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 1-6 of 6 results.

A280443 a(n) = A280442(n)/A223067(n) = A067624(n)*A046161(n)/A223068(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 11, 17, 1, 23, 1, 11, 1, 1, 1, 17, 11, 1, 1, 1, 23, 11, 43, 17, 1, 1, 121, 1, 1, 1, 1, 4301, 1, 1, 1, 73, 11, 1, 1, 17, 1, 11, 23, 43, 1, 1, 11, 17, 1, 1, 1, 11, 101, 23, 89, 17, 11, 1, 1, 83, 1, 11, 1
Offset: 0

Views

Author

Johannes W. Meijer and Joseph Abate, Jan 03 2017

Keywords

Comments

This sequence is related in a peculiar way to A223067 and A223068, sequences related to the complete elliptic integral of the first kind K(k), and to A280442 and A046161, sequences related to the unsigned Euler numbers A000364.
In this sequence certain prime numbers appear on a regular basis, either by itself or as a factor of a composite number, i.e., a(n)=11 if n=7+5*k, a(n)=17 if n=13+8*k, a(n)=23 if n=15+11*k, a(n)=43 if n=28+21*k, a(n)=73 if n=41+36*k, a(n)=101 if n=58+50*k, a(n)=89 if n=60+44*k, a(n)=83 if n=65+41*k, in all cases k >= 0. We observe that the period T of each prime is apparently T = (prime-1)/2.
Conjecture: The sequence A280443 will not have a(n)=1 after some point.

Crossrefs

Cf. A000364 (Euler numbers), A046161, A067624, A223067, A223068, A280442.

Programs

  • Maple
    nmax:=68: A067624 := n -> 2^(2*n)*(2*n)!: f := series((exp(add((-1)^n*euler(2*n) * x^n/(2*n), n=1..nmax+1))), x=0, nmax+1): for n from 0 to nmax do b(n) := coeff(f, x, n); a(n) := numer(b(n))/numer(b(n)/A067624(n)) od: seq(a(n), n=0..nmax);
  • Sage
    def A280443_list(prec):
        P. = PowerSeriesRing(QQ, default_prec=2*prec)
        g = lambda x: exp(sum((-1)^k*euler_number(2*k)*x^k/(2*k) for k in (1..prec+1)))
        R = P(g(x)).coefficients()
        d = lambda n: 2*n - sum(n.digits(2))
        return [(2^d(n)*R[n]/(numerator(R[n]/factorial(2*n)))) for n in (0..prec)]
    print(A280443_list(68)) # Peter Luschny, Jan 05 2017

Formula

a(n) = A280442(n)/A223067(n).
a(n) = A067624(n)*A046161(n)/A223068(n).
a(n) = A280442(n)/numer((A280442(n)/A046161(n))/A067624(n)).

A061549 Denominator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p=1/4.

Original entry on oeis.org

1, 8, 128, 1024, 32768, 262144, 4194304, 33554432, 2147483648, 17179869184, 274877906944, 2199023255552, 70368744177664, 562949953421312, 9007199254740992, 72057594037927936, 9223372036854775808, 73786976294838206464, 1180591620717411303424, 9444732965739290427392
Offset: 0

Views

Author

Leah Schmelzer (leah2002(AT)mit.edu), May 16 2001

Keywords

Comments

We observe that b(n) = log(a(n))/log(2) = A120738(n). Furthermore c(n+1) = b(n+1)-b(n) = A090739(n+1) and c(n+1)-3 = A007814(n+1) for n>=0. - Johannes W. Meijer, Jul 06 2009
Using WolframAlpha, it appears that 2*a(n) gives the coefficients of Pi in the denominators of 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

Examples

			For n=1, the binomial(2*n-1/2, -1/2) yields the term 3/8. The denominator of this term is 8, which is the second term of the sequence.
		

Crossrefs

Bisection of A046161.
Appears in A162448.

Programs

  • Magma
    A061549:= func< n | 2^(4*n-(&+Intseq(2*n, 2))) >;
    [A061549(n): n in [0..30]]; // G. C. Greubel, Oct 20 2024
  • Maple
    seq(denom(binomial(2*n-1/2, -1/2)), n=0..20);
  • Mathematica
    Table[Denominator[(4*n)!/(2^(4*n)*(2*n)!^2) ], {n, 0, 20}] (* Indranil Ghosh, Mar 11 2017 *)
  • PARI
    for(n=0, 20, print1(denominator((4*n)!/(2^(4*n)*(2*n)!^2)),", ")) \\ Indranil Ghosh, Mar 11 2017
    
  • Python
    import math
    f = math.factorial
    def A061549(n): return (2**(4*n)*f(2*n)**2) // math.gcd(f(4*n), (2**(4*n)*f(2*n)**2)) # Indranil Ghosh, Mar 11 2017
    
  • Sage
    # uses[A000120]
    def a(n): return 1 << (4*n - A000120(n))
    [a(n) for n in (0..19)]  # Peter Luschny, Dec 02 2012
    

Formula

a(n) = denominator of binomial(2*n-1/2, -1/2).
a(n) are denominators of coefficients of 1/(sqrt(1+x)-sqrt(1-x)) power series. - Benoit Cloitre, Mar 12 2002
a(n) = 16^n/A001316(n). - Paul Barry, Jun 29 2006
a(n) = denom((4*n)!/(2^(4*n)*(2*n)!^2)). - Johannes W. Meijer, Jul 06 2009
a(n) = abs(A067624(n)/A117972(n)). - Johannes W. Meijer, Jul 06 2009

Extensions

More terms from Asher Auel, May 20 2001

A120738 a(n) = 4*n - A000120(n).

