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 1-4 of 4 results.

A002802 a(n) = (2*n+3)!/(6*n!*(n+1)!).

Original entry on oeis.org

1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001
a(n) is half the number of (n+2) X 2 Young tableaux with a three 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], A000984 for one horizontal wall, and A002457 for two. - Michael Wallner, Jan 31 2022
From Robert Coquereaux, Feb 12 2024: (Start)
Call B(p,g) the number of genus g partitions of a set with p elements (genus-dependent Bell number). Up to an appropriate shift the given sequence counts the genus 1 partitions of a set: we have a(n) = B(n+4,1), with a(0)= B(4,1)=1.
When shifted with an offset 4 (i.e., defining b(p)=a(p-4), which starts with 0,0,0,1,10,70, etc., and b(4)=1), the given sequence reads b(p) = (1/( 2^4 3 )) * (1/( (2 p - 1) (2 p - 3))) * (1/(p - 4)!) * (2p)!/p!. In this form it appears as a generalization of Catalan numbers (that indeed count the genus 0 partitions).
Call C[p, [alpha], g] the number of partitions of a set with p elements, of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). Up to an appropriate shift the given sequence also counts the genus 1 partitions of p=2k into k parts of length 2, which is then called C[2k, [2^k], 1], and we have a(n) = C[2k, [2^k], 1] for k=n+2.
The two previous interpretations of this sequence, leading to a(n) = B(n+4, 1) and to a(n) = C[2(n+2), [2^(n+2)], 1] are not related in any obvious way. (End)

Examples

			G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...

References

  • C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
  • 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).

Links

Crossrefs

Cf. A035309, A000108 (for genus 0 maps), A046521 (third column).
Cf. A000984, A000911, A002457, A002697, A051133.
Column g=1 of A370235.

Programs

  • GAP
    F:=Factorial;; List([0..25], n-> F(2*n+3)/(6*F(n)*F(n+1)) ); # G. C. Greubel, Jul 20 2019
  • Magma
    F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // G. C. Greubel, Jul 20 2019
  • Maple
    seq(simplify(4^n*hypergeom([-n,-3/2], [1], 1)),n=0..25); # Peter Luschny, Apr 26 2016
  • Mathematica
    Table[(2*n+3)!/(6*n!*(n+1)!), {n, 0, 25}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
  • PARI
    {a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */
  • PARI
    {a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */
  • Sage
    f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # G. C. Greubel, Jul 20 2019

Formula

G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x).
Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48).
a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - Philippe Deléham, Jan 22 2004
a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004
a(n) = A051133(n+1)/3 = A000911(n)/6. - Zerinvary Lajos, Jun 02 2007
From Rui Duarte, Oct 08 2011: (Start)
Also convolution of A000984 with A002697, also convolution of A000302 with A002457.
a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n).
a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2).
a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2).
a(n) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. (End)
D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014
a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = (-4)^n*binomial(-5/2, n). - Peter Luschny, Oct 23 2018
Sum_{n>=0} 1/a(n) = 12 - 2*sqrt(3)*Pi. - Amiram Eldar, Oct 13 2020
E.g.f.: (1/12) exp(2 x) x^2 BesselI[2, 2 x]. - Robert Coquereaux, Feb 12 2024

A045896 Denominator of n/((n+1)*(n+2)) = A026741/A045896.

Original entry on oeis.org

1, 6, 6, 20, 15, 42, 28, 72, 45, 110, 66, 156, 91, 210, 120, 272, 153, 342, 190, 420, 231, 506, 276, 600, 325, 702, 378, 812, 435, 930, 496, 1056, 561, 1190, 630, 1332, 703, 1482, 780, 1640, 861, 1806, 946, 1980, 1035, 2162, 1128, 2352, 1225, 2550, 1326, 2756, 1431
Offset: 0

