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 1-5 of 5 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

A019675 Decimal expansion of Pi/8.

Original entry on oeis.org

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

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Comments

Equals Integral_{x>=0} sin(4*x)/(4*x) dx. - Jean-François Alcover, Feb 28 2013
Consider 4 circles inscribed in a square. Inscribe a square in each circle. And finally, inscribe 4 circles inside each four small squares. Totally we get 16 small circles. Pi/8 is the ratio of the area of the 16 small circles to the area of initial square. See the link. - Kirill Ustyantsev, Apr 30 2020

Examples

			Pi/8 = 0.392699081698724154807830422909937860524646174921888227621868... - _Vladimir Joseph Stephan Orlovsky_, Dec 02 2009
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.4, p. 492.

Crossrefs

Programs

  • Magma
    pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^100*(pi)/8))); // Vincenzo Librandi, Oct 07 2015
  • Mathematica
    RealDigits[N[Pi/8,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
  • PARI
    default(realprecision, 1002);
    eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+2)))), "3..-2"))  \\ Gheorghe Coserea, Oct 06 2015
    

Formula

From Peter Bala, Nov 15 2016: (Start)
Pi/8 = Sum_{k >= 1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)).
More generally, for n >= 0 we have 1/(2*n)! * Pi/4 = Sum_{k >= 1} (-1)^(k+n-1) * 1/Product_{j = -n..n} (2*k + 2*j - 1): when n = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For N divisible by 4, we have the asymptotic expansion Pi/8 - Sum_{k = 1..N/2} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)) ~ -1/2*(1/N^3 - 2/N^5 + 31/N^7 - 692/N^9 + ...), where the sequence of unsigned coefficients [1, 2, 31, 692, ...] equals A024235. (End)
Equals Integral_{x = 0..1} x*sqrt(1 - x^4) dx. - Peter Bala, Oct 27 2019
Equals Integral_{x = 0..oo} sin(x)^6/x^4 dx = Sum_{n >= 1} sin(n)^6/n^4, by the Abel-Plana formula. - Peter Bala, Nov 04 2019
From Amiram Eldar, Jul 12 2020: (Start)
Equals arctan(sqrt(2) - 1).
Equals Sum_{k>=0} (-1)^k/(4*k+2).
Equals Sum_{k>=0} 1/((4*k+1)*(4*k+3)) = Sum_{k>=0} 1/A001539(k).
Equals Integral_{x=0..oo} dx/(x^2 + 16).
Equals Integral_{x=0..oo} dx/(x^4 + 4) = Integral_{x=0..oo} x/(x^4 + 4) dx.
Equals Integral_{x=0..oo} x/(x^4 + 1)^2 dx = Integral_{x=0..1} x/(x^4 + 1) dx.
Equals Integral_{x=0..1} x * arcsin(x) dx. (End)
From Kritsada Moomuang, Jun 18 2025: (Start)
Equals Integral_{x=0..oo} (x*log(x + 1))/((x^2 + 1)^2) dx.
Equals Integral_{x=0..oo} (x^3 - 3*x + 3*arctan(x))/(3*x^5) dx. (End)

A162540 a(n) = (2*n+1)*(2*n+3)*(2*n+5)/3.

Original entry on oeis.org

5, 35, 105, 231, 429, 715, 1105, 1615, 2261, 3059, 4025, 5175, 6525, 8091, 9889, 11935, 14245, 16835, 19721, 22919, 26445, 30315, 34545, 39151, 44149, 49555, 55385, 61655, 68381, 75579, 83265, 91455, 100165, 109411, 119209, 129575, 140525, 152075, 164241
Offset: 0

Views

Author

Jacob Landon (jacoblandon(AT)aol.com), Jul 05 2009

Keywords

Crossrefs

Programs

  • Magma
    [(2*n+1)*(2*n+3)*(2*n+5)/3: n in [0..40]]; // Vincenzo Librandi, Nov 16 2011
    
  • Maple
    A162540:=n->(2*n+1)*(2*n+3)*(2*n+5)/3: seq(A162540(n), n=0..80); # Wesley Ivan Hurt, May 28 2016
  • Mathematica
    Table[((2n+1)(2n+3)(2n+5))/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{5,35,105,231},40] (* Harvey P. Dale, Nov 06 2011 *)
  • PARI
    Vec((5+15*x-5*x^2+x^3)/(x-1)^4 + O(x^100)) \\ Altug Alkan, Oct 26 2015

Formula

a(n) = A061550(n)/3 = A077415(2*n+3).
From R. J. Mathar, Jul 16 2009: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.
G.f.: (5 + 15*x - 5*x^2 + x^3)/(x-1)^4. (End)
a(n) = 5*Pochhammer(7/2,n)/Pochhammer(1/2,n). Hence e.g.f. is 5* 1F1(7/2;1/2;x), with 1F1 being the confluent hypergemetric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
E.g.f.: (8*x^3 + 60*x^2 + 90*x + 15)*exp(x)/3. - Robert Israel, May 27 2016
From Amiram Eldar, Jan 09 2021: (Start)
Sum_{n>=0} 1/a(n) = 1/4.
Sum_{n>=0} (-1)^n/a(n) = 3*Pi/8 - 1 = A093828 - 1. (End)

Extensions

Offset corrected, definition clarified by R. J. Mathar, Jul 16 2009

A196506 a(n) = 1*3*5 + 3*5*7 + 5*7*9 + ... (n terms).

Original entry on oeis.org

0, 15, 120, 435, 1128, 2415, 4560, 7875, 12720, 19503, 28680, 40755, 56280, 75855, 100128, 129795, 165600, 208335, 258840, 318003, 386760, 466095, 557040, 660675, 778128, 910575, 1059240, 1225395, 1410360, 1615503, 1842240, 2092035
Offset: 0

Views

Author

R. J. Mathar, Oct 03 2011

Keywords

Comments

All terms are multiples of 3.

References

  • Jolley, Summation of Series, Dover (1961), eq (43) page 8.

Crossrefs

Cf. A061550 (first differences).

Programs

  • Magma
    [((4*n^2-1)*(2*n+3)*(2*n+5)+15)/ 8 : n in [0..30]]; // Vincenzo Librandi, Oct 05 2011
  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{0,15,120,435,1128},40] (* or *) Accumulate[ Join[{0},Times@@@Partition[Range[1,111,2],3,1]]] (* or *) Table[2n^4-5n^2+3,{n,40}](* Harvey P. Dale, Mar 14 2015 *)

Formula

a(n) = ((4n^2 - 1)*(2n + 3)*(2n + 5) + 15)/ 8 = Sum_{i=1..n} (2i - 1)*(2i + 1)*(2i + 3).
G.f. -3*x*(5 + 15*x - 5*x^2 + x^3) / (x-1)^5 .
a(n) = 2 n^4 + 8 n^3 + 7 n^2 - 2 n. - Harvey P. Dale, Mar 14 2015, corrected by Eric Rowland, Aug 15 2017

A239545 Decimal expansion of Sum_{k>=0} (-1)^k/((2k+1)*(2k+3)*(2k+5)).

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Mar 21 2014

Keywords

Examples

			0.0593657483653908214744970895766045271913128415885548942885347407051...
		

Crossrefs

Cf. A000796, A001539, A005408 (odd numbers) A021016 (Sum_{k>=0} 1/((2k+1)*(2k+3)*(2k+5))), A061550.

Programs

Formula

Equals Pi/8 - 1/3 = A019675 - A010701.
Equals Sum_{k>=1} 1/((4*k+1)*(4*k+3)) = Sum_{k>=1} 1/A001539(k). - Amiram Eldar, Jul 04 2020
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