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.

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A206771 0 followed by the numerators of the reduced (A001803(n) + A001790(n)) / (2*A046161(n)).

Original entry on oeis.org

0, 1, 1, 9, 5, 175, 189, 1617, 429, 57915, 60775, 508079, 264537, 8788507, 9100525, 75218625, 9694845, 5109183315, 5250613995, 43106892675, 22090789875, 723694276305, 740104577355, 6049284520695, 1543768261425, 201547523019375
Offset: 0

Views

Author

Paul Curtz, Jan 10 2013

Keywords

Comments

We write the fractions a(n)/b(n) and higher order differences as a matrix:
0, 1, 1, 9/8, 5/4,...
1, 0, 1/8, 1/8, 15/128,... = (A001790(n)/A046161(n) +Lorbeta(n)) /2
-1, 1/8, 0, -1/128, -1/128,... = (Lorbeta(n+1) +A161200(n+1)/A046161(n+1)) / 2
9/8, -1/8, -1/128, 0, 1/1024,...
-5/4, 15/128, 1/128, 1/1024, 0,...
Here, Lorbeta(0)=1 and Lorbeta(n) = -A098597(n-1)/A046161(n) for n>0 is the inverse of the Lorentz factor.
The first line with numerators a(n) and denominators b(n) is 0, 1, 1, 9/8, 5/4, 175/128, 189/128, 1617/1024, 429/256, 57915/32768, 60775/32768,... It is an autosequence: Its inverse binomial transform is the signed sequence.
a(n+1)/(2*n-1)= 1, 3, 1, 25, 21, 147, 33, 3861, 3575, 26741,... .
a(n+1)/A146535(n) = 9, 5, 35, 27, 539, 39, 4455,... .
A001790(n)/A046161(n) yields the coefficients of the Lorentz factor (or Lorentz gamma factor). With b for beta and g for gamma:
g = (1-b^2)^-1 = 1 + (b^2)/2 + 3*(b^4)/8 + 5*(b^6)/16 + ... .
b = (1-g^-2)^-1 = 1 - (g^-2)/2 - (g^-4)/8 - (g^-6)/16 - ... .
Are the denominators of the first subdiagonal 1, 1/8, -1/128, 1/1024,... A061549(n) ?
a(n+1)/(A000108(n)*b(n)) = 1, 1, 9/16, 1/4, 25/256, 9/256, 49/4096, 1/256, 81/65536, 25/65536, 121/1048576,... = A191871(n+1)/ A084623(n+1)^2 ?

Examples

			From the first formula: a(1)=1*1, a(2)=1*1, a(3)=3*3, a(4)=1*5, a(5)=5*35, a(6)=3*63.
		

Crossrefs

Programs

  • Magma
    /* By definition: */ m:=25; R:=PowerSeriesRing(Rationals(), m); p:=Coefficients(R!(1/(1-x)^(1/2))); q:=Coefficients(R!((1-x)^(-3/2))); A001790:=[Numerator(p[i]): i in [1..m]]; A001803:=[Numerator(q[i]): i in [1..m]]; A046161:=[Denominator(Binomial(2*n,n)/4^n): n in [0..m-1]]; [0] cat [Numerator((A001803[n]+A001790[n])/(2*A046161[n])): n in [1..m]]; // Bruno Berselli, Mar 11 2013
  • Maple
    A206771 := proc(n)
            A001790(n)+A001803(n) ;
            %/2/A046161(n) ;
            numer(%) ;
    end proc: # R. J. Mathar, Jan 18 2013
  • Mathematica
    max = 25; A001803 = CoefficientList[Series[(1 - x)^(-3/2), {x, 0, max}], x] // Numerator; A001790 = CoefficientList[Series[1/Sqrt[(1 - x)], {x, 0, max}], x] // Numerator; A046161 = Table[Binomial[2n, n]/4^n, {n, 0, max}] // Denominator; a[n_] := (A001803[[n]] + A001790[[n]])/(2*A046161[[n]]) // Numerator; a[0] = 0; Table[a[n], {n, 0, max}]
    (* or (from 1st formula) : *) Table[ n*Numerator[4^(1-n)*Binomial[2n-2, n-1]]/2^IntegerExponent[n, 2], {n, 0, max}]
    (* or (from 2nd formula) : *) Table[ Numerator[ CatalanNumber[n-1]/2^(2n-1)]*Numerator[n^2/2^n], {n, 0, max}] (* Jean-François Alcover, Jan 31 2013 *)