Views

Author

Ralf W. Grosse-Kunstleve, Dec 11 1999

Keywords

Comments

Also period length divided by 2 of pairs (a,b), where a has period 2*n-2 and b has period n.
From Paul Curtz, Apr 17 2014: (Start)
Difference table of A026741/A045896:
0, 1/6, 1/6, 3/20, 2/15, 5/42, ...
1/6, 0, -1/60, -1/60, -1/70, -1/84, ... = 1/6, -A051712/A051713
-1/6, -1/60, 0, 1/420, 1/420, 1/504, ...
3/20, 1/60, 1/420, 0, -1/2520, -1/2520, ...
-2/15, -1/70, -1/420, -1/2520, 0, 1/13860, ...
5/42, 1/84, 1/504, 1/2520, -1/13860, 0, ...
Autosequence of the first kind. The main diagonal is A000004. The first two upper diagonals are equal. Their denominators are A000911. (End)

Links

Crossrefs

Factor of A160466.
Cf. A000004, A000384, A000911, A045895, A026741, A051714, A241269.

Programs

  • Haskell
    import Data.Ratio ((%), denominator)
a045896 n = denominator $ n % ((n + 1) * (n + 2))
-- Reinhard Zumkeller, Dec 12 2011
  • Maple
    seq((n+1)*(n+2)*(3-(-1)^n)/4, n=0..20); # C. Ronaldo
with(combinat): seq(lcm(n+1,binomial(n+2,n)), n=0..50); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[LCM[2*n + 2, n + 2]/2, {n, 0, 40}] (* corrected by Amiram Eldar, Sep 14 2022 *)
Denominator[#[[1]]/(#[[2]]#[[3]])&/@Partition[Range[0,60],3,1]] (* Harvey P. Dale, Aug 15 2013 *)
  • PARI
    Vec((2*x^3+3*x^2+6*x+1)/(1-x^2)^3+O(x^99)) \\ Charles R Greathouse IV, Mar 23 2016

Formula

G.f.: (2*x^3+3*x^2+6*x+1)/(1-x^2)^3.
a(n) = (n+1)*(n+2) if n odd; or (n+1)*(n+2)/2 if n even = (n+1)*(n+2)*(3-(-1)^n)/4. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 16 2004
a(2*n) = A000384(n+1); a(2*n+1) = A026741(n+1). - Reinhard Zumkeller, Dec 12 2011
Sum_{n>=0} 1/a(n) = 1 + log(2). - Amiram Eldar, Sep 11 2022
From Amiram Eldar, Sep 14 2022: (Start)
a(n) = lcm(2*n+2, n+2)/2.
a(n) = A045895(n+2)/2. (End)
E.g.f.: (2 + 8*x + x^2)*cosh(x)/2 + (2 + 2*x + x^2)*sinh(x). - Stefano Spezia, Apr 24 2024

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).

Original entry on oeis.org

3, 6, 15, 60, 105, 21, 126, 360, 495, 330, 429, 1092, 1365, 420, 1020, 2448, 2907, 1710, 1995, 4620, 5313, 759, 3450, 7800, 8775, 4914, 5481, 12180, 13485, 3720, 8184, 17952, 19635, 10710, 11655, 25308, 27417, 3705, 15990, 34440, 37023, 19866, 21285, 45540
Offset: 0

Views

Author

Paul Curtz, Apr 18 2014

Keywords

Comments

All terms are multiples of 3.
Difference table of c(n):
1/3, 1/6, 2/15, 7/60, 2/21,...
-1/6, -1/30, -1/60, -1/84, -1/105,...
2/15, 1/60, 1/210, 1/420, 1/630,...
-7/60, -1/84, -1/420, -1/1260, -1/2520,... .
This is an autosequence of the second kind; the inverse binomial transform is the signed sequence. The main diagonal is the first upper diagonal multiplied by 2.
Denominators of the main diagonal: A051133(n+1).
Denominators of the first upper diagonal; A000911(n).
c(n) is a companion to A026741(n)/A045896(n).
Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/A027642(n).
Are the numerators of the main diagonal (-1)^n? If yes, what is the value of 1/3 - 1/30 + 1/210,... or 1 - 1/10 + 1/70 - 1/420, ... , from A002802(n)?
Is a(n+40) - a(n) divisible by 10?
No: a(5) = 21 but a(45) = 12972. # Robert Israel, Jul 17 2023
Are the common divisors to A014206(n) and A007531(n+3) of period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2?
Reduce c(n) = f(n) = b(n)/a(n) = 1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, ... .
Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind
1, 1/2, 1/3, 1/4, 1/5, 1/6, ...
0, 1/6, 1/6, 3/20, 2/15, 5/42, ...
1/3, 1/6, 2/15, 7/60, 11/105, 2/21, ...
0, 1/10, 1/10, 13/140, 3/35, 5/63, ...
1/5, 1/10, 3/35, 11/140, 23/315, 43/630, ...
0, 1/14, 1/14, 17/252, 4/63, ...
This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.
Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).
First row: see A003506, Leibniz's Harmonic Triangle.
Second row: A026741/A045896.
a(n) is the denominator of the third row f(n).
The first column is 1, 0, 1/3, 0, 1/5, 0, 1/7, 0, ... . Numerators: A093178(n+1). This incites, considering tan(1), to introduce before the first row
Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .

