A117972 -id:A117972 - cp's OEIS frontend

A048896 a(n) = 2^(A000120(n+1) - 1), n >= 0.

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4

Offset: 0

Wolfdieter Lang

a(n) = 2^A048881 = 2^{maximal power of 2 dividing the n-th Catalan number (A000108)}. [Comment corrected by N. J. A. Sloane, Apr 30 2018]
Row sums of triangle A128937. - Philippe Deléham, May 02 2007
a(n) = sum of (n+1)-th row terms of triangle A167364. - Gary W. Adamson, Nov 01 2009
a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011
For n > 0: a(n) = A007954(A007931(n)). - Reinhard Zumkeller, Oct 26 2012
a(n) = A261363(2*(n+1), n+1). - Reinhard Zumkeller, Aug 16 2015
From Gus Wiseman, Oct 30 2022: (Start)
Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4).
Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4.
(End)

			From _Omar E. Pol_, Jul 21 2009: (Start)
If written as a triangle:
  1;
  1,2;
  1,2,2,4;
  1,2,2,4,2,4,4,8;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16;
  1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32;
  ...,
the first half-rows converge to Gould's sequence A001316.
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Neil J. Calkin, Eunice Y. S. Chan, and Robert M. Corless, Some Facts and Conjectures about Mandelbrot Polynomials, Maple Transactions (2021) Vol. 1, No. 1 Article 1.
Neil J. Calkin, Eunice Y. S. Chan, Robert M. Corless, David J. Jeffrey, and Piers W. Lawrence, A Fractal Eigenvector, arXiv:2104.01116 [math.DS], 2021.
Emeric Deutsch and B. E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, arXiv:math/0407326 [math.CO], 2004; J. Num. Theory 117 (2006), 191-215.
OEIS Wiki, Montgomery's pair correlation conjecture
Gus Wiseman, Statistics, classes, and transformations of standard compositions

This is Guy Steele's sequence GS(3, 5) (see A135416).
Equals first right hand column of triangle A160468.
Equals A160469(n+1)/A002425(n+1).
Cf. A000079, A001316, A117972, A160476, A167364, A220466, A261363.
Standard compositions are listed by A066099.
The opposite version (counting refinements) is A080100.
The version for Heinz numbers of partitions is A317141.
Cf. A000120, A001511, A029931, A048793, A058891, A070939, A272919, A357134.

a048896 n = a048896_list !! n
a048896_list = f [1] where f (x:xs) = x : f (xs ++ [x,2*x])
-- Reinhard Zumkeller, Mar 07 2011

import Data.List (transpose)
a048896 = a000079 . a000120
a048896_list = 1 : concat (transpose
   [zipWith (-) (map (* 2) a048896_list) a048896_list,
    map (* 2) a048896_list])
-- Reinhard Zumkeller, Jun 16 2013

[Numerator(2^n / Factorial(n+1)): n in [0..100]]; // Vincenzo Librandi, Apr 12 2014

a := n -> 2^(add(i,i=convert(n+1,base,2))-1): seq(a(n), n=0..97); # Peter Luschny, May 01 2009

NestList[Flatten[#1 /. a_Integer -> {a, 2 a}] &, {1}, 4] // Flatten (* Robert G. Wilson v, Aug 01 2012 *)
Table[Numerator[2^n / (n + 1)!], {n, 0, 200}] (* Vincenzo Librandi, Apr 12 2014 *)
Denominator[Table[BernoulliB[2*n] / (Zeta[2*n]/Pi^(2*n)), {n, 1, 100}]] (* Terry D. Grant, May 29 2017 *)
Table[Denominator[((2 n)!/2^(2 n + 1)) (-1)^n], {n, 1, 100}]/4 (* Terry D. Grant, May 29 2017 *)
2^IntegerExponent[CatalanNumber[Range[0,100]],2] (* Harvey P. Dale, Apr 30 2018 *)

a(n)=if(n<1,1,if(n%2,a(n/2-1/2),2*a(n-1)))

