial(2*n, n). - Kolosov Petro, Apr 16 2018
a(n) = (-4)^n*binomial(-3/2, n). - Peter Luschny, Oct 23 2018
a(n) = 1 / Sum_{s=0..n} (-1)^s * binomial(n, s) / (n+s+1). - Kolosov Petro, Jan 22 2019
a(n) = Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
4^n/a(n) = Integral_{x=0..1} (1 - x^2)^n. - Michael Somos, Jun 13 2019
D-finite with recurrence: 0 = a(n)*(6 + 4*n) - a(n+1)*(n + 1) for all n in Z. - Michael Somos, Jun 13 2019
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)/sqrt(5). - Amiram Eldar, Sep 10 2020
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*arcsin(sqrt(x)/2) / sqrt(x*(4-x)).
E.g.f. for {1/a(n)}: exp(x/4)*sqrt(Pi/x)*erf(sqrt(x)/2). (End)
G.f. for {1/a(n)}: 4*arctan(sqrt(x/(4-x))) / sqrt(x*(4-x)). - Michael Somos, Jun 17 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k + 1)*binomial(n+k, k). This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (n + 2*k + 1) * binomial(n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+1)*n+1, n). Cf. A090816 and A306290. - Peter Bala, Nov 02 2024
a(n) = (1/Pi)*(2*n + 1)*(2^(2*n + 1))*Integral_{x=0..oo} 1/(x^2 + 1)^(n + 1) dx. - Velin Yanev, Jan 28 2025

A001803 Numerators in expansion of (1 - x)^(-3/2).

Original entry on oeis.org

1, 3, 15, 35, 315, 693, 3003, 6435, 109395, 230945, 969969, 2028117, 16900975, 35102025, 145422675, 300540195, 9917826435, 20419054425, 83945001525, 172308161025, 1412926920405, 2893136075115, 11835556670925
Offset: 0

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Author

Keywords

Comments

a(n) is the denominator of the integral from 0 to Pi of (sin(x))^(2*n+1). - James R. Buddenhagen, Aug 17 2008
a(n) is the denominator of (2n)!!/(2*n + 1)!! = 2^(2*n)*n!*n!/(2*n + 1)! (see Andersson). - N. J. A. Sloane, Jun 27 2011
a(n) = (2*n + 1)*A001790(n). A046161(n)/a(n) = 1, 2/3, 8/15, 16/35, 128/315, 256/693, ... is binomial transform of Madhava-Gregory-Leibniz series for Pi/4 (i.e., 1 - 1/3 + 1/5 - 1/7 + ... ). See A173384 and A173396. - Paul Curtz, Feb 21 2010
a(n) is the denominator of Integral_{x=-oo..oo} sech(x)^(2*n+2) dx. The corresponding numerator is A101926(n). - Mohammed Yaseen, Jul 25 2023

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. 798.
  • G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:9 at page 51.

Crossrefs

The denominator is given in A046161.
Largest odd divisors of A001800, A002011, A002457, A005430, A033876, A086228.
Bisection of A004731, A004735, A086116.
Second column of triangle A100258.
Cf. A002596 (numerators in expansion of (1-x)^(1/2)).
Cf. A161198 (triangle related to the series expansions of (1-x)^((-1-2*n)/2)).
A163590 is the odd part of the swinging factorial, A001790 at even indices. - Peter Luschny, Aug 01 2009

Programs

  • Julia
    A001803(n) = sum(<<(A001790(k), A005187(n) - A005187(k)) for k in 0:n) # Peter Luschny, Oct 03 2019
    
  • Magma
    A001803:= func< n | Numerator(Binomial(n+2,2)*Catalan(n+1)/4^n) >;
    [A001803(n): n in [0..30]]; // G. C. Greubel, Apr 27 2025
    
  • Maple
    swing := proc(n) option remember; if n = 0 then 1 elif irem(n, 2) = 1 then swing(n-1)*n else 4*swing(n-1)/n fi end:
    sigma := n -> 2^(add(i,i= convert(iquo(n,2),base,2))):
    a := n -> swing(2*n+1)/sigma(2*n+1); # Peter Luschny, Aug 01 2009
    A001803 := proc(n) (2*n+1)*binomial(2*n,n)/4^n ; numer(%) ; end proc: # R. J. Mathar, Jul 06 2011
    a := n -> denom(Pi*binomial(n, -1/2)): seq(a(n), n = 0..22); # Peter Luschny, Dec 06 2024
  • Mathematica
    Numerator/@CoefficientList[Series[(1-x)^(-3/2),{x,0,25}],x]  (* Harvey P. Dale, Feb 19 2011 *)
    Table[Denominator[Beta[1, n + 1, 1/2]], {n, 0, 22}] (* Gerry Martens, Nov 13 2016 *)
  • PARI
    a(n) = numerator((2*n+1)*binomial(2*n,n)/(4^n)); \\ Altug Alkan, Sep 06 2018
    
