Displaying 1-10 of 13 results found.
Numerator of the n-th harmonic number H(n) divided by (n+1); a(n) = A001008(n) / ((n+1)* A002805(n)).
(Formerly M4765 N2036)
+0
6
1, 1, 11, 5, 137, 7, 363, 761, 7129, 671, 83711, 6617, 1145993, 1171733, 1195757, 143327, 42142223, 751279, 275295799, 55835135, 18858053, 830139, 444316699, 269564591, 34052522467, 34395742267, 312536252003, 10876020307, 9227046511387, 300151059037
COMMENTS
Numerators of coefficients for numerical differentiation.
REFERENCES
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables).
A. N. Lowan, H. E. Salzer and A. Hillman, A table of coefficients for numerical differentiation, Bull. Amer. Math. Soc., 48 (1942), 920-924.
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).
FORMULA
G.f.: (-log(1-x))^2 (for fractions A002547(n)/ A002548(n)). - Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002 A002547(n)/ A002548(n) = 2*Stirling_1(n+2, 2)(-1)^n/(n+2)! - Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002
Numerator of u(n) = Sum_{k=1..n-1} 1/(k*(n-k)) (u(n) is asymptotic to 2*log(n)/n). - Benoit Cloitre, Apr 12 2003; corrected by Istvan Mezo, Oct 29 2012 a(n) = numerator of 2*Integral_{0..1} x^(n+1)*log(x/(1-x)) dx. - Groux Roland, May 18 2011
EXAMPLE
H(n) = Sum_{k=1..n} 1/k, begins 1, 3/2, 11/6, 25/12, ... so H(n)/(n+1) begins 1/2, 1/2, 11/24, 5/12, ....
a(4) = numerator(H(4)/(4+1)) = 5.
MAPLE
H := proc(a, b) option remember; local m, p, q, r, s;
if b - a <= 1 then return 1, a fi; m := iquo(a + b, 2);
p, q := H(a, m); r, s := H(m, b); p*s + q*r, q*s; end:
A002547 := proc(n) H(1, n+1); numer(%[1]/(%[2]*(n+1))) end:
MATHEMATICA
a[n_]:= Numerator[HarmonicNumber[n]/(n+1)])]; Table[a[n], {n, 35}] (* modified by G. C. Greubel, Jul 03 2019 *)
PROG
(PARI) h(n) = sum(k=1, n, 1/k);
vector(35, n, numerator(h(n)/(n+1))) \\ G. C. Greubel, Jul 03 2019 (Magma) [Numerator(HarmonicNumber(n)/(n+1)): n in [1..35]]; // G. C. Greubel, Jul 03 2019 (Sage) [numerator(harmonic_number(n)/(n+1)) for n in (1..35)] # G. C. Greubel, Jul 03 2019 (GAP) List([1..35], n-> NumeratorRat(Sum([1..n], k-> 1/k)/(n+1))) # G. C. Greubel, Jul 03 2019
EXTENSIONS
More terms from Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002
Denominators of coefficients for numerical differentiation.
(Formerly M4822 N2063)
+0
9
1, 1, 12, 6, 180, 10, 560, 1260, 12600, 1260, 166320, 13860, 2522520, 2702700, 2882880, 360360, 110270160, 2042040, 775975200, 162954792, 56904848, 2586584, 1427794368, 892371480, 116008292400, 120470149800, 1124388064800
COMMENTS
Denominator of u(n) = sum( k=1, n-1, 1/(k(n-k)) ) (u(n) is asymptotic to 2*log(n)/n). - Benoit Cloitre, Apr 12 2003; corrected by Istvan Mezo, Oct 29 2012 Expected area of the convex hull of n points picked at random inside a triangle with unit area. - Eric W. Weisstein, Apr 15 2004
REFERENCES
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).
FORMULA
A002547(n)/a(n) = 2*Stirling_1(n+2, 2)(-1)^n/(n+2)!.
EXAMPLE
0, 0, 1/12, 1/6, 43/180, 3/10, 197/560, 499/1260, 5471/12600, ...
MAPLE
seq(denom(Stirling1(j+2, 2)/(j+2)!*2!*(-1)^j), j=0..50);
MATHEMATICA
Table[Denominator[1 - 2*HarmonicNumber[n - 1]/n], {n, 2, 30}] (* Wesley Ivan Hurt, Mar 24 2014 *)
EXTENSIONS
More terms, GF, formula, Maple code from Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002
a(n) = (2n+1)!/n!^2.
(Formerly M4198 N1752)
+0
144
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
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 Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 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. 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
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]
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) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009 a(n) = f(n, n-3) where f is given in A034261. 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.: 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.: 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*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) = 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 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
EXAMPLE
G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...
