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 1-8 of 8 results.

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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Author

Keywords

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

A007406 Wolstenholme numbers: numerator of Sum_{k=1..n} 1/k^2.

Original entry on oeis.org

1, 5, 49, 205, 5269, 5369, 266681, 1077749, 9778141, 1968329, 239437889, 240505109, 40799043101, 40931552621, 205234915681, 822968714749, 238357395880861, 238820721143261, 86364397717734821, 17299975731542641, 353562301485889, 354019312583809, 187497409728228241
Offset: 1

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Author

Keywords

Comments

By Wolstenholme's theorem, p divides a(p-1) for prime p > 3. - T. D. Noe, Sep 05 2002
Also p divides a( (p-1)/2 ) for prime p > 3. - Alexander Adamchuk, Jun 07 2006
The rationals a(n)/A007407(n) converge to Zeta(2) = (Pi^2)/6 = 1.6449340668... (see the decimal expansion A013661).
For the rationals a(n)/A007407(n), n >= 1, see the W. Lang link under A103345 (case k=2).
See the Wolfdieter Lang link under A103345 on Zeta(k, n) with the rationals for k=1..10, g.f.s and polygamma formulas. - Wolfdieter Lang, Dec 03 2013
Denominator of the harmonic mean of the first n squares. - Colin Barker, Nov 13 2014
Conjecture: for n > 3, gcd(n, a(n-1)) = A089026(n). Checked up to n = 10^5. - Amiram Eldar and Thomas Ordowski, Jul 28 2019
True if n is prime, by Wolstenholme's theorem. It remains to show that gcd(n, a(n-1)) = 1 if n > 3 is composite. - Jonathan Sondow, Jul 29 2019
From Peter Bala, Feb 16 2022: (Start)
Sum_{k = 1..n} 1/k^2 = 1 + (1 - 1/2^2)*(n-1)/(n+1) - (1/2^2 - 1/3^2)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3^2 - 1/4^2)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) - (1/4^2 - 1/5^2)*(n-1)*(n-2)*(n-3)*(n-4)/((n+1)*(n+2)*(n+3)*(n+4)) + .... Cf. A082687 and A120778.
This identity allows us to extend the definition of Sum_{k = 1..n} 1/k^2 to non-integral values of n. (End)
Numerators of the Eulerian numbers T(-2,k) for k = 0,1..., if T(n,k) is extended to negative n by the recurrence T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) (indexed as in A173018). - Michael J. Collins, Oct 10 2024

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A001008, A007407 (denominators), A000290, A082687, A120778.
Numbers n such that a(n) is prime are listed in A111354. Primes in {a(n)} are listed in A123751. - Alexander Adamchuk, Oct 11 2006

Programs

  • Haskell
    import Data.Ratio ((%), numerator)
    a007406 n = a007406_list !! (n-1)
    a007406_list = map numerator $ scanl1 (+) $ map (1 %) $ tail a000290_list
    -- Reinhard Zumkeller, Jul 06 2012
    
  • Magma
    [Numerator(&+[1/k^2:k in [1..n]]):n in [1..23]]; // Marius A. Burtea, Aug 02 2019
  • Maple
    a:= n-> numer(add(1/i^2, i=1..n)): seq(a(n), n=1..24);  # Zerinvary Lajos, Mar 28 2007
  • Mathematica
    a[n_] := If[ n<1, 0, Numerator[HarmonicNumber[n, 2]]]; Table[a[n], {n, 100}]
    Numerator[HarmonicNumber[Range[20],2]] (* Harvey P. Dale, Jul 06 2014 *)
  • PARI
    {a(n) = if( n<1, 0, numerator( sum( k=1, n, 1 / k^2 ) ) )} /* Michael Somos, Jan 16 2011 */
    

Formula

Sum_{k=1..n} 1/k^2 = sqrt(Sum_{j=1..n} Sum_{i=1..n} 1/(i*j)^2). - Alexander Adamchuk, Oct 26 2004
G.f. for rationals a(n)/A007407(n), n >= 1: polylog(2,x)/(1-x).
a(n) = Numerator of (Pi^2)/6 - Zeta(2,n). - Artur Jasinski, Mar 03 2010

A120996 Numerators of partial sums of Catalan numbers scaled by powers of 1/9.