Formula

a(n) = A000265(n) * A001790(n-1).
a(n) = A098597(n-1) * A191871(n). See also A181318
a(n) = numerator of n*binomial(2n-2,n-1)/4^(n-1). - Nathaniel Johnston, Dec 16 2022

Extensions

a(11)-a(25) from Jean-François Alcover, Jan 13 2013

A173396 a(n) = A046161(n) + A001803(n).

Original entry on oeis.org

2, 5, 23, 51, 443, 949, 4027, 8483, 142163, 296481, 1232113, 2552405, 21095279, 43490633, 178977107, 367649059, 12065310083, 24714021721, 101124870709, 206667899393, 1687804827349, 3442891889003, 14034579926477, 28583749273429
Offset: 0

Views

Author

Paul Curtz, Feb 17 2010

Keywords

Crossrefs

Programs

  • GAP
    List([0..30], n-> (NumeratorRat((2*n+1)*Binomial(2*n, n)/(4^n)) + DenominatorRat(Binomial(2*n, n)/(4^n)))); # G. C. Greubel, Dec 09 2018
  • Magma
    [Numerator((2*n+1)*Binomial(2*n, n)/(4^n)) + Denominator(Binomial(2*n, n)/(4^n)): n in [0..30]]; // G. C. Greubel, Dec 09 2018
    
  • Maple
    A001803 := proc(n) (2*n+1)*binomial(2*n,n)/4^n ; numer(%) ; end proc:
    A046161 := proc(n) binomial(2*n,n)/4^n ; denom(%) ; end proc:
    A173386 := proc(n) A001803(n)+A046161(n) ; end proc: # R. J. Mathar, Jul 06 2011
  • Mathematica
    Table[Numerator[(2*n+1)*Binomial[2*n, n]/(4^n)] + Denominator[Binomial[2*n, n]/(4^n)], {n,0,30}] (* G. C. Greubel, Dec 09 2018 *)
  • PARI
    for(n=0,30, print1(numerator((2*n+1)*binomial(2*n, n)/(4^n)) + denominator(binomial(2*n, n)/4^n), ", ")) \\ G. C. Greubel, Dec 09 2018
    
  • Sage
    [(numerator((2*n+1)*binomial(2*n, n)/(4^n)) + denominator(binomial(2*n, n)/(4^n))) for n in range(30)] # G. C. Greubel, Dec 09 2018
    

Formula

a(n) = A101926(n) + A173384(n).

A224270 Absolute values of the numerators of the third column of ( 0 followed by (interleave 0 , A001803(n))/A060818(n) ) and its successive differences.

Original entry on oeis.org

1, 1, 5, 11, 95, 203, 861, 1815, 30459, 63635, 264979, 550069, 4555915, 9412543, 38816525, 79898895, 2627302995, 5392044675, 22104436695, 45256266825, 370241638305, 756514878405, 3088866211275, 6300861570705, 102746354288175, 209286947903319
Offset: 0

Views

Author

Paul Curtz, Apr 02 2013

Keywords

Comments

The array is
0, 0, 1, 0, 3/2, 0, 15/8, 0,...
0, 1, -1, 3/2, -3/2, 15/8, -15/8,...
1, -2, 5/2, -3, 27/8, -15/4,...
-3, 9/2, -11/2, 51/8, -57/8,...
15/2, -10, 95/8, -27/2,...
-35/2, 175/8, -203/8,...
315/8, -189/4,...
-693/8,...
Note A001803 in the first column and a variant of A206771(n) in the second column.
Now consider a(n)/A046161(n) and its differences:
1, 1/2, 5/8, 11/16, 95/128, 203/256, 861/1024,...
-1/2, 1/8, 1/16, 7/128, 13/256, 49/1024,... =b(n)/A046161(n)
5/8, -1/16, -1/128, -1/256, -3/1024,...
-11/16, 7/128, 1/256, 1/1024,...
95/128, -13/256, -3/1024,...
-203/256, 49/1024,...
861/1024,...
This an autosequence of second kind. The first column is the signed sequence.
(Its companion, the corresponding autosequence of first kind, is 0, 1, 1, 9/8, 5/4,... in A206771).
Main diagonal: 1, 1/8, -1/128,... = A002596(n)/A061549(n) ?
b(n) = a(n+1) - A171977*a(n). Also for two successive rows (with shifted A171977).

