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.

Showing 1-10 of 18 results. Next

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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Author

Keywords

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

A112302 Decimal expansion of quadratic recurrence constant sqrt(1 * sqrt(2 * sqrt(3 * sqrt(4 * ...)))).

Original entry on oeis.org

1, 6, 6, 1, 6, 8, 7, 9, 4, 9, 6, 3, 3, 5, 9, 4, 1, 2, 1, 2, 9, 5, 8, 1, 8, 9, 2, 2, 7, 4, 9, 9, 5, 0, 7, 4, 9, 9, 6, 4, 4, 1, 8, 6, 3, 5, 0, 2, 5, 0, 6, 8, 2, 0, 8, 1, 8, 9, 7, 1, 1, 1, 6, 8, 0, 2, 5, 6, 0, 9, 0, 2, 9, 8, 2, 6, 3, 8, 3, 7, 2, 7, 9, 0, 8, 3, 6, 9, 1, 7, 6, 4, 1, 1, 4, 6, 1, 1, 6, 7, 1, 5, 5, 2, 8
Offset: 1

Views

Author

Michael Somos, Sep 02 2005

Keywords

Comments

From Johannes W. Meijer, Jun 27 2016: (Start)
With Phi(z, p, q) the Lerch transcendent, define LP(n) = (1/n) * sum(Phi(1/2, n-k, 1) * LP(k), k=0..n-1), with LP(0) = 1. Conjecture: Lim_{n -> infinity} LP(n) = A112302.
For similar formulas, see A090998 and A135002. For background information, see A274181.
The structure of the n! * LP(n) formulas leads to the multinomial coefficients A036039. (End)

Examples

			1.6616879496335941212958189227499507499644186350250682081897111680...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.
  • S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., AMS Chelsea 2000. See Appendix I. p. 348.

Crossrefs

Programs

  • Mathematica
    RealDigits[ Fold[ N[ Sqrt[ #2*#1], 128] &, Sqrt@ 351, Reverse@ Range@ 350], 10, 111][[1]] (* Robert G. Wilson v, Nov 05 2010 *)
    Exp[-Derivative[1, 0][PolyLog][0, 1/2]] // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Apr 07 2014, after Jonathan Sondow *)
  • PARI
    {a(n) = if( n<-1, 0, n++; default( realprecision, n+2); floor( prodinf( k=1, k^2^-k)* 10^n) % 10)};
    
  • PARI
    prodinf(n=1,n^2^-n) \\ Charles R Greathouse IV, Apr 07 2013
    
  • Python
    from mpmath import polylog, diff, exp, mp
    mp.dps = 120
    somos_const = exp(-diff(lambda n: polylog(n, 1/2), 0))
    A112302 = [int(d) for d in mp.nstr(somos_const, n=mp.dps)[:-1] if d != '.']  # Jwalin Bhatt, Nov 23 2024

Formula

Equals Product_{n>=1} n^(1/2^n). - Jonathan Sondow, Apr 07 2013
Equals exp(A114124) = A188834/2 = sqrt(A259235). - Hugo Pfoertner, Nov 23 2024
From Jwalin Bhatt, Apr 02 2025: (Start)
Equals exp(-PolyLog'(0,1/2)), where PolyLog'(x,y) represents the derivative of the polylogarithm w.r.t. x.
Equals Product_{n>=1} (1+1/n)^(1/2^n).
Equals exp(Sum_{n>=2} log(n)/2^n).
Equals 2*exp(Sum_{n>=1} (log(1+1/n)-1/n)/2^n). (End)

A052129 a(0) = 1; thereafter a(n) = n*a(n-1)^2.