MAPLE
A002457:=n->(n+1) * binomial(2*(n+1), (n+1)) / 2; seq( A002457(n), n=0..50);
seq((2*n)!*coeff(series(HeunC(0, 0, -2, -1/4, 7/4, 4*x^2), x, 2*n+1), x, 2*n), n=0..22); # Peter Luschny, Nov 22 2013
PROG
(PARI) {a(n) = if( n<0, 0, (2*n + 1)! / n!^2)}; /* Michael Somos, Dec 09 2002 */ (PARI) a(n) = (2*n+1)*binomial(2*n, n); \\ Altug Alkan, Apr 16 2018 (Haskell)
a002457 n = a116666 (2 * n + 1) (n + 1)
(Sage)
A002457 = lambda n: binomial(n+1/2, 1/2)<<2*n
(Magma) [Factorial(2*n+1)/Factorial(n)^2: n in [0..25]]; // Vincenzo Librandi, Oct 12 2015
CROSSREFS
Cf. A000531 (Banach's original match problem). The rightmost diagonal of the triangle A331431.
Numerators in expansion of 1/sqrt(1-x).
(Formerly M2508 N0992)
+0
76
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
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 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 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} (-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
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.
FORMULA
a(n) = numerator( binomial(2*n,n)/4^n ) (cf. A046161). 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 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)
EXAMPLE
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, ...
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;
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)
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 *)
PROG
(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 @CachedFunction
def swing(n):
if n == 0: return 1
return swing(n-1)*n if is_odd(n) else 4*swing(n-1)/n
(Magma)
A001790:= func< n | Numerator((n+1)*Catalan(n)/4^n) >;
CROSSREFS
Cf. A060818 (denominator of binomial(2*n,n)/2^n), A061549 (denominators). Cf. A161198 (triangle of coefficients for (1-x)^((-1-2*n)/2)). Cf. A163590 (odd part of the swinging factorial). First column and diagonal 1 of triangle A100258.
KEYWORD
nonn,easy,nice,frac,changed
Denominators of coefficients in Taylor series expansion of log(1+x)^2/sqrt(1+x).
(Formerly M2133 N0846)
+0
3
1, 1, 2, 24, 48, 5760, 11520, 35840, 215040, 51609600, 103219200, 13624934400, 5449973760, 1322526965760, 3606891724800, 158703235891200, 105802157260800, 14800210341396480, 29600420682792960, 3749386619820441600
COMMENTS
Old title: Denominators of coefficients for numerical differentiation.
REFERENCES
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).
MATHEMATICA
Denominator[CoefficientList[Series[Log[1 + x]^2/Sqrt[1 + x]/x, {x, 0, 40}], x]] (* Vincenzo Librandi, Mar 25 2014 *)
Denominators of coefficients of log(1+x)/sqrt(1+x).
(Formerly M5139 N2229)
+0
2
1, 1, 24, 12, 640, 1920, 107520, 1792, 2064384, 10321920, 43253760, 64880640, 5398069248, 4198498304, 503819796480, 62977474560, 16610786017280, 4271344975872, 649244436332544, 41618233098240, 27967452642017280
REFERENCES
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).
MATHEMATICA
Denominator[CoefficientList[Series[Log[1 + x]/Sqrt[1 + x]/x, {x, 0, 40}], x]] (* Vincenzo Librandi, Mar 25 2014 *)
Numerators of coefficients for numerical differentiation.
(Formerly M4034 N1676)
+0
2
1, -5, 259, -3229, 117469, -7156487, 2430898831, -60997921, 141433003757, -25587296781661, 51270597630767, -6791120985104747, 3400039831130408821, -15317460638921852507, 25789165074168004597399, -1550286106708510672406629, 24823277118070193095631689
REFERENCES
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).
FORMULA
a(n) is the numerator of (-1)^(n-1)*Cn-1{1^2..(2n-1)^2}/((2n)!*2^(2n-3)), where Cn{1^2..(2n+1)^2} equals 1 when n=0, otherwise it is the sum of the products of all possible combinations, of size n, of the numbers (2k+1)^2 with k=0,1,...,n. - Ruperto Corso, Dec 15 2011
MAPLE
with(combinat):
a:=n->add(mul(k, k=j), j=choose([seq((2*i-1)^2, i=1..n)], n-1))*(-1)^(n-1)/(2^(2*n-3)*(2*n)!):
a(n) = binomial(2*n+1,n)*(n+1)^2.
(Formerly M4855 N2075)
+0
16
1, 12, 90, 560, 3150, 16632, 84084, 411840, 1969110, 9237800, 42678636, 194699232, 878850700, 3931426800, 17450721000, 76938289920, 337206098790, 1470171918600, 6379820115900, 27569305764000, 118685861314020, 509191949220240, 2177742427450200, 9287309860732800
COMMENTS
Coefficients for numerical differentiation.
Take the first n integers 1,2,3..n and find all combinations with repetitions allowed for the first n of them. Find the sum of each of these combinations to get this sequence. Example for 1 and 2: 1,2,1+1,1+2,2+2 gives sum of 12=a(2). - J. M. Bergot, Mar 08 2016 Let cos(x) = 1 -x^2/2 +x^4/4!-x^6/6!.. = Sum_i (-1)^i x^(2i)/(2i)! be the standard power series of the cosine, and y = 2*(1-cos(x)) = 4*sin^2(x/2) = x^2 -x^4/12 +x^6/360 ...= Sum_i 2*(-1)^(i+1) x^(2i)/(2i)! be a closely related series. Then this sequence represents the reversion x^2 = Sum_i 1/a(i) *y^(i+1). - R. J. Mathar, May 03 2022
REFERENCES
C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
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).