Examples

			a(n)=numerators of 0+1=1, 0+1/2=1/2, 1/4+3/8=5/8, 3/8+5/16=11/16, 15/32+35/128=95/128,... .
		

Crossrefs

Programs

  • Mathematica
    nmax = 25; t1 = Table[ Numerator[ (2*n+1)*(Binomial[2*n, n]/4^n)] / Denominator[ Binomial[2*n, n]/4^n], {n, 0, Ceiling[nmax/2]}]; t2 = Join[{0}, Table[ If[ OddQ[n], 0, t1[[n/2]] ], {n, 1, nmax+2}] ]; t3 = Table[ Differences[t2, n], {n, 0, nmax}]; t3[[All, 3]] // Numerator // Abs (* Jean-François Alcover, Apr 02 2013 *)

Formula

Numerators of (0, 0 followed by A001803(n)/(4*A046161(n))) + A001790(n)/A046161(n).

Extensions

More terms from Jean-François Alcover, Apr 02 2013

A242735 Array read by antidiagonals: form difference table of the sequence of rationals 0, 0 followed by A001803(n)/A046161(n), then extract numerators.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, -3, -1, 1, 3, 15, 3, -1, 3, 15, -35, -5, 1, -1, 5, 35, 315, 35, -5, 3, -5, 35, 315, -693, -63, 7, -3, 3, -7, 63, 693, 3003, 231, -21, 7, -5, 7, -21, 231, 3003, -6435, -429, 33, -9, 5, -5, 9, -33, 429, 6435
Offset: 0

Views

Author

Paul Curtz, May 21 2014

Keywords

Comments

Difference table of c(n)/d(n) = 0, 0, followed by A001803(n)/A046161(n):
0, 0, 1, 3/2, 15/8, 35/16, 315/128, ...
0, 1, 1/2, 3/8, 5/16, 35/128, 63/256, ...
1, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, ...
-3/2, 3/8, 1/16, 3/128, 3/256, 7/1024, 9/2048, ...
15/8, -5/16, -5/128, -3/256, -5/1024, -5/2048, -45/32768, ...
-35/16, 35/128, 7/256, 7/1024, 5/2048, 35/32768, 35/65536, ... etc.
d(n) = 1, 1, followed by A046161(n).
c(n)/d(n) is an autosequence (a sequence whose inverse binomial transform is the signed sequence) of the second kind (the main diagonal is equal to the first upper diagonal multiplied by 2). See A187791.
Antidiagonal denominators: repeat n+1 times d(n).
Second row without 0: Lorentz (gamma) factor = A001790(n)/A046161(n).
Third row: Lorentz beta factor = 1 followed by -A098597(n). Lorbeta(n) in A206771.

Examples

			a(n) as a triangle:
   0;
   0,  0;
   1,  1,  1;
  -3, -1,  1,  3;
  15,  3, -1,  3, 15;
  etc.
		

Programs

  • Mathematica
    c[n_] := (2*n-3)*Binomial[2*(n-2), n-2]/4^(n-2) // Numerator; d[n_] := Binomial[2*(n-2), n-2]/4^(n-2) // Denominator; Clear[a]; a[0, k_] := c[k]/d[k]; a[n_, k_] := a[n, k] = a[n-1, k+1] - a[n-1, k]; Table[a[n-k, k] // Numerator, {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 17 2014 *)

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

Views

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

A046161 a(n) = denominator of binomial(2n,n)/4^n.