Original entry on oeis.org

1, 1, 2, 12, 576, 1658880, 16511297126400, 1908360529573854283038720000, 29134719286683212541013468732221146917416153907200000000
Offset: 0

Views

Author

Reinhard Zumkeller, Feb 12 2002

Keywords

Comments

Somos's quadratic recurrence sequence.
Iff n is prime (n>2), the n-adic valuation of a(2n) is 3*A001045(n) (three times the values at the prime indices of Jacobsthal numbers), which is 2^n+1. For example: the 11-adic valuation at a(22) = 2049 = 3*A001045(11)= 683. 3*683 = 2^11+1 = 2049. True because: When n is prime, n-adic valuation is 1 at A052129(n), then doubles as n-increases to 2n, at which point 1 is added; thus A052129(2n) = 2^n+1. Since 3*A001045(n) = 2^n+1, n-adic valuation of A052129(2n) = 3*A001045(n) when n is prime. - Bob Selcoe, Mar 06 2014
Unreduced denominators of: f(1) = 1, f(n) = f(n-1) + f(n-1)/(n-1). - Daniel Suteu, Jul 29 2016

Examples

			a(3) = 3*a(2)^2 = 3*(2*a(1)^2)^2 = 3*(2*(1*a(0)^2)^2)^2 = 3*(2*(1*1^2)^2)^2 = 3*(2*1)^2 = 3*4 = 12.
G.f. = 1 + x + 2*x^2 + 12*x^3 + 576*x^4 + 1658880*x^5 + 16511297126400*x^6 + ...
		

References

  • S. R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • Mathematica
    Join[{1},RecurrenceTable[{a[1]==1,a[n]==n a[n-1]^2},a,{n,10}]]  (* Harvey P. Dale, Apr 26 2011 *)
    a[ n_] := If[ n < 1, Boole[n == 0], Product[ (n - k)^2^k, {k, 0, n - 1}]]; (* Michael Somos, May 24 2013 *)
    a[n_] := Product[ k^(2^(n - k)), {k,1,n}] (* Jonathan Sondow, Mar 17 2014 *)
    NestList[{#[[1]]+1,#[[1]]*#[[2]]^2}&,{1,1},10][[All,2]] (* Harvey P. Dale, Jul 30 2018 *)
  • PARI
    {a(n) = if( n<1, n==0, prod(k=0, n-1, (n - k)^2^k))}; /* Michael Somos, May 24 2013 */

Formula

a(n) ~ s^(2^n) / (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...) where s = 1.661687949633... (see A112302) and A116603. - Michael Somos, Apr 02 2006
a(n) = n * A030450(n - 1) if n>0. - Michael Somos, Oct 22 2006
a(n) = (a(n-1) + a(n-2)^2) * (a(n-1) / a(n-2))^2. - Michael Somos, Mar 20 2012
a(n) = product_{k=1..n} k^(2^(n-k)). - Jonathan Sondow, Mar 17 2014
A088679(n+1)/a(n) = n+1. -Daniel Suteu, Jul 29 2016

A088802 Denominators of the coefficients of powers of n^(-1) in the Romanovsky series expansion of the mean of the standard deviation from a normal population.

Original entry on oeis.org

1, 4, 32, 128, 2048, 8192, 65536, 262144, 8388608, 33554432, 268435456, 1073741824, 17179869184, 68719476736, 549755813888, 2199023255552, 140737488355328, 562949953421312, 4503599627370496, 18014398509481984
Offset: 0

Views

Author

Eric W. Weisstein, Oct 16 2003

Keywords

Comments

Is this the same sequence as A123854? - N. J. A. Sloane, Mar 21 2007
Almost certainly this is the same as A123854. - Michael Somos, Aug 23 2007
Asymptotic expansion of Gamma(N/2) / Gamma((N-1)/2) = (N/2)^(1/2) * (c(0) + c(1)/N + c(2)/N^2 + ... ). a(n) = denominator(c(n)). - Michael Somos, Aug 23 2007

References

  • V. Romanovsky, On the Moments of the Standard Deviation and of the Correlation Coefficient in Samples from Normal, Metron 5(4) (1925), 3-46.