a(n) = 1 << (hammingweight(n+1)-1); \\ Kevin Ryde, Feb 19 2022

a(n) = 2^A048881(n).
a(n) = 2^k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n).
It appears that a(n) = Sum_{k=0..n} binomial(2*(n+1), k) mod 2. - Christopher Lenard (c.lenard(AT)bendigo.latrobe.edu.au), Aug 20 2001
a(0) = 1; a(2*n) = 2*a(2*n-1); a(2*n+1) = a(n).
a(n) = (1/2) * A001316(n+1). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004
It appears that a(n) = Sum_{k=0..2n} floor(binomial(2n+2, k+1)/2)(-1)^k = 2^n - Sum_{k=0..n+1} floor(binomial(n+1, k)/2). - Paul Barry, Dec 24 2004
a(n) = Sum_{k=0..n} (T(n,k) mod 2) where T = A039598, A053121, A052179, A124575, A126075, A126093. - Philippe Deléham, May 02 2007
a(n) = numerator(b(n)), where sin(x)^2/x = Sum_{n>0} b(n)*(-1)^n x^(2*n-1). - Vladimir Kruchinin, Feb 06 2013
a((2*n+1)*2^p-1) = A001316(n), p >= 0 and n >= 0. - Johannes W. Meijer, Feb 12 2013
a(n) = numerator(2^n / (n+1)!). - Vincenzo Librandi, Apr 12 2014
a(2n) = (2n+1)!/(n!n!)/A001803(n). - Richard Turk, Aug 23 2017
a(2n-1) = (2n-1)!/(n!(n-1)!)/A001790(n). - Richard Turk, Aug 23 2017

New definition from N. J. A. Sloane, Mar 01 2008

A061549 Denominator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p=1/4.

Original entry on oeis.org

1, 8, 128, 1024, 32768, 262144, 4194304, 33554432, 2147483648, 17179869184, 274877906944, 2199023255552, 70368744177664, 562949953421312, 9007199254740992, 72057594037927936, 9223372036854775808, 73786976294838206464, 1180591620717411303424, 9444732965739290427392

Offset: 0

Leah Schmelzer (leah2002(AT)mit.edu), May 16 2001

We observe that b(n) = log(a(n))/log(2) = A120738(n). Furthermore c(n+1) = b(n+1)-b(n) = A090739(n+1) and c(n+1)-3 = A007814(n+1) for n>=0. - Johannes W. Meijer, Jul 06 2009

Using WolframAlpha, it appears that 2*a(n) gives the coefficients of Pi in the denominators of the residues of f(z) = 2z choose z at odd negative half integers. E.g., the residues of f(z) at z = -1/2, -3/2, -5/2 are 1/(2*Pi), 1/(16*Pi), and 3/(256*Pi) respectively. - Nicholas Juricic, Mar 31 2022

			For n=1, the binomial(2*n-1/2, -1/2) yields the term 3/8. The denominator of this term is 8, which is the second term of the sequence.

Indranil Ghosh, Table of n, a(n) for n = 0..500
Robert M. Kozelka, Grade Point Averages and the Central Limit Theorem, American Mathematical Monthly. Nov. 1979 (86:9) pp. 773-7.
Eric Weisstein's World of Mathematics, Circle Line Picking
Eric Weisstein's World of Mathematics, Gamma Function

Cf. A007814, A061548, A090739.
Bisection of A046161.
Appears in A162448.

A061549:= func< n | 2^(4*n-(&+Intseq(2*n, 2))) >;
[A061549(n): n in [0..30]]; // G. C. Greubel, Oct 20 2024

seq(denom(binomial(2*n-1/2, -1/2)), n=0..20);

Table[Denominator[(4*n)!/(2^(4*n)*(2*n)!^2) ], {n, 0, 20}] (* Indranil Ghosh, Mar 11 2017 *)

for(n=0, 20, print1(denominator((4*n)!/(2^(4*n)*(2*n)!^2)),", ")) \\ Indranil Ghosh, Mar 11 2017

import math
f = math.factorial
def A061549(n): return (2**(4*n)*f(2*n)**2) // math.gcd(f(4*n), (2**(4*n)*f(2*n)**2)) # Indranil Ghosh, Mar 11 2017