Links

Programs

  • Maple
    seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))),n=0..1000);
  • Mathematica
    Denominator[Table[(n^2+n+2)/Times@@(n+{1,2,3}),{n,0,50}]] (* Harvey P. Dale, Mar 27 2015 *)
  • PARI
    for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014

Formula

c(n) = A014206(n)/A007531(n+3).
The sum of the difference table main diagonal is 1/3 - 1/30 + 1/210 - ... = 10*A086466-4 = 4*(sqrt(5)*log(phi)-1) = 0.3040894... - Jean-François Alcover, Apr 22 2014
a(n) = (n+1)*(n+2)*(n+3)/gcd(4*n - 4, n^2 + n + 2), where gcd(4*n - 4, n^2 + n + 2) is periodic with period 16. - Robert Israel, Jul 17 2023

Extensions

More terms from Colin Barker, Apr 18 2014

A051133 a(n) = binomial(2n,n)*n*(2n+1)/2.

Original entry on oeis.org

0, 3, 30, 210, 1260, 6930, 36036, 180180, 875160, 4157010, 19399380, 89237148, 405623400, 1825305300, 8143669800, 36064823400, 158685222960, 694247850450, 3022020054900, 13095420237900, 56517076816200, 243023430309660, 1041528987041400
Offset: 0

Views

Author

Wouter Meeussen

Keywords

Examples

			G.f. = 3*x + 30*x^2 + 210*x^3 + 1260*x^4 + 6930*x^5 + 36036*x^6 + ...

Links

Crossrefs

Cf. A000911, A000984, A000217, A002802.

Programs

  • Magma
    [Binomial(2*n,n)*n*(2*n+1)/2: n in [0..25]]; // G. C. Greubel, Feb 10 2019
  • Maple
    seq(binomial(2*n,n)*binomial(n,(n-2))/2, n=1..23); # Zerinvary Lajos, May 05 2007
  • Mathematica
    a[ n_]:= SeriesCoefficient[ 3x(1-4x)^(-5/2), {x, 0, n}]; (* Michael Somos, Sep 09 2013 *)
Table[Binomial[2*n, n]*n*(2*n + 1)/2, {n, 0, 22}] (* Amiram Eldar, Oct 22 2020 *)
  • PARI
    {a(n) = if( n<1, 0, (2*n + 1)! / (2 * n! *(n-1)!))}; /* Michael Somos, Sep 09 2013 */
  • PARI
    {a(n) = 2^(n+2) * polcoeff( pollegendre( n+3), n-1)}; /* Michael Somos, Sep 09 2013 */
  • Sage
    [binomial(2*n,n)*n*(2*n+1)/2 for n in (0..25)] # G. C. Greubel, Feb 10 2019

Formula

a(n) = (1/2) * A000911(n-1).
a(n) = (1/2)*A000984(n+1)*A000217(n). - Zerinvary Lajos, May 05 2007
a(n) = 3*A002802(n-1). - Zerinvary Lajos, Jun 02 2007
(-n+1)*a(n) + 2*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
G.f.: 3*x * (1 - 4*x)^(-5/2). - Michael Somos, Sep 09 2013
Sum_{n>=1} 1/a(n) = 4 - 2*Pi/sqrt(3). - Amiram Eldar, Oct 22 2020
Showing 1-4 of 4 results.

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