Crossrefs

Programs

  • GAP
    List([0..25], n-> DenominatorRat(Sum([0..n], k-> Binomial(2*k, k)/8^k))); # G. C. Greubel, Jan 29 2020
  • Magma
    [Denominator( &+[Binomial(2*k, k)/8^k: k in [0..n]] ): n in [0..25]]; // G. C. Greubel, Jan 29 2020
    
  • Maple
    seq(denom(add(binomial(2*k, k)/8^k, k = 0 .. n)), n = 0..25); # G. C. Greubel, Jan 29 2020
  • Mathematica
    Table[Denominator[Sum[Binomial[2*k, k]/8^k, {k,0,n}]], {n,0,25}] (* G. C. Greubel, Jan 29 2020 *)
  • PARI
    {a(n) = if( n<0, 0, 2^(3*n - subst( Pol( binary( n ) ), x, 1) ) ) } /* Michael Somos, Aug 23 2007 */
    
  • Sage
    [denominator( binomial(1/4, n) ) for n in (0..25)] # G. C. Greubel, Jan 29 2020
    

Formula

From G. C. Greubel, Jan 29 2020: (Start)
a(n) = denominator(Sum_{k=0..n} binomial(2*k, k)/8^k).
a(n) = denominator(binomial(1/4, n)). (End)

A116603 Coefficients in asymptotic expansion of sequence A052129.

Original entry on oeis.org

1, 2, -1, 4, -21, 138, -1091, 10088, -106918, 1279220, -17070418, 251560472, -4059954946, 71250808916, -1351381762990, 27552372478592, -601021307680207, 13969016314470386, -344653640328891233, 8997206549370634644, -247772400254700937149, 7178881153198162073002
Offset: 0

Views

Author

Michael Somos, Feb 18 2006

Keywords

Examples

			G.f. = 1 + 2*x - x^2 + 4*x^3 - 21*x^4 + 138*x^5 - 1091*x^6 + 10088*x^7 + ...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • Mathematica
    terms = 20; A[] = 1; Do[A[x] = -A[x] + 2/A[x/(1+x)]^(-1/2)*(1+x) + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jul 28 2011, updated Jan 12 2018 *)
  • PARI
    {a(n) = my(A); if( n<0, 0, A=1; for( k=1, n, A = truncate( A + O(x^k)) + x * O(x^k); A = -A + 2 / subst(A^(-1/2), x, x/(1 + x)) * (1 + x);); polcoeff(A, n))};

Formula

a(0) = 1; thereafter, a(n) = (1/n)*Sum_{j=1..n} (-1)^(j-1)*2*b(j)*a(n-j), where b(j) = A000670(j) [Nemes]. - N. J. A. Sloane, Sep 11 2017
G.f. A(x) satisfies (1 + x)^2 = A(x)^2 / A(x/(1 + x)).
A003504(n+1) ~ C^(2^n) * (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...) where C = 1.04783144757... (see A115632).
A052129(n) ~ s^(2^n) / (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...) where s = 1.661687949633... (see A112302).
From Seiichi Manyama, May 26 2025: (Start)
G.f.: Product_{k>=1} (1 + k*x)^(1/2^k).
G.f.: exp(2 * Sum_{k>=1} (-1)^(k-1) * A000670(k) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A084785. (End)
a(n) ~ (-1)^(n+1) * (n-1)! / log(2)^(n+1). - Vaclav Kotesovec, May 27 2025

A123851 A cubic recurrence: a(0) = 1, a(n) = n*a(n-1)^3 for n >= 1.

