A100071 a(n) = n * binomial(n-1, floor((n-1)/2)) = n * max_{i=0..n} binomial(n-1, i).

0, 1, 2, 6, 12, 30, 60, 140, 280, 630, 1260, 2772, 5544, 12012, 24024, 51480, 102960, 218790, 437580, 923780, 1847560, 3879876, 7759752, 16224936, 32449872, 67603900, 135207800, 280816200, 561632400, 1163381400, 2326762800

Offset: 0

Author

Paul Barry, Nov 02 2004

Keywords

• easy
• nonn

Comments

Old name: An inverse Chebyshev transform of n.

Hankel transform is (-1)^n*n*2^(n-1), A085750. This is the inverse binomial transform of -n. - Paul Barry, Jan 11 2007

Corollary 3 of the Farhi reference mentions this sequence. - Roger L. Bagula, Nov 08 2009

Number of UDUD's in all length n+3 left factors of Dyck paths (here U=(1,1) and D=(1,-1)). Example: a(2)=2 because in (UDUD)U, UDUUD, UDUUU, UUDDU, U(UDUD), UUDUU, UUUDD, UUUDU, UUUUD, and UUUUU we have a total of two UDUDs (shown between parentheses). Also number of UUDD's in all length n+3 left factors of Dyck paths (here U=(1,1) and D=(1,-1)). Example: a(2)=2 because in UDUDU, UDUUD, UDUUU, (UUDD)U, UUDUD, UUDUU, U(UUDD), UUUDU, UUUUD, and UUUUU we have a total of two UUDDs (shown between parentheses). - Emeric Deutsch, Jun 19 2011

Apparently the number of long ascents in all symmetric Dyck (n+1)-paths. - David Scambler, Aug 17 2012

Beginning with the least positive term multiple of an odd prime p (which is a(p)), we have exactly p+1 consecutive terms multiple of p. - Vladimir Shevelev, Aug 17 2012

Apparently also the count of 'unmatched symbols' in the binary strings of length n (see A008314). - Wouter Meeussen, May 26 2013

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• Ruggero Bandiera and Florian Schaetz, Eulerian idempotent, pre-Lie logarithm and combinatorics of trees, arXiv:1702.08907 [math.CO], 2017. See p. 34.
• F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq., Vol. 14 (2011), Article 11.3.5.
• Bakir Farhi, An identity involving the least common multiple of binomial coefficients and its application, The American Mathematical Monthly, Vol. 116, No. 9 (2009), pp. 836-839, see p. 838; arXiv preprint, arXiv:0906.2295 [math.NT], 2009.
• Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
• D. Nečas, Elementary models of low-pressure plasma polymerisation into nanofibrous mats, Phys. Scr. (2025) Vol. 100, 055601. See p. 10.
• Helmut Prodinger, Dispersed Dyck paths revisited, arXiv:2402.13026 [math.CO], 2024.

Crossrefs

Cf. A134757, A008314.

Programs

Magma

[n*Binomial(n-1, Floor((n-1)/2)): n in [0..35]]; // Vincenzo Librandi, Sep 14 2015

Maple

swing := n -> n!/iquo(n,2)!^2:
A100071 := n -> swing(n)*(n/2)^(n-1 mod 2):
seq(A100071(i),i=0..30); # Peter Luschny, Aug 31 2011

Mathematica

Table[(Floor[n/2] + Ceiling[n/2] + 1)!/(Floor[n/2]!*Ceiling[n/2]!), {n, 1, 40}] (* Stefan Steinerberger, Nov 04 2008 *)
Table[If[n == 0, 0, n*Binomial[n - 1, Floor[(n - 1)/2]]], {n, 0, 30}] (* Roger L. Bagula, Nov 08 2009 *);
Table[ Tr[ Table[Count[match[-1 + 2*IntegerDigits[n, 2, k]], 0], {n, 2^(k - 1), 2^k - 1}]], {k, 16}] (* function 'match' see A008314; Wouter Meeussen, May 26 2013 *)

PARI

a(n) = n * binomial(n-1, (n-1)\2); \\ Michel Marcus, Sep 14 2015

Sage

def A100071(n):
    f = factorial(n)/factorial(n//2)^2
    return f if is_odd(n) else f*(n/2)
[A100071(n) for n in (0..50)]  # Peter Luschny, Aug 17 2012

Formula

G.f.: 2*x*(1 - sqrt(1 - 4*x^2))/(sqrt(1 - 4*x^2)*(sqrt(1 - 4*x^2) + 2*x - 1)^2).

G.f.: (1/sqrt(1 - 4*x^2))*x*c(x^2)/(1 - x*c(x^2))^2.

a(n) = Sum_{k = 0..floor(n/2)} binomial(n,k)*(n - 2*k).

Sum_{k = 0..floor(n/2)} binomial(n-k,k)*(-1)^k*a(n-2k) = 1.

From Paul Barry, Jan 11 2007: (Start)

a(n) = n*binomial(n-1, floor((n-1)/2));

a(n) = Sum_{k = 0..n} binomial(n,k)*2^(n-k)*binomial(2*k-2, k-1)*(-1)^(k-1). (End)

Starting (1, 2, 6, 12, ...), = inverse binomial transform of A134757: (1, 3, 11, 37, 123, 401, ...). - Gary W. Adamson, Nov 08 2007

a(n) = a(n-1)*n/floor(n/2) for n > 0. - Reinhard Zumkeller, Jan 20 2008

G.f.: x/((1 - 2*x)*sqrt(1 - 4*x^2)). - Paul Barry, Apr 25 2008

a(n) = (floor(n/2) + ceiling(n/2) + 1)!/(floor(n/2)! * ceiling(n/2)!). - Stefan Steinerberger, Nov 04 2008

a(n) = A056040(n)*(n/2)^((n-1) mod 2). - Peter Luschny, Aug 31 2011

Asymptotic: a(n) ~ b(n) where b(n) = ceiling(2^(n-1)*sqrt(2*n-(-1)^n)/sqrt(Pi)). b(n) is also a lower bound of a(n) and an upper bound of 2^(n-1). With corollary 3 from Bakir Farhi (see reference) lcm(1,2,...,n) >= a(n) >= b(n) >= 2^(n-1). - Peter Luschny, Aug 17 2012

a(n) = n for n < 3, a(n) = 4*a(n-2) + 2*a(n-1)/(n-1) for n >= 3. - Alexander R. Povolotsky, Aug 17 2012

E.g.f.: x*(BesselI(0,2*x) + BesselI(1,2*x)). - Peter Luschny, Aug 19 2012

a(n) = (-1)^(n*(n+1)/2) * Sum_{k = 0..n} (-1)^k*k*binomial(n,k)^2. - Peter Bala, Jul 25 2016

a(n) = n!/(floor((n-1)/2)!*ceiling((n-1)/2)!). See the Bandiera link. - Michel Marcus, Feb 28 2017

D-finite with recurrence (-n+1)*a(n) + 2*a(n-1) + 4*(n-1)*a(n-2) = 0. - R. J. Mathar, Aug 09 2017

From Amiram Eldar, Mar 10 2022: (Start)

Sum_{n>=1} 1/a(n) = Pi/sqrt(3).

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(3)). (End)

Extensions

Name changed, using part of a comment from Paul Barry, by Peter Luschny, Aug 17 2012