  • SageMath
    def A001803(n): return numerator((n+1)*binomial(2*n+2,n+1)/2^(2*n+1))
    print([A001803(n) for n in range(31)]) # G. C. Greubel, Apr 27 2025

Formula

a(n) = (2*n + 1)! /(n!^2*2^A000120(n)) = (n + 1)*binomial(2*n+2,n+1)/2^(A000120(n)+1). - Ralf Stephan, Mar 10 2004
From Johannes W. Meijer, Jun 08 2009: (Start)
a(n) = numerator( (2*n+1)*binomial(2*n,n)/(4^n) ).
(1 - x)^(-3/2) = Sum_{n>=0} ((2*n+1)*binomial(2*n,n)/4^n)*x^n. (End)
Truncations of rational expressions like those given by the numerator or denominator operators are artifacts in integer formulas and have many disadvantages. A pure integer formula follows. Let n$ denote the swinging factorial and sigma(n) = number of '1's in the base-2 representation of floor(n/2). Then a(n) = (2*n+1)$ / sigma(2*n+1) = A056040(2*n+1) / A060632(2*n+2). Simply said: This sequence gives the odd part of the swinging factorial at odd indices. - Peter Luschny, Aug 01 2009
a(n) = denominator(Pi*binomial(n, -1/2)). - Peter Luschny, Dec 06 2024

A005430 Apéry numbers: n*C(2*n,n).

Original entry on oeis.org

0, 2, 12, 60, 280, 1260, 5544, 24024, 102960, 437580, 1847560, 7759752, 32449872, 135207800, 561632400, 2326762800, 9617286240, 39671305740, 163352435400, 671560012200, 2756930576400, 11303415363240, 46290177201840, 189368906734800, 773942488394400
Offset: 0

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Author

Keywords

Comments

Appears as diagonal in A003506. - Zerinvary Lajos, Apr 12 2006
The aerated sequence 1,0,2,0,12,0,60,0,... has e.g.f. 1+x*Bessel_I(1,2x). - Paul Barry, Mar 29 2010
Conjecture: the terms of the inverse binomial transform are 2*A132894(n). - R. J. Mathar, Oct 21 2012

References

  • Frank Harary and Edgar M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 78, (3.5.25).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A002011, A002457, A002736, A005258, A005259, A005429, 1/beta(n, n+1) in A061928.

Programs

  • GAP
    List([0..30], n-> n*Binomial(2*n,n)); # G. C. Greubel, Dec 09 2018
  • Magma
    [n*Binomial(2*n,n): n in [0..30]]; // G. C. Greubel, Dec 09 2018
    
  • Maple
    A005430 := n -> n*binomial(2*n, n);
  • Mathematica
    Table[n*Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, May 29 2015 *)
  • PARI
    a(n)=-(-1)^n*real(polcoeff(serlaplace(x^2*besselh1(1,2*x)),2*n)) \\ Ralf Stephan
    
  • Sage
    [n*binomial(2*n,n) for n in range(30)] # G. C. Greubel, Dec 09 2018
    

Formula

a(n) = A002011(n-1)/2 = 2 * A002457(n-1).
Sum_{n >= 1} 1/a(n) = Pi*sqrt(3)/9. - Benoit Cloitre, Apr 07 2002
G.f.: 2*x/sqrt((1-4*x)^3). - Marco A. Cisneros Guevara, Jul 25 2011
E.g.f.: a(n) = n!* [x^n] exp(2*x)*2*x*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
D-finite with recurrence (-n+1)*a(n) + 2*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 03 2012
G.f.: 2*x*(1-4*x)^(-3/2) = -G(0)/2 where G(k) = 1 - (2*k+1)/(1 - 2*x/(2*x - (k+1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
a(n-1) = Sum_{k=0..floor(n/2)} k*C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(phi)/sqrt(5) = A086466, where phi is the golden ratio. (End)
1/a(n) = (-1)^n*Sum_{j=0..n-1} binomial(n-1,j)*Bernoulli(j+n)/(j+n) for n >= 1. See the Amdeberhan & Cohen link. - Peter Luschny, Jun 20 2017
1/a(n) = Sum_{k=0..n} (-1)^(k+1)*binomial(n,k)*HarmonicNumber(n+k) for n >= 1. - Peter Luschny, Aug 15 2017
Sum_{n>=1} x^n/a(n) = 2*sqrt(x/(4-x))*arcsin(sqrt(x)/2), for abs(x) < 4 (Adegoke et al., 2022, section 6, p. 11). - Amiram Eldar, Dec 07 2024

Extensions

More terms from James Sellers, May 01 2000

A134400 M * A007318, where M = triangle with (1, 1, 2, 3, ...) in the main diagonal and the rest zeros.