Original entry on oeis.org

1, 10, 92, 833, 7511, 22547, 202967, 1826846, 49326272, 443941310, 3995488586, 35959456060, 323635312552, 2912718555868, 2912718853028, 26214470754457, 235930240718743, 6370116542620991, 57331049042801819
Offset: 0

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Author

Wolfdieter Lang, Aug 16 2006

Keywords

Comments

Denominators are given under A120997.
This is the second member (p=1) of the first p-family of partial sums of normalized scaled Catalan series CsnI(p):=sum(C(k)/L(2*p)^(2*k),k=0..infinity) with limit L(2*p)*(F(2*p+1) - F(2*p)*phi) = L(2*p)/phi^(2*p), with C(n)=A000108(n) (Catalan), F(n)= A000045(n) (Fibonacci), L(n) = A000032(n) (Lucas) and phi:=(1+sqrt(5))/2 (golden section).
Other members of this rational p-family are: A120778(n)/A120777(n) (p=0), ......
From the expansion of sqrt(1-x) = 1 -(x/2)*sum(C(k)*(x/4)^k,k=0..infinity), for |x|<=1, one has sum(C(k)/q^k,k=0..infinity) = (q - sqrt(q*(q-4)))/2, for |q|>=4.
The CsnI(1) series value is lim_{n->infinity}(r(n)) = 3*(2-phi) = 3/phi^2 = 1.145898034 (maple10, 10 digits).
The partial sums of the above mentioned first p-family are rI(p;n):=sum(C(k)/L(2*p)^(2*k), k=0..n), n>=0, for p=0,1,...
For special q values q(n):=2 + L(2*n), n>=0, one finds the above given limits for the p-family members for n=2*p (even), p>=0. The Pell equation x^2 -5*y^2 = +4 with general (nonnegative) solutions (x,y)=(L(2*n),F(2*n)) and well known identities for Fibonacci and Lucas numbers were used to derive the above given p-family result. See the W. Lang link.

Examples

			Rationals r(n): [1, 10/9, 92/81, 833/729, 7511/6561, 22547/19683,
202967/177147, 1826846/1594323,...].
		

Formula

a(n)=numerator(r(n)) with r(n) := rI(p=1,n) = sum(C(k)/L(2)^(2*k),k=0..n), with Lucas L(2)=3 and C(k):=A000108(k) (Catalan). The rationals r(n) are given in lowest terms.

A082687 Numerator of Sum_{k=1..n} 1/(n+k).

Original entry on oeis.org

1, 7, 37, 533, 1627, 18107, 237371, 95549, 1632341, 155685007, 156188887, 3602044091, 18051406831, 7751493599, 225175759291, 13981692518567, 14000078506967, 98115155543129, 3634060848592973, 3637485804655193
Offset: 1

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Author

Benoit Cloitre, Apr 12 2003

Keywords

Comments

Numerator of Sum_{k=0..n-1} 1/((k+1)(2k+1)) (denominator is A111876). - Paul Barry, Aug 19 2005
Numerator of the sum of all matrix elements of n X n Hilbert matrix M(i,j) = 1/(i+j-1) (i,j = 1..n). - Alexander Adamchuk, Apr 11 2006
Numerator of the 2n-th alternating harmonic number H'(2n) = Sum ((-1)^(k+1)/k, k=1..2n). H'(2n) = H(2n) - H(n), where H(n) = Sum_{k=1..n} 1/k is the n-th Harmonic Number. - Alexander Adamchuk, Apr 11 2006
a(n) almost always equals A117731(n) = numerator(n*Sum_{k=1..n} 1/(n+k)) = numerator(Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1)) but differs for n = 14, 53, 98, 105, 111, 114, 119, 164. - Alexander Adamchuk, Jul 16 2006
Sum_{k=1..n} 1/(n+k) = n!^2 *Sum_{j=1..n} (-1)^(j+1) /((n+j)!(n-j)!j). - Leroy Quet, May 20 2007
Seems to be the denominator of the harmonic mean of the first n hexagonal numbers. - Colin Barker, Nov 19 2014
Numerator of 2*n*binomial(2*n,n)*Sum_{k = 0..n-1} (-1)^k* binomial(n-1,k)/(n+k+1)^2. Cf. A049281. - Peter Bala, Feb 21 2017
From Peter Bala, Feb 16 2022: (Start)
2*Sum_{k = 1..n} 1/(n+k) = 1 + 1/(1*2)*(n-1)/(n+1) - 1/(2*3)*(n-1)*(n-2)/((n+1)*(n+2)) + 1/(3*4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) - 1/(4*5)*(n-1)*(n-2)*(n-3)*(n-4)/((n+1)*(n+2)*(n+3)*(n+4)) + - .... Cf. A101028.
2*Sum_{k = 1..n} 1/(n+k) = n - (1 + 1/2^2)*n*(n-1)/(n+1) + (1/2^2 + 1/3^2)*n*(n-1)*(n-2)/((n+1)*(n+2)) - (1/3^2 + 1/4^2)*n*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + (1/4^2 + 1/5^2)*n*(n-1)*(n-2)*(n-3)*(n-4)/((n+1)*(n+2)*(n+3)*(n+4)) - + .... Cf. A007406 and A120778.
These identities allow us to extend the definition of Sum_{k = 1..n} 1/(n+k) to non-integral values of n. (End)