Original entry on oeis.org

1, 2, 8, 16, 128, 256, 1024, 2048, 32768, 65536, 262144, 524288, 4194304, 8388608, 33554432, 67108864, 2147483648, 4294967296, 17179869184, 34359738368, 274877906944, 549755813888, 2199023255552, 4398046511104, 70368744177664, 140737488355328, 562949953421312
Offset: 0

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Comments

Also denominator of e(0,n) (see Maple line). - N. J. A. Sloane, Feb 16 2002
Denominator of coefficient of x^n in (1+x)^(k/2) or (1-x)^(k/2) for any odd integer k. - Michael Somos, Sep 15 2004
Numerator of binomial(2n,n)/4^n = A001790(n).
Denominators in expansion of sqrt(c(x)), c(x) the g.f. of A000108. - Paul Barry, Jul 12 2005
Denominator of 2^m*Gamma(m+3/4)/(Gamma(3/4)*Gamma(m+1)). - Stephen Crowley, Mar 19 2007
Denominator in expansion of Jacobi_P(n,1/2,1/2,x). - Paul Barry, Feb 13 2008
This sequence equals the denominators of the coefficients of the series expansions of (1-x)^((-1-2*n)/2) for all integer values of n; see A161198 for detailed information. - Johannes W. Meijer, Jun 08 2009
Numerators of binomial transform of 1, -1/3, 1/5, -1/7, 1/9, ... (Madhava-Gregory-Leibniz series for Pi/4): 1, 2/3, 8/15, 16/35, 128/315, 256/693, .... First differences are -1/3, -2/15, -8/105, -16/315, -128/3465, -256/9009, ... which contain the same numerators, negated. The second differences are 1/5, 2/35, 8/315, 16/1155, 128/15015, ... again with the same numerators. Second column: 2/3, -2/15, 2/35, -2/63, 2/99; see A000466(n+1) = A005563(2n+1). Third column: 8*(1/15, -1/105, 1/315, -1/693, ...), see A061550. See A173294 and A173296. - Paul Curtz, Feb 16 2010
0, 1, 5/3, 11/5, 93/35, 193/63, 793/231, ... = (0 followed by A120778(n))/A001790(n) is the binomial transform of 0, 1, -1/3, 1/5, -1/7, 1/9, ... . See A173755 and formula below. - Paul Curtz, Mar 13 2013
Numerator of power series of arcsin(x)/sqrt(1-x^2), centered at x=0. - John Molokach, Aug 02 2013
Denominators of coefficients in the Taylor series expansion of Sum_{n>=0} exp((-1)^n * Euler(2*n)*x^n/(2*n)), see A280442 for the numerators. - Johannes W. Meijer, Jan 05 2017
Denominators of Pochhammer(n+1, -1/2)/sqrt(Pi). - Adam Hugill, Sep 11 2022
a(n) is the denominator of the mean value of cos(x)^(2*n) from x = 0 to 2*Pi. - Stefano Spezia, May 16 2023
4^n/binomial(2n,n) is the expected value of the number of socks that are randomly drawn out of a drawer of n different pairs of socks, when one sock is drawn out at a time until a matching pair is found (King and King, 2005). - Amiram Eldar, Jul 02 2023
a(n) is the denominator of (1/Pi) * Integral_{x=-oo..+oo} sech(x)^(2*n+1) dx. The corresponding numerator is A001790(n). - Mohammed Yaseen, Jul 29 2023
a(n) is the numerator of Integral_{x=0..Pi/2} sin(x)^(2*n+1) dx. The corresponding denominator is A001803(n). - Mohammed Yaseen, Sep 22 2023

Examples

			sqrt(1+x) = 1 + (1/2)*x - (1/8)*x^2 + (1/16)*x^3 - (5/128)*x^4 + (7/256)*x^5 - (21/1024)*x^6 + (33/2048)*x^7 + ...
binomial(2n,n)/4^n => 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, 6435/32768, ...
The sequence e(0,n) begins 1, 3/2, 21/8, 77/16, 1155/128, 4389/256, 33649/1024, 129789/2048, 4023459/32768, ...
		

References

  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, 1968; Chap. III, Eq. 4.1.
  • B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 513, Eq. (7.282).
  • Eli Maor, e: The Story of a Number. Princeton, New Jersey: Princeton University Press (1994), p. 72.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equations 6:14:5 - 6:14:9 at pages 50-51.

Crossrefs

Cf. A161198 triangle related to the series expansions of (1-x)^((-1-2*n)/2) for all values of n.