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 3, 9, 9, 3, 4, 16, 24, 16, 4, 5, 25, 50, 50, 25, 5, 6, 36, 90, 120, 90, 36, 6, 7, 49, 147, 245, 245, 147, 49, 7, 8, 64, 224, 448, 560, 448, 224, 64, 8, 9, 81, 324, 756, 1134, 1134, 756, 324, 81, 9, 10, 100, 450, 1200, 2100, 2520, 2100, 1200, 450, 100, 10
Offset: 0

Views

Author

Gary W. Adamson, Oct 23 2007

Keywords

Comments

Row sums = A134401: (1, 2, 8, 24, 64, 160, 384, ...).
Triangle T(n,k), read by rows, given by [1,1,-1,1,0,0,0,0,0,...] DELTA [1,1,-1,1,0,0,0,0,0,...] where DELTA is the operator defined in A084938. A134402*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 26 2007
For n > 0, from n athletes, select a team of k players and then choose a coach who is allowed to be on the team or not. - Geoffrey Critzer, Mar 13 2010
Row sums are A036289 if first term changed to zero. Diagonal sums are A023610, starting with the 2nd diagonal. Partial sums of diagonals are A002940 if first term changed to zero. - John Molokach, Jul 06 2013
For n > 0, T(n,k) is the number of states in Sokoban puzzle with n non-obstacles cells and k boxes (see Russell and Norvig at page 157). - Stefano Spezia, Dec 03 2023

Examples

			First few rows of the triangle:
  1;
  1,  1;
  2,  4,   2;
  3,  9,   9,   3;
  4, 16,  24,  16,   4;
  5, 25,  50,  50,  25,   5;
  6, 36,  90, 120,  90,  36,  6;
  7, 49, 147, 245, 245, 147, 49, 7;
  ...
		

References

  • Stuart Russell and Peter Norvig, Artificial Intelligence: A Modern Approach, Fourth Edition, Hoboken: Pearson, 2021.

Crossrefs

T(2n,n) give A002011(n-1) for n>=1.

Programs

  • Maple
    with(combstruct): for n from 0 to 10 do seq(`if`(n=0, 1, n)* count( Combination(n), size=m), m=0..n) od; # Zerinvary Lajos, Apr 09 2008
  • Mathematica
    Join[{1},Table[Table[n*Binomial[n, k], {k,0, n}], {n, 10}]] //Flatten (* Geoffrey Critzer, Mar 13 2010 adapted by Stefano Spezia, Dec 03 2023 *)

Formula

From Geoffrey Critzer, Mar 13 2010: (Start)
T(0,0) = 1 and T(n,k) = n * binomial(n,k) for n > 0.
E.g.f. for column k is: (x^k/k!)*exp(x)*(x+k). (End)
T(n,k) = A003506(n,k) + A003506(n,k-1). - Geoffrey Critzer, Mar 13 2010
G.f.: (1-x-x*y+x^2+x^2*y+x^2*y^2)/(1-2*x-2*x*y+x^2+2*x^2*y+x^2*y^2). - Philippe Deléham, Nov 14 2013
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(0,0)=T(1,0)=T(1,1)=1, T(2,0)=T(2,2)=2, T(2,1)=4, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 14 2013
E.g.f.: 1 + exp(y*x)*exp(x)*(y*x + x). - Geoffrey Critzer, Mar 15 2015

Extensions

a(55)-a(65) from Stefano Spezia, Dec 03 2023

A002012 Almost trivalent maps.

Original entry on oeis.org

4, 32, 200, 1120, 5880, 29568, 144144, 686400, 3208920, 14780480, 67251184, 302865472, 1352078000, 5990745600, 26369978400, 115407434880, 502503206040, 2178032472000, 9401840170800, 40434981787200, 173319035569680, 740642835229440, 3156148445580000
Offset: 0

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References

  • R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
  • 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

Formula

a(n) = 2*(n+3)*(2*(n+1))! / (3*n!*(n+1)!). [Mathai & Rathie, Eq. (22)] - Andrey Zabolotskiy, Jun 24 2024

Extensions

Terms a(7) and beyond from Andrey Zabolotskiy, Jun 24 2024
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