# uses[A000120]
def a(n): return 1 << (4*n - A000120(n))
[a(n) for n in (0..19)]  # Peter Luschny, Dec 02 2012

a(n) = denominator of binomial(2*n-1/2, -1/2).
a(n) are denominators of coefficients of 1/(sqrt(1+x)-sqrt(1-x)) power series. - Benoit Cloitre, Mar 12 2002
a(n) = 16^n/A001316(n). - Paul Barry, Jun 29 2006
a(n) = denom((4*n)!/(2^(4*n)*(2*n)!^2)). - Johannes W. Meijer, Jul 06 2009
a(n) = abs(A067624(n)/A117972(n)). - Johannes W. Meijer, Jul 06 2009

More terms from Asher Auel, May 20 2001

A156769 a(n) = denominator(2^(2n-2)/factorial(2n-1)).

Original entry on oeis.org

1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 49308808782358125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875, 122529844256906551386796875, 4043484860477916195764296875

Offset: 1

Johannes W. Meijer, Feb 15 2009

Resembles A036279, the denominators in the Taylor series for tan(x). The first difference occurs at a(12).
The numerators of the two formulas for this sequence lead to A001316, Gould's sequence.
Stephen Crowley indicated on Aug 25 2008 that a(n) = denominator(Zeta(2*n)/Zeta(1-2*n)) and here numerator((Zeta(2*n)/Zeta(1-2*n))/(2*(-1)^(n)*(Pi)^(2*n))) leads to Gould's sequence.
This sequence appears in the Eta and Zeta triangles A160464 and A160474. Its resemblance to the sequence of the denominators of the Taylor series for tan(x) led to the conjecture A156769(n) = A036279(n)*A089170(n-1). - Johannes W. Meijer, May 24 2009

G. C. Greubel, Table of n, a(n) for n = 1..250

Cf. A036279 Denominators in Taylor series for tan(x).
Cf. A001316 Gould's sequence appears in the numerators.
Cf. A000265, A036279, A089170, A117972, A160464, A160469 (which resembles the numerators of the Taylor series for tan(x)), A160474. - Johannes W. Meijer, May 24 2009

[Denominator(4^(n-1)/Factorial(2*n-1)): n in [1..25]]; // G. C. Greubel, Jun 19 2021

a := n ->(2*n-1)!*2^(add(i,i=convert(n-1,base,2))-2*n+2); # Peter Luschny, May 02 2009

a[n_] := Denominator[4^(n-1)/(2n-1)!];
Array[a, 15] (* Jean-François Alcover, Jun 20 2018 *)

[denominator(4^(n-1)/factorial(2*n-1)) for n in (1..25)] # G. C. Greubel, Jun 19 2021

a(n) = denominator( Product_{k=1..n-1} 2/(k*(2*k+1)) ).
G.f.: (1/2)*z^(1/2)*sinh(2*z^(1/2)).
From Johannes W. Meijer, May 24 2009: (Start)
a(n) = abs(A117972(n))/A000265(n).
a(n) = A036279(n)*A089170(n-1). (End)
a(n) = A049606(2*n-1). - Zhujun Zhang, May 29 2019

A120738 a(n) = 4*n - A000120(n).

Original entry on oeis.org

0, 3, 7, 10, 15, 18, 22, 25, 31, 34, 38, 41, 46, 49, 53, 56, 63, 66, 70, 73, 78, 81, 85, 88, 94, 97, 101, 104, 109, 112, 116, 119, 127, 130, 134, 137, 142, 145, 149, 152, 158, 161, 165, 168, 173, 176, 180, 183, 190, 193, 197, 200, 205, 208, 212, 215, 221, 224, 228

Offset: 0

Paul Barry, Jun 29 2006

Partial sums of A090739.
a(n) is also the increasing sequence of exponents of x in Product_{k > 1} (1 + x^(2^k - 1)). - Paul Pearson (ppearson(AT)rochester.edu), Aug 06 2008
Related to partial sums of the Ruler sequence A001511 by a(n) = A005187(2n), therefore {a(n)+1} are the indices of 1's in A252488. - M. F. Hasler, Jan 22 2015

Michel Marcus, Table of n, a(n) for n = 0..10000
Keith Johnson, and Kira Scheibelhut, Rational Polynomials That Take Integer Values at the Fibonacci Numbers, American Mathematical Monthly 123.4 (2016): 338-346. See p. 340.