Programs

  • Magma
    [Denominator(Binomial(2*n,n)/4^n): n in [0..30]]; // Vincenzo Librandi, Jul 18 2015
    
  • Maple
    e := proc(l,m) local k; add(2^(k-2*m)*binomial(2*m-2*k,m-k)* binomial(m+k, m) *binomial(k,l), k=l..m); end: seq(denom(e(0,n)), n = 0..24);
    Z[0]:=0: for k to 30 do Z[k]:=simplify(1/(2-z*Z[k-1])) od: g:=sum((Z[j]-Z[j-1]), j=1..30): gser:=series(g, z=0, 27): seq(denom(coeff(gser, z, n)), n=-1..23); # Zerinvary Lajos, May 21 2008
    A046161 := proc(n) option remember: if n = 0 then 1 else 2^A001511(n) * procname(n-1) fi: end: A001511 := proc(n): padic[ordp](2*n, 2) end: seq(A046161(n), n = 0..24); # Johannes W. Meijer, Nov 04 2012
    A046161 := n -> 4^n/2^add(i,i=convert(n, base, 2)):
    seq(A046161(n), n=0..24); # Peter Luschny, Apr 08 2014
  • Mathematica
    a[n_, m_] := Binomial[n - m/2 + 1, n - m + 1] - Binomial[n - m/2, n - m + 1]; s[n_] := Sum[ a[n, k], {k, 0, n}]; Table [Denominator[s[n]], {n, 0, 26}] (* Michele Dondi (bik.mido(AT)tiscalinet.it), Jul 11 2002 *)
    Denominator[Table[Binomial[2n,n]/4^n,{n,0,30}]] (* Harvey P. Dale, Oct 29 2012 *)
    Table[Denominator@LegendreP[2n,0],{n,0,24}] (* Andres Cicuttin, Jan 22 2018 *)
  • Maxima
    a(n) := denom(binomial(-1/2,n));
    makelist(a(n),n,0,24); /* Peter Luschny, Nov 21 2012 */
    
  • PARI
    a(n)=if(n<0,0,denominator(binomial(2*n,n)/4^n)) /* Michael Somos, Sep 15 2004 */
    
  • PARI
    a(n)=my(s=n);while(n>>=1,s+=n);2^s \\ Charles R Greathouse IV, Apr 07 2012
    
  • PARI
    a(n)=denominator(I^-n*pollegendre(n,I/2)) \\ Charles R Greathouse IV, Mar 18 2017
    
  • Python
    def A046161(n): return 1<<(n<<1)-n.bit_count() # Chai Wah Wu, Nov 15 2022
  • Sage
    def A046161(n):
        A005187 = lambda n: A005187(n//2) + n if n > 0 else 0
        return 2^A005187(n)
    [A046161(n) for n in (0..24)]  # Peter Luschny, Nov 16 2012
    

Formula

a(n) = 2^(2*n - 1 - A048881(n-1)), if n > 0.
a(n) = 2^A005187(n).
a(n) = 4^n/2^A000120(n). - Michael Somos, Sep 15 2004
a(n) = 2^A001511(n)*a(n-1) with a(0) = 1. - Johannes W. Meijer, Nov 04 2012
a(n) = denominator(binomial(-1/2,n)). - Peter Luschny, Nov 21 2012
a(n) = (0 followed by A120778(n)) + A001790(n). - Paul Curtz, Mar 13 2013
a(n) = 2^n*A060818(n). - Johannes W. Meijer, Jan 05 2017
a(n)/A001790(n) ~ sqrt(Pi*n) (King and King, 2005). - Amiram Eldar, Jul 02 2023

A001790 Numerators in expansion of 1/sqrt(1-x).

Original entry on oeis.org

1, 1, 3, 5, 35, 63, 231, 429, 6435, 12155, 46189, 88179, 676039, 1300075, 5014575, 9694845, 300540195, 583401555, 2268783825, 4418157975, 34461632205, 67282234305, 263012370465, 514589420475, 8061900920775, 15801325804719
Offset: 0