LINKS
R. Shenton and A. W. Kemp, An S-fraction and ln^2(1+x), Journal of Computational and Applied Mathematics, 26 (1989) 367-370 North-Holland.
FORMULA
G.f.: (1 + 2x)/(1 - 4x)^(5/2).
a(n-1) = sum(i_1 + i_2 + ... + i_n) where the sum is over 0 <= i_1 <= i_2 <= ... <= i_n <= n; a(n) = (n+1)^2 C(2n+1, n). - David Callan, Nov 20 2003 Asymptotics: a(n)-> (1/64) * (128*n^2+176*n+41) * 4^n * n^(-1/2)/(sqrt(Pi)), for n->infinity. - Karol A. Penson, Aug 05 2013 E.g.f.: ((1 + 2*x)*(1 + 8*x)*BesselI(0,2*x) + 2*x*(3 + 8*x)*BesselI(1,2*x))*exp(2*x).
Sum_{n>=0} 1/a(n) = Pi^2/9 = A100044. (End) With x = y^2/(1 + y) we have log^2(1 + y) = Sum_{n >= 0} (-1)^n*x^(n+1)/a(n). See Shenton and Kemp.
Series reversion ( Sum_{n >= 0} (-1)^n*x^(n+1)/a(n) ) = Sum_{n >= 1} 2*x^n/(2*n)! = Sum_{n >= 1} x^n/ A002674(n). (End) D-finite with recurrence n^2*a(n) -2*(n+1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
MAPLE
seq((n+1)^2*(binomial(2*n+2, n+1))/2, n=0..29); # Zerinvary Lajos, May 31 2006
MATHEMATICA
Table[Binomial[2n+1, n](n+1)^2, {n, 0, 20}] (* Harvey P. Dale, Mar 23 2011 *)
PROG
(PARI) a(n)=binomial(2*n+1, n)*(n+1)^2
(PARI) x='x+O('x^99); Vec((1+2*x)/(1-4*x)^(5/2)) \\ Altug Alkan, Jul 09 2016 (Python)
from sympy import binomial
def a(n): return binomial(2*n + 1, n)*(n + 1)**2 # Indranil Ghosh, Apr 18 2017
Denominators of coefficients for numerical differentiation.
(Formerly M5177 N2249)
+0
3
1, 24, 5760, 322560, 51609600, 13624934400, 19837904486400, 2116043145216, 20720294477955072, 15747423803245854720, 131978409017679544320, 72852081777759108464640, 151532330097738945606451200, 2828603495157793651320422400, 19687080326298243813190139904000
REFERENCES
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).
FORMULA
a(n) is the denominator of (-1)^(n-1)*Cn-1{1^2..(2n-1)^2}/((2n)!*2^(2n-3)), where Cn{1^2..(2n+1)^2} is equal to 1 when n=0, otherwise, it is the sum of the products of all possible combinations, of size n, of the numbers (2k+1)^2 with k=0,1,..,n. - Ruperto Corso, Dec 15 2011
MAPLE
with(combinat): a:=n->add(mul(k, k=j), j=choose([seq((2*i-1)^2, i=1..n)], n-1))*(-1)^(n-1)/(2^(2*n-3)*(2*n)!): seq(denom(a(n)), n=1..20); # Ruperto Corso, Dec 15 2011
Denominator of Sum_{i+j+k=n; i,j,k > 0} 1/(i*j*k).
(Formerly M1110 N0424)
+0
2
1, 2, 4, 8, 15, 240, 15120, 672, 8400, 100800, 69300, 4950, 17199000, 22422400, 33633600, 201801600, 467812800, 102918816000, 410646075840, 3555377280, 215100325440, 5162407810560, 30920671782000, 190281057120, 1085315579548200, 562756226432400, 22969641895200
COMMENTS
Denominators of coefficients for numerical differentiation.
REFERENCES
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables).
A. N. Lowan, H. E. Salzer and A. Hillman, A table of coefficients for numerical differentiation, Bull. Amer. Math. Soc., 48 (1942), 920-924.
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).
FORMULA
G.f.: (-log(1-x))^3 (for fractions A002545(n)/ A002546(n)). - Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002 A002545(n)/ A002546(n) = 6*Stirling_1(n+3, 3)(-1)^n/(n+3)!. - Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002
MAPLE
seq(denom(-Stirling1(j, 3)/j!*3!*(-1)^j), j=3..50); # Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002
MATHEMATICA
Denominator[Table[Sum[1/i/j/(n-i-j), {i, n-2}, {j, n-i-1}], {n, 3, 100}]] (* Ryan Propper *)
EXTENSIONS
More terms from Barbara Margolius (b.margolius(AT)math.csuohio.edu), Jan 19 2002
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