Examples

			H'(2n) = H(2n) - H(n) = {1/2, 7/12, 37/60, 533/840, 1627/2520, 18107/27720, 237371/360360, 95549/144144, 1632341/2450448, 155685007/232792560, ...}, where H(n) = A001008/A002805.
n=2: HilbertMatrix(n,n)
   1  1/2
  1/2 1/3
so a(2) = Numerator(1 + 1/2 + 1/2 + 1/3) = Numerator(7/3) = 7.
The n X n Hilbert matrix begins:
   1   1/2  1/3  1/4  1/5  1/6  1/7  1/8  ...
  1/2  1/3  1/4  1/5  1/6  1/7  1/8  1/9  ...
  1/3  1/4  1/5  1/6  1/7  1/8  1/9  1/10 ...
  1/4  1/5  1/6  1/7  1/8  1/9  1/10 1/11 ...
  1/5  1/6  1/7  1/8  1/9  1/10 1/11 1/12 ...
  1/6  1/7  1/8  1/9  1/10 1/11 1/12 1/13 ...
		

Crossrefs

Bisection of A058313, A082688 (denominators).

Programs

  • Magma
    [Numerator((HarmonicNumber(2*n) -HarmonicNumber(n))): n in [1..40]]; // G. C. Greubel, Jul 24 2023
    
  • Maple
    a := n -> numer(harmonic(2*n) - harmonic(n)):
    seq(a(n), n=1..20); # Peter Luschny, Nov 02 2017
  • Mathematica
    Numerator[Sum[1/k,{k,1,2*n}] - Sum[1/k,{k,1,n}]] (* Alexander Adamchuk, Apr 11 2006 *)
    Table[Numerator[Sum[1/(i + j - 1), {i, n}, {j, n}]], {n, 20}] (* Alexander Adamchuk, Apr 11 2006 *)
    Table[HarmonicNumber[2 n] - HarmonicNumber[n], {n, 20}] // Numerator (* Eric W. Weisstein, Dec 14 2017 *)
  • PARI
    a(n) = numerator(sum(k=1, n, 1/(n+k))); \\ Michel Marcus, Dec 14 2017
    
  • SageMath
    [numerator(harmonic_number(2*n,1) - harmonic_number(n,1)) for n in range(1,41)] # G. C. Greubel, Jul 24 2023

Formula

Limit_{n -> oo} Sum_{k=1..n} 1/(n+k) = log(2).
Numerator of Psi(2*n+1) - Psi(n+1). - Vladeta Jovovic, Aug 24 2003
a(n) = numerator((Sum_{k=1..2*n} 1/k) - Sum_{k=1..n} 1/k). - Alexander Adamchuk, Apr 11 2006
a(n) = numerator(Sum_{j=1..n} (Sum_{i=1..n} 1/(i+j-1))). - Alexander Adamchuk, Apr 11 2006
The o.g.f for Sum_{k=1..n} 1/(n+k) is f(x) = (sqrt(x)*log((1+sqrt(x))/(1-sqrt(x))) + log(1-x))/(2*x*(1-x)).

A120777 a(n) = 2^(2*n - valuation(CatalanNumber(n), 2)).

Original entry on oeis.org

1, 4, 8, 64, 128, 512, 1024, 16384, 32768, 131072, 262144, 2097152, 4194304, 16777216, 33554432, 1073741824, 2147483648, 8589934592, 17179869184, 137438953472, 274877906944, 1099511627776, 2199023255552, 35184372088832, 70368744177664, 281474976710656, 562949953421312
Offset: 0

Views

Author

Wolfdieter Lang, Jul 20 2006

Keywords

Comments

Previous name: One half of denominators of partial sums of a series for sqrt(2).
Also denominators of partial sums Sum_{k=0..n} (C(k)/(-4)^k) = A120788(n)/A120777(n).
One half of denominators of partial sums which involve Catalan numbers A000108(k) divided by 4^k with alternating signs.
The listed numbers coincide with the denominators of sum(C(k)/4^k, k=0..n). See numerators A120778. In general these denominators may be different. See e.g. A120783 versus A120793 and A120787 versus A120796.