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Comments

Also numerator of e(n-1,n-1) (see Maple line).
Leading coefficient of normalized Legendre polynomial.
Common denominator of expansions of powers of x in terms of Legendre polynomials P_n(x).
Also the numerator of binomial(2*n,n)/2^n. - T. D. Noe, Nov 29 2005
This sequence gives the numerators of the Maclaurin series of the Lorentz factor (see Wikipedia link) of 1/sqrt(1-b^2) = dt/dtau where b=u/c is the velocity in terms of the speed of light c, u is the velocity as observed in the reference frame where time t is measured and tau is the proper time. - Stephen Crowley, Apr 03 2007
Truncations of rational expressions like those given by the numerator operator 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)$ / sigma(2*n) = A056040(2*n) / A060632(2*n+1). Simply said: this sequence is the odd part of the swinging factorial at even indices. - Peter Luschny, Aug 01 2009
It appears that a(n) = A060818(n)*A001147(n)/A000142(n). - James R. Buddenhagen, Jan 20 2010
The convolution of sequence binomial(2*n,n)/4^n with itself is the constant sequence with all terms = 1.
a(n) equals the denominator of Hypergeometric2F1[1/2, n, 1 + n, -1] (see Mathematica code below). - John M. Campbell, Jul 04 2011
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2-2*x+2)^n dx. - Leonid Bedratyuk, Nov 17 2012
a(n) = numerator of the mean value of cos(x)^(2*n) from x = 0 to 2*Pi. - Jean-François Alcover, Mar 21 2013
Constant terms for normalized Legendre polynomials. - Tom Copeland, Feb 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums:
a(n)/A060818(n) = (-2)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A060818(k) = -i.
Sum_{k>=0} (-1)^k*a(k)/A060818(k) = 1/sqrt(3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A060818(k) = -1/sqrt(3).
a(n)/A046161(n) = (-1)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} (-1)^k*a(k)/A046161(k) = 1/sqrt(2).
Sum_{k>=0} (-1)^(k+1)*a(k)/A046161(k) = -1/sqrt(2). (End)
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2+1)^n dx. (n=1 is the Cauchy distribution.) - Harry Garst, May 26 2017
Let R(n, d) = (Product_{j prime to d} Pochhammer(j / d, n)) / n!. Then the numerators of R(n, 2) give this sequence and the denominators are A046161. For d = 3 see A273194/A344402. - Peter Luschny, May 20 2021
Using WolframAlpha, it appears a(n) gives the numerator in 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
a(n) is the numerator of (1/Pi) * Integral_{x=-oo..+oo} sech(x)^(2*n+1) dx. The corresponding denominator is A046161. - Mohammed Yaseen, Jul 29 2023
a(n) is the numerator of (1/Pi) * Integral_{x=0..Pi/2} sin(x)^(2*n) dx. The corresponding denominator is A101926(n). - Mohammed Yaseen, Sep 19 2023

Examples

			1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, 12155/128, 46189/256, ...
binomial(2*n,n)/4^n => 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, 6435/32768, ...
		

References

  • P. J. Davis, Interpolation and Approximation, Dover Publications, 1975, p. 372.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, 1968; Chap. III, Eq. 4.1.
  • 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 6:14:6 at page 51.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 102.

Crossrefs

Cf. A060818 (denominator of binomial(2*n,n)/2^n), A061549 (denominators).
Cf. A123854 (denominators).
Cf. A161198 (triangle of coefficients for (1-x)^((-1-2*n)/2)).
Cf. A163590 (odd part of the swinging factorial).
Cf. A001405.
First column and diagonal 1 of triangle A100258.
Bisection of A036069.
Bisections give A061548 and A063079.
Inverse Moebius transform of A180403/A046161.
Numerators of [x^n]( (1-x)^(p/2) ): A161202 (p=5), A161200 (p=3), A002596 (p=1), this sequence (p=-1), A001803 (p=-3), A161199 (p=-5), A161201 (p=-7).