Original entry on oeis.org

1, 1, 2, 24, 55296, 845378412871680, 3624972460853492659595005581182702601633792000
Offset: 0

Views

Author

Keywords

Comments

A cubic analog of Somos's quadratic recurrence sequence A052129.
Terms a(7) onward are too big to include in data section. - G. C. Greubel, Aug 10 2019

Examples

			a(3) = 3*a(2)^3 = 3*(2*a(1)^3)^3 = 3*(2*(1*a(0)^3)^3)^3 = 3*(2*(1*1^3)^3)^3 = 3*(2*1)^3 = 3*8 = 24.
G.f. = 1 + x + 2*x^2 + 24*x^3 + 55296*x^4 + 845378412871680*x^5 + ...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • GAP
    List([0..8], n-> Product([0..n-1], k-> (n-k)^(3^k)) ); # G. C. Greubel, Aug 10 2019
  • Magma
    [n eq 0 select 1 else (&*[(n-k)^(3^k): k in [0..n-1]]):n in [0..8]]; // G. C. Greubel, Aug 10 2019
    
  • Mathematica
    a[n_]:= If[n==0, 1, n*a[n-1]^3]; Table[a[n], {n,0,7}]
    nxt[{n_,a_}]:={n+1, (n+1)a^3}; NestList[nxt,{0,1},7][[All,2]] (* Harvey P. Dale, May 25 2019 *)
  • PARI
    {a(n) = if( n<1, n==0, prod(k=0, n-1, (n - k)^3^k))}; /* Michael Somos, Aug 07 2016 */
    
  • Sage
    [1]+[prod((n-k)^(3^k) for k in (0..n-1)) for n in (1..8)] # G. C. Greubel, Aug 10 2019
    

Formula

a(n) ~ c^(3^n)*n^(-1/2)/(1 + 3/(4*n) - 15/(32*n^2) + 113/(128*n^3) + ...) where c = 1.1563626843322... is the cubic recurrence constant A123852.

Extensions

Corrected by Harvey P. Dale, May 25 2019

A220002 Numerators of the coefficients of an asymptotic expansion in even powers of the Catalan numbers.

Original entry on oeis.org

1, 5, 21, 715, -162877, 19840275, -7176079695, 1829885835675, -5009184735027165, 2216222559226679575, -2463196751104762933637, 1679951011110471133453965, -5519118103058048675551057049, 5373485053345792589762994345215, -12239617587594386225052760043303511
Offset: 0

Views

Author

Peter Luschny, Dec 27 2012

Keywords

Comments

Let N = 4*n+3 and A = sum_{k>=0} a(k)/(A123854(k)*N^(2*k)) then
C(n) ~ 8*4^n*A/(N*sqrt(N*Pi)), C(n) = (4^n/sqrt(Pi))*(Gamma(n+1/2)/ Gamma(n+2)) the Catalan numbers A000108.
The asymptotic expansion of the Catalan numbers considered here is based on the Taylor expansion of square root of the sine cardinal. This asymptotic series involves only even powers of N, making it more efficient than the asymptotic series based on Stirling's approximation to the central binomial which involves all powers (see for example: D. E. Knuth, 7.2.1.6 formula (16)). The series is discussed by Kessler and Schiff but is included as a special case in the asymptotic expansion given by J. L. Fields for quotients Gamma(x+a)/Gamma(x+b) and discussed by Y. L. Luke (p. 34-35), apparently overlooked by Kessler and Schiff.

Examples

			With N = 4*n+3 the first few terms of A are A = 1 + 5/(4*N^2) + 21/(32*N^4) + 715/(128*N^6) - 162877/(2048*N^8) + 19840275/(8192*N^10). With this A C(n) = round(8*4^n*A/(N*sqrt(N*Pi))) for n = 0..39 (if computed with sufficient numerical precision).
		

References

  • Donald E. Knuth, The Art of Computer Programming, Volume 4, Fascicle 4: Generating All Trees—History of Combinatorial Generation, 2006.
  • Y. L. Luke, The Special Functions and their Approximations, Vol. 1. Academic Press, 1969.

Crossrefs

The logarithmic version is A220422. Appears in A193365 and A220466.
Cf. A220412.

