login
a(n) = n * binomial(n-1, floor((n-1)/2)) = n * max_{i=0..n} binomial(n-1, i).
25

%I #116 Feb 21 2024 08:28:06

%S 0,1,2,6,12,30,60,140,280,630,1260,2772,5544,12012,24024,51480,102960,

%T 218790,437580,923780,1847560,3879876,7759752,16224936,32449872,

%U 67603900,135207800,280816200,561632400,1163381400,2326762800

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

%C Old name: An inverse Chebyshev transform of n.

%C Hankel transform is (-1)^n*n*2^(n-1), A085750. This is the inverse binomial transform of -n. - _Paul Barry_, Jan 11 2007

%C Corollary 3 of the Farhi reference mentions this sequence. - _Roger L. Bagula_, Nov 08 2009

%C 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

%C Apparently the number of long ascents in all symmetric Dyck (n+1)-paths. - _David Scambler_, Aug 17 2012

%C 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

%C Apparently also the count of 'unmatched symbols' in the binary strings of length n (see A008314). - _Wouter Meeussen_, May 26 2013

%H T. D. Noe, <a href="/A100071/b100071.txt">Table of n, a(n) for n = 0..1000</a>

%H Ruggero Bandiera and Florian Schaetz, <a href="https://arxiv.org/abs/1702.08907">Eulerian idempotent, pre-Lie logarithm and combinatorics of trees</a>, arXiv:1702.08907 [math.CO], 2017. See p. 34.

%H F. Disanto, A. Frosini, and S. Rinaldi, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Rinaldi/square.html">Square involutions</a>, J. Int. Seq., Vol. 14 (2011), Article 11.3.5.

%H Bakir Farhi, <a href="https://www.jstor.org/stable/40391302">An identity involving the least common multiple of binomial coefficients and its application</a>, The American Mathematical Monthly, Vol. 116, No. 9 (2009), pp. 836-839, see p. 838; <a href="http://arxiv.org/abs/0906.2295">arXiv preprint</a>, arXiv:0906.2295 [math.NT], 2009.

%H Nikita Gogin and Mika Hirvensalo, <a href="https://pca-pdmi.ru/2020/files/10/GoHi2020ExtAbstract.pdf">On the Moments of Squared Binomial Coefficients</a>, (2020).

%H Helmut Prodinger, <a href="https://arxiv.org/abs/2402.13026">Dispersed Dyck paths revisited</a>, arXiv:2402.13026 [math.CO], 2024.

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

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

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

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

%F From _Paul Barry_, Jan 11 2007: (Start)

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

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

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

%F a(n) = a(n-1)*n/floor(n/2) for n > 0. - _Reinhard Zumkeller_, Jan 20 2008

%F G.f.: x/((1 - 2*x)*sqrt(1 - 4*x^2)). - _Paul Barry_, Apr 25 2008

%F a(n) = (floor(n/2) + ceiling(n/2) + 1)!/(floor(n/2)! * ceiling(n/2)!). - _Stefan Steinerberger_, Nov 04 2008

%F a(n) = A056040(n)*(n/2)^((n-1) mod 2). - _Peter Luschny_, Aug 31 2011

%F 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

%F 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

%F E.g.f.: x*(BesselI(0,2*x) + BesselI(1,2*x)). - _Peter Luschny_, Aug 19 2012

%F a(n) = (-1)^(n*(n+1)/2) * Sum_{k = 0..n} (-1)^k*k*binomial(n,k)^2. - _Peter Bala_, Jul 25 2016

%F a(n) = n!/(floor((n-1)/2)!*ceiling((n-1)/2)!)). See the Banderia link. - _Michel Marcus_, Feb 28 2017

%F 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

%F From _Amiram Eldar_, Mar 10 2022: (Start)

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

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

%p swing := n -> n!/iquo(n,2)!^2:

%p A100071 := n -> swing(n)*(n/2)^(n-1 mod 2):

%p seq(A100071(i),i=0..30); # _Peter Luschny_, Aug 31 2011

%t Table[(Floor[n/2] + Ceiling[n/2] + 1)!/(Floor[n/2]!*Ceiling[n/2]!), {n, 1, 40}] (* _Stefan Steinerberger_, Nov 04 2008 *)

%t Table[If[n == 0, 0, n*Binomial[n - 1, Floor[(n - 1)/2]]], {n, 0, 30}] (* _Roger L. Bagula_, Nov 08 2009 *);

%t 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 *)

%o (Sage)

%o def A100071(n):

%o f = factorial(n)/factorial(n//2)^2

%o return f if is_odd(n) else f*(n/2)

%o [A100071(n) for n in (0..50)] # _Peter Luschny_, Aug 17 2012

%o (Magma) [n*Binomial(n-1, Floor((n-1)/2)): n in [0..35]]; // _Vincenzo Librandi_, Sep 14 2015

%o (PARI) a(n) = n * binomial(n-1, (n-1)\2); \\ _Michel Marcus_, Sep 14 2015

%Y Cf. A134757, A008314.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Nov 02 2004

%E Name changed, using part of a comment from _Paul Barry_, by _Peter Luschny_, Aug 17 2012