Programs

  • Magma
    A001790:= func< n | Numerator((n+1)*Catalan(n)/4^n) >;
    [A001790(n): n in [0..40]]; // G. C. Greubel, Sep 23 2024
  • Maple
    e := proc(l,m) local k; add(2^(k-2*m)*binomial(2*m-2*k,m-k)*binomial(m+k,m)*binomial(k,l),k=l..m); end;
    # From Peter Luschny, Aug 01 2009: (Start)
    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)/sigma(2*n); # (End)
    A001790 := proc(n) binomial(2*n, n)/4^n ; numer(%) ; end proc : # R. J. Mathar, Jan 18 2013
  • Mathematica
    Numerator[ CoefficientList[ Series[1/Sqrt[(1 - x)], {x, 0, 25}], x]]
    Table[Denominator[Hypergeometric2F1[1/2, n, 1 + n, -1]], {n, 0, 34}]   (* John M. Campbell, Jul 04 2011 *)
    Numerator[Table[(-2)^n*Sqrt[Pi]/(Gamma[1/2 - n]*Gamma[1 + n]),{n,0,20}]] (* Ralf Steiner, Apr 07 2017 *)
    Numerator[Table[Binomial[2n,n]/2^n, {n, 0, 25}]] (* Vaclav Kotesovec, Apr 07 2017 *)
    Table[Numerator@LegendreP[2 n, 0]*(-1)^n, {n, 0, 25}] (* Andres Cicuttin, Jan 22 2018 *)
    A = {1}; Do[A = Append[A, 2^IntegerExponent[n, 2]*(2*n - 1)*A[[n]]/n], {n, 1, 25}]; Print[A] (* John Lawrence, Jul 17 2020 *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( pollegendre(n), n) * 2^valuation((n\2*2)!, 2))};
    
  • PARI
    a(n)=binomial(2*n,n)>>hammingweight(n); \\ Gleb Koloskov, Sep 26 2021
    
  • Sage
    # uses[A000120]
    @CachedFunction
    def swing(n):
        if n == 0: return 1
        return swing(n-1)*n if is_odd(n) else 4*swing(n-1)/n
    A001790 = lambda n: swing(2*n)/2^A000120(2*n)
    [A001790(n) for n in (0..25)]  # Peter Luschny, Nov 19 2012
    

Formula

a(n) = numerator( binomial(2*n,n)/4^n ) (cf. A046161).
a(n) = A000984(n)/A001316(n) where A001316(n) is the highest power of 2 dividing C(2*n, n) = A000984(n). - Benoit Cloitre, Jan 27 2002
a(n) = denominator of (2^n/binomial(2*n,n)). - Artur Jasinski, Nov 26 2011
a(n) = numerator(L(n)), with rational L(n):=binomial(2*n,n)/2^n. L(n) is the leading coefficient of the Legendre polynomial P_n(x).
L(n) = (2*n-1)!!/n! with the double factorials (2*n-1)!! = A001147(n), n >= 0.
Numerator in (1-2t)^(-1/2) = 1 + t + (3/2)t^2 + (5/2)t^3 + (35/8)t^4 + (63/8)t^5 + (231/16)t^6 + (429/16)t^7 + ... = 1 + t + 3*t^2/2! + 15*t^3/3! + 105*t^4/4! + 945*t^5/5! + ... = e.g.f. for double factorials A001147 (cf. A094638). - Tom Copeland, Dec 04 2013
From Ralf Steiner, Apr 08 2017: (Start)
a(n)/A061549(n) = (-1/4)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A061549(k) = 2/sqrt(3).
Sum_{k>=0} (-1)^k*a(k)/A061549(k) = 2/sqrt(5).
Sum_{k>=0} (-1)^(k+1)*a(k)/A061549(k) = -2/sqrt(5).
a(n)/A123854(n) = (-1/2)^n*sqrt(Pi)/(gamma(1/2 - n)*gamma(1 + n)).
Sum_{k>=0} a(k)/A123854(k) = sqrt(2).
Sum_{k>=0} (-1)^k*a(k)/A123854(k) = sqrt(2/3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A123854(k) = -sqrt(2/3). (End)
a(n) = 2^A007814(n)*(2*n-1)*a(n-1)/n. - John Lawrence, Jul 17 2020
Sum_{k>=0} A086117(k+3)/a(k+2) = Pi. - Antonio Graciá Llorente, Aug 31 2024
a(n) = A001803(n)/(2*n+1). - G. C. Greubel, Sep 23 2024

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

Original entry on oeis.org

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

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Comments

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)

Examples

			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)
		

Crossrefs

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).
Standard compositions are listed by A066099.
The opposite version (counting refinements) is A080100.
The version for Heinz numbers of partitions is A317141.