Cf. A000120, A000225, A001316, A001511, A005187, A011371, A030308, A061549, A067624, A086343, A090739, A117972, A252488.

A120738:= func< n | 4*n-(&+Intseq(n, 2)) >;
[A120738(n): n in [0..100]]; // G. C. Greubel, Oct 20 2024

a:=n->simplify(log[2](16^n/(add(modp(binomial(n,k),2),k=0..n))));
a:=n->simplify(log[2](16^n/(2^(n-(padic[ordp](n!,2)))))); # Note: n-(padic[ordp](n!,2)) is the number of 1's in the binary expansion of n. - Paul Pearson (ppearson(AT)rochester.edu), Aug 06 2008

Table[4 n - DigitCount[n, 2, 1], {n, 0, 58}] (* Michael De Vlieger, Nov 06 2016 *)

{a(n) = if( n < 0, 0, 4*n - subst( Pol( binary( n ) ), x, 1) ) } /* Michael Somos, Aug 28 2007 */

a(n) = 4*n - hammingweight(n); \\ Michel Marcus, Nov 06 2016

# Python 3.10
def A120738(n): return (n<<2)-n.bit_count() # Chai Wah Wu, Jul 12 2022

A120738 = lambda n: 4*n - sum(n.digits(2))
print([A120738(n) for n in (0..58)]) # Peter Luschny, Nov 06 2016

a(n) = log_2(16^n/A001316(n)). [This was the original definition.]
a(n) = 2n + A005187(n).
a(n) = 3n + A011371(n).
a(n) = 4n - log_2(A001316(n)).
a(n) = log_2(A061549(n)).
2^a(n) = 16^n/A001316(n) = A061549(n).
a(n) = A086343(n) + A001511(n) for n>0. - Alford Arnold, Mar 23 2009
2^a(n) = abs(A067624(n)/A117972(n)). - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_{k>=0} (A030308(n,k)*A000225(k+2)). - Philippe Deléham, Oct 16 2011
a(n) = A005187(2n). - M. F. Hasler, Jan 22 2015

Definition simplified by M. F. Hasler, Dec 29 2012

A160469 The left hand column of the triangle A160468.

Original entry on oeis.org

1, 1, 2, 17, 62, 1382, 21844, 929569, 6404582, 443861162, 18888466084, 1936767361654, 58870668456604, 8374643517010684, 689005380505609448, 129848163681107301953, 1736640792209901647222, 418781231495293038913922

Offset: 1

Johannes W. Meijer, May 24 2009

Resembles A002430, the numerators of the Taylor series for tan(x). The first difference occurs at a(12). (Its resemblance to this sequence led to the conjecture A160469(n) = A002430(n)*A089170(n-1).)

Equals the first left hand column of A160468.
Equals A002430(n)*A089170(n-1).
Equals (A002430(n)/A036279(n))*(A117972(n)/A000265(n)).
Equals A048896(n-1)*A002425(n).
Cf. A156769 (which resembles the denominators of the Taylor series for tan(x)).

a(n) = A002430(n)*A089170(n-1) with A002430 (n) = numer((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)* bernoulli(2*n)/(2*n)!) and A089170 (n-1) = numer(2*bernoulli(2*n)* (4^n-1)/(2*n))/ numer((4^n-1)*bernoulli(2*n)/(2*n)!) for n = 1, 2, 3, ....

A117973 a(n) = 2^(wt(n)+1), where wt() = A000120().

Original entry on oeis.org

2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32, 32, 64, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32, 32, 64, 8, 16, 16, 32, 16, 32, 32, 64, 16, 32, 32, 64, 32, 64, 64, 128, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32

Offset: 0

Eric W. Weisstein, Apr 06 2006

Denominator of Zeta'(-2n).
If Gould's sequence A001316 is written as a triangle, this is what the rows converge to. In other words, let S_0 = [2], and construct S_{n+1} by following S_n with 2*S_n. Then this is S_{oo}. - N. J. A. Sloane, May 30 2009
In A160464 the coefficients of the ES1 matrix are defined. This matrix led to the discovery that the successive differences of the ES1[1-2*m,n] coefficients for m = 1, 2, 3, ..., are equal to the values of Zeta'(-2n), see also A094665 and A160468. - Johannes W. Meijer, May 24 2009

			-zeta(3)/(4*Pi^2), (3*zeta(5))/(4*Pi^4), (-45*zeta(7))/(8*Pi^6), (315*zeta(9))/(4*Pi^8), (-14175*zeta(11))/(8*Pi^10), ...