Crossrefs

Programs

  • Maple
    a := n -> denom(binomial(2*n+2, n+1) / 2^(2*n+1)):
    seq(a(n), n=0..22); # Johannes W. Meijer, Sep 23 2012
    Conjecture: The following Maple program appears to generate this sequence! 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))/2, n=0..22); # Zerinvary Lajos, May 21 2008
    a := proc(n) option remember: if n = 0 then b(0):=0 else b(n) := b(n-1) + A001511(n+1) fi: a(n) := 2^b(n) end proc: A001511 := proc(n) option remember: if n = 1 then 1 else procname(n-1) + (-1)^n * procname(floor(n/2)) fi: end proc:
    seq(a(n), n=0..22); # Johannes W. Meijer, Jul 06 2009, revised Sep 23 2012
  • Mathematica
    Table[Denominator[CatalanNumber[k]/(-4)^k], {k, 0, 22}] (* Jean-François Alcover, Jun 21 2013 *)
    (* Alternative: *)
    A120777[n_] := 2^(2*n - IntegerExponent[CatalanNumber[n], 2]);
    Table[A120777[n], {n, 0, 26}]  (* Peter Luschny, Apr 16 2024 *)

Formula

a(n) = denominator(r(n)), with the rationals r(n) defined under A120088.
From Johannes W. Meijer, Jul 06 2009: (Start)
a(n) = denominator(C(2*n+2,n+1)/2^(2*n+1)).
If b(n) = log(a(n))/log(2) then c(n) = b(n+1)-b(n) = A001511(n+1) i.e. the ruler function. (End)
a(n) = 2^(2*n- A048881(n)) = 2^A283208(n). - Amiram Eldar, Apr 18 2024

Extensions

New name by Peter Luschny, Apr 16 2024

A141244 Numerators in the expansion of (1-sqrt(1-x^2))/(1-x).

Original entry on oeis.org

0, 0, 1, 1, 5, 5, 11, 11, 93, 93, 193, 193, 793, 793, 1619, 1619, 26333, 26333, 53381, 53381, 215955, 215955, 436109, 436109, 3518265, 3518265, 7088533, 7088533, 28539857, 28539857, 57414019
Offset: 0

Views

Author

Paul Barry, Jun 17 2008

Keywords

Comments

The denominators in the expansion of (1-sqrt(1-x^2))/(1-x) are 1,1,2,2,8,8,16,16,... or 2^A005187(n) doubled. The sequence 0,1/2,1/2,5/8,5/8,... is the image of n under the Chebyshev related (rational) Riordan array c((x/2)^2),(x/2)c((x/2)^2)) with c(x) the g.f. of A000108. The image of n+1 under this array is 1,1,1,....

Crossrefs

Programs

  • PARI
    a(n) = my(y=x+O(x^(n+1))); numerator(polcoeff((1-sqrt(1-y^2))/(1-y), n)); \\ Michel Marcus, Aug 17 2014

Formula

a(2n) = a(2n+1) = A120778(n-1) (conjectural). - Greg Martin, Aug 16 2014, corrected by M. F. Hasler, Aug 18 2014

A160481 Row sums of the Beta triangle A160480.

Original entry on oeis.org

-1, -10, -264, -13392, -1111680, -137030400, -23500108800, -5351202662400, -1562069156659200, -568747270103040000, -252681700853514240000, -134539938778433126400000, -84573370199475510312960000, -61972704966344777143418880000, -52361960516341326660973363200000
Offset: 2

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Author

Johannes W. Meijer, May 24 2009, Sep 19 2012

Keywords

Comments

It is conjectured that the row sums of the Beta triangle depend on three different sequences. Two Maple algorithms are given. The first one gives the row sums according to the Beta triangle A160480 and the second one gives the row sums according to our conjecture.

Crossrefs

A160480 is the Beta triangle.
Row sum factors A120778, A000165 and A049606.

Programs

  • Maple
    nmax := 14; mmax := nmax: for n from 1 to nmax do BETA(n, n) := 0 end do: m := 1: for n from m+1 to nmax do BETA(n,m) := (2*n-3)^2*BETA(n-1, m)-(2*n-4)! od: for m from 2 to mmax do for n from m+1 to nmax do BETA(n, m) := (2*n-3)^2*BETA(n-1, m) - BETA(n-1, m-1) od: od: for n from 2 to nmax do s1(n) := 0: for m from 1 to n-1 do s1(n) := s1(n) + BETA(n, m) od: od: seq(s1(n), n=2..nmax);
    # End first program
    nmax := nmax; A120778 := proc(n): numer(sum(binomial(2*k1, k1)/(k1+1) / 4^k1, k1=0..n)) end proc: A000165 := proc(n): 2^n*n! end proc: A049606 := proc(n): denom(2^n/n!) end proc: for n from 2 to nmax do s2(n) := (-1)*A120778(n-2)*A000165(n-2)*A049606(n-1) end do: seq(s2(n), n=2..nmax);
    # End second program
  • Mathematica
    BETA[2, 1] = -1; BETA[n_, 1] := BETA[n, 1] = (2*n - 3)^2*BETA[n - 1, 1] - (2*n - 4)!; BETA[n_ /; n > 2, m_ /; m > 0] /; 1 <= m <= n := BETA[n, m] = (2*n - 3)^2*BETA[n - 1, m] - BETA[n - 1, m - 1]; BETA[, ] = 0;
    Table[Sum[BETA[n, m], {m, 1, n - 1}], {n, 2, 14}] (* Jean-François Alcover, Dec 13 2017 *)