Programs

  • Haskell
    a048896 n = a048896_list !! n
    a048896_list = f [1] where f (x:xs) = x : f (xs ++ [x,2*x])
    -- Reinhard Zumkeller, Mar 07 2011
    
  • Haskell
    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
    
  • Magma
    [Numerator(2^n / Factorial(n+1)): n in [0..100]]; // Vincenzo Librandi, Apr 12 2014
  • Maple
    a := n -> 2^(add(i,i=convert(n+1,base,2))-1): seq(a(n), n=0..97); # Peter Luschny, May 01 2009
  • Mathematica
    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 *)
  • PARI
    a(n)=if(n<1,1,if(n%2,a(n/2-1/2),2*a(n-1)))
    
  • PARI
    a(n) = 1 << (hammingweight(n+1)-1); \\ Kevin Ryde, Feb 19 2022
    

Formula

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

Extensions

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

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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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

A100258 Triangle of coefficients of normalized Legendre polynomials, with increasing exponents.

Original entry on oeis.org

1, 0, 1, -1, 0, 3, 0, -3, 0, 5, 3, 0, -30, 0, 35, 0, 15, 0, -70, 0, 63, -5, 0, 105, 0, -315, 0, 231, 0, -35, 0, 315, 0, -693, 0, 429, 35, 0, -1260, 0, 6930, 0, -12012, 0, 6435, 0, 315, 0, -4620, 0, 18018, 0, -25740, 0, 12155, -63, 0, 3465, 0, -30030, 0, 90090, 0, -109395, 0, 46189
Offset: 0

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Author

Ralf Stephan, Nov 13 2004

Keywords

Comments

For a relation to Jacobi quartic elliptic curves, see the MathOverflow link. For a self-convolution of the polynomials relating them to the Chebyshev and Fibonacci polynomials, see A049310 and A053117. For congruences and connections to other polynomials (Jacobi, Gegenbauer, and Chebyshev) see the Allouche et al. link. For relations to elliptic cohomology and modular forms, see references in Copeland link.- Tom Copeland, Feb 04 2016

Examples

			Triangle begins:
   1;
   0,   1;
  -1,   0,     3;
   0,  -3,     0,   5;
   3,   0,   -30,   0,   35;
   0,  15,     0, -70,    0,   63;
  -5,   0,   105,   0, -315,    0,    231;
   0, -35,     0, 315,    0, -693,      0, 429;
  35,   0, -1260,   0, 6930,    0, -12012,   0, 6435;
  ...
		

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.

Crossrefs

Without zeros: A008316. Row sums are A060818.
Columns (with interleaved zeros and signs) include A001790, A001803, A100259. Diagonals include A001790, A001800, A001801, A001802.

Programs

  • Mathematica
    row[n_] := CoefficientList[ LegendreP[n, x], x]*2^IntegerExponent[n!, 2]; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jan 15 2015 *)
  • PARI
    a(k,n)=polcoeff(pollegendre(k,x),n)*2^valuation(k!,2)
    
  • Python
    from mpmath import *
    mp.dps=20
    def a007814(n):
        return 1 + bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')
    for n in range(11):
        y=2**sum(a007814(i) for i in range(2, n+1))
        l=chop(taylor(lambda x: legendre(n, x), 0, n))
        print([int(i*y) for i in l]) # Indranil Ghosh, Jul 02 2017

Formula

The n-th normalized Legendre polynomial is generated by 2^(-n-a(n)) (d/dx)^n (x^2-1)^n / n! with a(n) = A005187(n/2) for n even and a(n) = A005187((n-1)/2) for n odd. The non-normalized polynomials have the o.g.f. 1 / sqrt(1 - 2xz + z^2). - Tom Copeland, Feb 07 2016
The consecutive nonzero entries in the m-th row are, in order, (c+b)!/(c!(m-b)!(2b-m)!*A048896(m-1)) with sign (-1)^b where c = m/2-1, m/2, m/2+1, ..., (m-1) and b = c+1 if m is even and sign (-1)^c with c = (m-1)/2, (m-1)/2+1, (m-1)/2+2, ..., (m-1) with b = c+1 if m is odd. For the 9th row the 5 consecutive nonzero entries are 315, -4620, 18018, -25740, 12155 given by c = 4,5,6,7,8 and b = 5,6,7,8,9. - Richard Turk, Aug 22 2017
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