J. Sondow and E. W. Weisstein, MathWorld: Riemann Zeta Function

Cf. A001316, A117972, A160464, A094665, A160468, A220466.

S := [2]; S := [op(S), op(2*S)]; # repeat ad infinitum! - N. J. A. Sloane, May 30 2009
a := n -> 2^(add(i,i=convert(n,base,2))+1); # Peter Luschny, May 02 2009

Mathematica
```
Denominator[(2*n)!/2^(2*n + 1)]
```

For n>=0, a(n) = 2 * A001316(n). - N. J. A. Sloane, May 30 2009
For n>0, a(n) = 4 * A048896(n). - Peter Luschny, May 02 2009
a(0) = 2; for n>0, write n = 2^i + j where 0 <= j < 2^i; then a(n) = 2*a(j).
a((2*n+1)*2^p-1) = 2^(p+1) * A001316(n), p >= 0. - Johannes W. Meijer, Jan 28 2013

Entry revised by N. J. A. Sloane, May 30 2009

A160468 Triangle of polynomial coefficients related to the o.g.f.s of the RES1 polynomials.

Original entry on oeis.org

1, 1, 2, 1, 17, 26, 2, 62, 192, 60, 1, 1382, 7192, 5097, 502, 2, 21844, 171511, 217186, 55196, 2036, 2, 929569, 10262046, 20376780, 9893440, 1089330, 16356, 4, 6404582, 94582204, 271154544, 215114420, 48673180, 2567568, 16376, 1

Offset: 1

Johannes W. Meijer, May 24 2009

In A160464 we defined the ES1 matrix by ES1[2*m-1,n=1] and in A094665 it was shown that the n-th term of the coefficients of matrix row ES1[1-2*m,n] for m >= 1 can be generated with the RES1(1-2*m,n) polynomials.
We define the o.g.f.s. of these polynomials by GFRES1(z,1-2*m) = sum(RES1(1-2*m,n)*z^(n-1), n=1..infinity) for m >= 1. The general expression of the o.g.f.s. is GFRES1(z,1-2*m) = (-1)*RE(z,1-2*m)/(2*p(m-1)*(z-1)^(m)). The p(m-1), m >= 1, sequence is Gould's sequence A001316.
The coefficients of the RE(z,1-2*m) polynomials lead to the triangle given above.
The E(z,n) = numer(sum((-1)^(n+1)*k^n*z^(k-1), k=1..infinity)) polynomials with n >= 1, see the Maple algorithm, lead to the Eulerian numbers A008292.
Some of our results are conjectures based on numerical evidence.

			The first few rows are:
[1]
[1]
[2, 1]
[17, 26, 2]
[62, 192, 60, 1]
The first few polynomials RE(z,m) are:
RE(z,-1) = 1
RE(z,-3) = 1
RE(z,-5) = 2+z
RE(z,-7) = 17+26*z+2*z^2
The first few GFRES1(z,m) are:
GFRES1(z,-1) = -(1/1)*(1)/(2*(z-1)^1)
GFRES1(z,-3) = -(1/2)*(1)/(2*(z-1)^2)
GFRES1(z,-5) = -(1/2)*(2+z)/(2*(z-1)^3)
GFRES1(z,-7) = -(1/4)*(17+26*z+2*z^2)/(2*(z-1)^4)

Grzegorz Rzadkowski, M Urlinska, A Generalization of the Eulerian Numbers, arXiv preprint arXiv:1612.06635, 2016

Cf. A160464, A094665 and A083061.
For the Eulerian numbers E(n, k) see A008292.
The p(n) sequence equals Gould's sequence A001316.
The first right hand column of the triangle equals A048896.
The first left hand column equals A160469.
The row sums equal the absolute values of A117972.