Programs

  • Maple
    A220002 := proc(n) local s; s := n -> `if`(n > 0, s(iquo(n,2))+n, 0);
    (-1)^n*mul(4*i+2, i = 1..2*n)*2^s(iquo(n,2))*coeff(taylor(sqrt(sin(x)/x), x,2*n+2), x, 2*n) end: seq(A220002(n), n = 0..14);
    # Second program illustrating J. L. Fields expansion of gamma quotients.
    A220002 := proc(n) local recF, binSum, swing;
    binSum := n -> add(i,i=convert(n,base,2));
    swing := n -> n!/iquo(n, 2)!^2;
    recF := proc(n, x) option remember; `if`(n=0, 1, -2*x*add(binomial(n-1,2*k+1)*bernoulli(2*k+2)/(2*k+2)*recF(n-2*k-2,x),k=0..n/2-1)) end: recF(2*n,-1/4)*2^(3*n-binSum(n))*swing(4*n+1) end:
  • Mathematica
    max = 14; CoefficientList[ Series[ Sqrt[ Sinc[x]], {x, 0, 2*max+1}], x^2][[1 ;; max+1]]*Table[ (-1)^n*Product[ (2*k+1), {k, 1, 2*n}], {n, 0, max}] // Numerator (* Jean-François Alcover, Jun 26 2013 *)
  • Sage
    length = 15; T = taylor(sqrt(sin(x)/x),x,0,2*length+2)
    def A005187(n): return A005187(n//2) + n if n > 0 else 0
    def A220002(n):
        P = mul(4*i+2 for i in (1..2*n)) << A005187(n//2)
        return (-1)^n*P*T.coefficient(x, 2*n)
    [A220002(n) for n in range(length)]
    
  • Sage
    # Second program illustrating the connection with the Euler numbers.
    def A220002_list(n):
        S = lambda n: sum((4-euler_number(2*k))/(4*k*x^(2*k)) for k in (1..n))
        T = taylor(exp(S(2*n+1)),x,infinity,2*n-1).coefficients()
        return [t[0].numerator() for t in T][::-1]
    A220002_list(15)

Formula

Let [x^n]T(f(x)) denote the coefficient of x^n in the Taylor expansion of f(x) then r(n) = (-1)^n*prod_{i=1..2n}(2i+1)*[x^(2*n)]T(sqrt(sin(x)/x)) is the rational coefficient of the asymptotic expansion (in N=4*n+3) and a(n) = numerator(r(n)) = r(n)*2^(3*n-bs(n)), where bs(n) is the binary sum of n (A000120).
Also a(n) = numerator([x^(2*n)]T(exp(S))) where S = sum_{k>=1}((4-E(2*k))/ (4*k)*x^(2*k)) and E(n) the Euler numbers A122045.
Also a(n) = sf(4*n+1)*2^(3*n-bs(n))*F_{2*n}(-1/4) where sf(n) is the swinging factorial A056040, bs(n) the binary sum of n and F_{n}(x) J. L. Fields' generalized Bernoulli polynomials A220412.
In terms of sequences this means
r(n) = (-1)^n*A103639(n)*A008991(n)/A008992(n),
a(n) = (-1)^n*A220371(n)*A008991(n)/A008992(n).
Note that a(n) = r(n)*A123854(n) and A123854(n) = 2^A004134(n) = 8^n/2^A000120(n).
Formula from Johannes W. Meijer:
a(n) = d(n+1)*A098597(2*n+1)*(A008991(n)/A008992(n)) with d(1) = 1 and
d(n+1) = -4*(2*n+1)*A161151(n)*d(n),
d(n+1) = (-1)^n*2^(-1)*(2*(n+1))!*A060818(n)*A048896(n).

A123852 Decimal expansion of (1*(2*(3*...)^(1/3))^(1/3))^(1/3).