Formula

Rowsums(n) = (-1)*A120778(n-2)*A000165(n-2)*A049606(n-1) for n >= 2.
Conjecture: a(n) = (2*n-3)! - 2^(2*n-3)*(n-1)!*(n-2)!, for n >= 2 (gives the first 13 terms). - Christopher P. Herzog, Nov 25 2014
Meijer's and Herzog's conjectures can also be written as: a(n) = -A129890(n-2)*A000165(n-2) = A009445(n-2) - A002474(n-2). - Peter Luschny, Dec 01 2014

Extensions

a(15)-a(16) from Stefano Spezia, Jun 28 2024

A362534 Numerators of the ratio of the symmetry-constrained bound to the adiabatic bound on polarization transfer in AXn spin-1/2 systems.

Original entry on oeis.org

1, 1, 6, 6, 15, 15, 140, 140, 315, 315, 1386, 1386, 3003, 3003, 51480, 51480, 109395, 109395, 92378, 92378, 969969, 969969, 2704156, 2704156, 16900975, 16900975, 70204050, 70204050, 145422675, 145422675, 4808643120, 4808643120, 9917826435, 9917826435, 40838108850, 40838108850
Offset: 1

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Author

Mohamed Sabba, Apr 24 2023

Keywords

Comments

In spin physics and NMR, these numbers appear as the numerators of the ratio of different classes of upper bounds on the transfer of z-magnetization in AXn spin systems from a group of spin-1/2 nuclei Xn to a single spin-1/2 nucleus A.
The symmetry-constrained upper bounds are given by the function f(n):
(1) for even n, f(n) = (2^(1-n))*n*binomial(n-1, n/2)
(2) for odd n, f(n) = (2^(1-n))*n*binomial(n-1, (n-1)/2)
The adiabatic bounds are given by the function g(n):
(3) for even n, g(n) = 2*(1-(2^(-n))*binomial(n, n/2))
(4) for odd n, g(n) = 2*(1-(2^(-n))*binomial(n, (n-1)/2))
Where we have the relation:
(5) g(n) = 2*(1 - f(n+1)/(n+1))
The sequence a(n) is defined as the numerator of f(n)/g(n):
(6) a(n) = numerator(f(n)/g(n))
(7) for even n, f(n)/g(n) = (n/2)/(2^(n)*binomial(n, n/2)^(-1) - 1)
(8) for odd n, f(n)/g(n) = ((n+1)/2)/(2^(n)*binomial(n, (n+1)/2)^(-1) - 1)
The first few values of the upper symmetry-constrained bounds f(n) are {1, 1, 3/2, 3/2, 15/8, 15/8, 35/16, 35/16, 315/128, 315/128, ...} which appears to be related to A086116 and A001803.
The first few values of the upper adiabatic bounds g(n) are {1, 1, 5/4, 5/4, 11/8, 11/8, 93/64, 93/64, 193/128, 193/128, ...} which appears to be related to A141244 and A120778.
The first few values of f(n)/g(n) are {1, 1, 6/5, 6/5, 15/11, 15/11, 140/93, 140/93, 315/193, 315/193, ...}
Conjecture: the numerator of g(n) is the denominator of f(n)/g(n).

Crossrefs

Cf. A001803, A086116, A120778, A141244 (denominators but shifted).

Programs

  • Mathematica
    Table[Numerator[Ceiling[n/2]  (2^n Binomial[n, Ceiling[n/2]]^-1 - 1 )^-1], {n, 1, 20}]
  • PARI
    a(n) = numerator(ceil(n/2)/(2^(n)*binomial(n,ceil(n/2))^(-1) - 1)); \\ Michel Marcus, Apr 25 2023

Formula

a(n) = numerator(ceiling(n/2)/(2^(n)*binomial(n,ceiling(n/2))^(-1) - 1)).
Showing 1-8 of 8 results.