nmax := 8; mmax := nmax: T(0, x) := 1: for i from 1 to nmax do dgr := degree(T(i-1, x), x): for na from 0 to dgr do c(na) := coeff(T(i-1, x), x, na) od: T(i-1, x+1) := 0: for nb from 0 to dgr do T(i-1, x+1) := T(i-1, x+1) + c(nb)*(x+1)^nb od: for nc from 0 to dgr do ECGP(i-1, nc+1) := coeff(T(i-1, x), x, nc) od: T(i, x) := expand((2*x+1)*(x+1)*T(i-1, x+1) - 2*x^2*T(i-1, x)) od: dgr := degree (T(nmax, x), x): kmax := nmax: for k from 1 to kmax do p := k: for m from 1 to k do E(m, k) := sum((-1)^(m-q)*(q^k)*binomial(k+1, m-q), q=1..m) od: fx(p) := (-1)^(p+1) * (sum(E(r, k)*z^(k-r), r=1..k))/(z-1)^(p+1): GF(-(2*p+1)) := sort(simplify(((-1)^p* 1/2^(p+1)) * sum(ECGP(k-1, k-s)*fx(k-s), s=0..k-1)), ascending): NUMGF(-(2*p+1)) := -numer(GF(-(2*p+1))): for n from 1 to mmax+1 do A(k+1, n) := coeff(NUMGF(-(2*p+1)), z, n-1) od: od: for m from 2 to mmax do A(1, m) := 0 od: A(1, 1) := 1: FT(1) := 1: for n from 1 to nmax do for m from 1 to n do FT((n)*(n-1)/2+m+1) := A(n+1, m) end do end do: a := n-> FT(n): seq(a(n), n = 1..(nmax+1)*(nmax)/2+1);

T[ n_, k_] := Coefficient[a[2 n]/2^IntegerExponent[(2 n)!, 2], x, n + k];
a[0] = a[1] = 1; a[ m_] := a[m] = With[{n = m - 1}, x Sum[ a[k] a[n - k] Binomial[n, k], {k, 0, n}]]; Join[{1}, Flatten@Table[T[n, k], {n, 1, 8}, {k, 0, n - 1}]] (* Michael Somos, Apr 22 2020 *)

Edited by Johannes W. Meijer, Sep 23 2012

A162445 A sequence related to the Beta function.

Original entry on oeis.org

1, 8, 384, 46080, 2064384, 3715891200, 392398110720, 1428329123020800, 274239191619993600, 1678343852714360832000, 102043306245033138585600, 4714400748520531002654720000, 160144566965128191597871104000

Offset: 0

Johannes W. Meijer, Jul 06 2009

We define F(z) = Beta(1/2-z/2,1/2+z/2)/Beta(1/2,1/2) = 1/sin(Pi*(1+z)/2) with Beta(z,w) the Beta function. See A008956 for a closely related function.
For the Taylor series expansion of F(z) we can write F(z) = sum(b(n)*(Pi*z)^(2*n)/a(n), n=0..infinity) with b(n) = A046976(n) and a(n) the sequence given above.
We can also write F(z) = sum(c(n)*(Pi*z)^(2*n)/d(n), n=0..infinity) with c(n) = A000364(n) and d(n) = A067624(n).
If p(n) is the exponent of the prime factor 2 in a(n) than p(n) = A120738(n) and 2^p(n) = A061549(n) = abs((4*n)!!/A117972(n)).

Bisection of A050971
Equals 2^(2*n)*A046977(n)
Cf. A008956, A046976, A000364, A067624, A120738, A061549 and A117972.

Denominator[Table[EulerE[2n]/(4n)!!,{n,0,20}]] (* Harvey P. Dale, Jun 23 2013 *)

a(n) = denom(euler(2*n)/(4*n)!!)

cp's OEIS Frontend

A048896 a(n) = 2^(A000120(n+1) - 1), n >= 0.

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A061549 Denominator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p=1/4.

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A156769 a(n) = denominator(2^(2*n-2)/factorial(2*n-1)).

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A120738 a(n) = 4*n - A000120(n).

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A160469 The left hand column of the triangle A160468.

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A117973 a(n) = 2^(wt(n)+1), where wt() = A000120().

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A156769 a(n) = denominator(2^(2n-2)/factorial(2n-1)).