Original entry on oeis.org

1, 1, 5, 6, 3, 6, 2, 6, 8, 4, 3, 3, 2, 2, 6, 9, 7, 1, 6, 8, 5, 3, 3, 7, 0, 3, 2, 2, 8, 8, 7, 3, 6, 9, 3, 5, 6, 5, 1, 3, 0, 1, 4, 5, 4, 3, 8, 9, 1, 8, 8, 8, 6, 3, 7, 9, 9, 9, 2, 5, 9, 5, 9, 8, 9, 8, 3, 1, 7, 7, 8, 1, 6, 0, 7, 2, 8, 2, 6, 1, 9, 4, 6, 0, 7, 9, 0, 8, 1, 3, 3, 8, 2, 0, 3, 7, 8, 3, 1, 7
Offset: 1

Views

Author

Keywords

Comments

Cubic recurrence constant (see A123851): a cubic analog of Somos's quadratic recurrence constant A112302.

Examples

			1.156362684332269716853370322887369356513014543891888637999259598983177816...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Programs

  • Mathematica
    Take[RealDigits[Product[N[n^3^-n,200], {n,400}]][[1]], 100]
    RealDigits[Exp[-D[PolyLog[n, 1/3], n]/.n->0], 10, 100][[1]] (* Jean-François Alcover, Jan 28 2014 *)
  • PARI
    prodinf(n=1, n^(1/3^n)) \\ Michel Marcus, Aug 03 2019

Formula

Equals Product_{n>=1} n^(1/3^n).

Extensions

References updated by R. J. Mathar, Aug 12 2010

A123853 Numerators in an asymptotic expansion for the cubic recurrence sequence A123851.

Original entry on oeis.org

1, 3, -15, 113, -5397, 84813, -3267755, 74391561, -15633072909, 465681118929, -31041303829713, 1145088996404679, -185348722911971841, 8165727090278785521, -778296382754673737187, 39898888480559205453945, -35033447016186321707305533
Offset: 0

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Author

Keywords

Comments

A cubic analog of the asymptotic expansion A116603 of Somos's quadratic recurrence sequence A052129. Denominators are A123854.

Examples

			A123851(n) ~ c^(3^n)*n^(-1/2)/(1 + 3/(4*n) - 15/(32*n^2) + 113/(128*n^3) - 5397/(2048*n^4) + ...) where c = 1.1563626843322... is the cubic recurrence constant A123852.
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Crossrefs

Cf. A052129, A112302, A116603, A123851, A123852, A123854 (denominators).

Programs

  • Maple
    f:=proc(t,x) exp(sum(ln(1+m*x)/t^m,m=1..infinity)); end; for j from 0 to 29 do numer(coeff(series(f(3,x),x=0,30),x,j)); od;
  • PARI
    {a(n) = local(A); if(n < 0, 0, A = 1 + O(x) ; for( k = 1, n, A = truncate(A) + x * O(x^k); A += x^k * polcoeff( 3/4 * (subst(1/A, x, x^2/(1-x^2))^2/(1-x^2) - 1/subst(A, x, x^2)^(2/3)), 2*k ) ); numerator( polcoeff( A, n ) ) ) } /* Michael Somos, Aug 23 2007 */

A067623 Consider the power series (x+1)^(1/3)=1+x/3-x^2/9+5x^3/81+...; sequence gives denominators of coefficients.

Original entry on oeis.org

1, 3, 9, 81, 243, 729, 6561, 19683, 59049, 1594323, 4782969, 14348907, 129140163, 387420489, 1162261467, 10460353203, 31381059609, 94143178827, 2541865828329, 7625597484987, 22876792454961, 205891132094649, 617673396283947
Offset: 0

Views

Author

Benoit Cloitre, Feb 02 2002

Keywords

Comments

All terms are powers of 3.

Crossrefs

Cf. A004128, A046161, A067622 (numerators), A123854.

Programs

Formula

a(n) = 3^A004128(n).
a(n) = 3^n*a(floor(n/3)). - Vladeta Jovovic, Mar 01 2004
a(n) = denominator(binomial(1/3, n)). - Peter Luschny, Apr 07 2016
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