Original entry on oeis.org

0, 3, 7, 10, 15, 18, 22, 25, 31, 34, 38, 41, 46, 49, 53, 56, 63, 66, 70, 73, 78, 81, 85, 88, 94, 97, 101, 104, 109, 112, 116, 119, 127, 130, 134, 137, 142, 145, 149, 152, 158, 161, 165, 168, 173, 176, 180, 183, 190, 193, 197, 200, 205, 208, 212, 215, 221, 224, 228
Offset: 0

Views

Author

Paul Barry, Jun 29 2006

Keywords

Comments

Partial sums of A090739.
a(n) is also the increasing sequence of exponents of x in Product_{k > 1} (1 + x^(2^k - 1)). - Paul Pearson (ppearson(AT)rochester.edu), Aug 06 2008
Related to partial sums of the Ruler sequence A001511 by a(n) = A005187(2n), therefore {a(n)+1} are the indices of 1's in A252488. - M. F. Hasler, Jan 22 2015

Crossrefs

Programs

  • Magma
    A120738:= func< n | 4*n-(&+Intseq(n, 2)) >;
    [A120738(n): n in [0..100]]; // G. C. Greubel, Oct 20 2024
  • Maple
    a:=n->simplify(log[2](16^n/(add(modp(binomial(n,k),2),k=0..n))));
    a:=n->simplify(log[2](16^n/(2^(n-(padic[ordp](n!,2)))))); # Note: n-(padic[ordp](n!,2)) is the number of 1's in the binary expansion of n. - Paul Pearson (ppearson(AT)rochester.edu), Aug 06 2008
  • Mathematica
    Table[4 n - DigitCount[n, 2, 1], {n, 0, 58}] (* Michael De Vlieger, Nov 06 2016 *)
  • PARI
    {a(n) = if( n < 0, 0, 4*n - subst( Pol( binary( n ) ), x, 1) ) } /* Michael Somos, Aug 28 2007 */
    
  • PARI
    a(n) = 4*n - hammingweight(n); \\ Michel Marcus, Nov 06 2016
    
  • Python
    # Python 3.10
    def A120738(n): return (n<<2)-n.bit_count() # Chai Wah Wu, Jul 12 2022
    
  • Sage
    A120738 = lambda n: 4*n - sum(n.digits(2))
    print([A120738(n) for n in (0..58)]) # Peter Luschny, Nov 06 2016
    

Formula

a(n) = log_2(16^n/A001316(n)). [This was the original definition.]
a(n) = 2n + A005187(n).
a(n) = 3n + A011371(n).
a(n) = 4n - log_2(A001316(n)).
a(n) = log_2(A061549(n)).
2^a(n) = 16^n/A001316(n) = A061549(n).
a(n) = A086343(n) + A001511(n) for n>0. - Alford Arnold, Mar 23 2009
2^a(n) = abs(A067624(n)/A117972(n)). - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_{k>=0} (A030308(n,k)*A000225(k+2)). - Philippe Deléham, Oct 16 2011
a(n) = A005187(2n). - M. F. Hasler, Jan 22 2015

Extensions

Definition simplified by M. F. Hasler, Dec 29 2012

A320959 The exponential limit of arctanh (odd indices only).

Original entry on oeis.org

1, 10, 1248, 631440, 852647040, 2462394816000, 13241729554099200, 120563962753538304000, 1733764260005567741952000, 37343395325946891151466496000, 1155311729350231354981936496640000, 49626886713956000390638096497377280000, 2878005957927359237424925417166882734080000
Offset: 0

Views

Author

Peter Luschny, Nov 08 2018

Keywords

Comments

See A320956 for definitions and comments.

Examples

			Illustration of the convergence in the sense of A320956:
   [0] 0, 0, 0,  0, 0,    0, 0,      0, 0,         0, ...
   [1] 0, 1, 0,  2, 0,   24, 0,    720, 0,     40320, ... A005359
   [2] 0, 1, 0,  8, 0,  384, 0,  46080, 0,  10321920, ... A067624
   [3] 0, 1, 0, 10, 0,  984, 0, 262800, 0, 132289920, ...
   [4] 0, 1, 0, 10, 0, 1224, 0, 514800, 0, 445576320, ...
   [5] 0, 1, 0, 10, 0, 1248, 0, 615600, 0, 725840640, ...
   [6] 0, 1, 0, 10, 0, 1248, 0, 630720, 0, 832527360, ...
   [7] 0, 1, 0, 10, 0, 1248, 0, 631440, 0, 851155200, ...
   [8] 0, 1, 0, 10, 0, 1248, 0, 631440, 0, 852606720, ...
   [9] 0, 1, 0, 10, 0, 1248, 0, 631440, 0, 852647040, ...
		

Crossrefs

Cf. A320955 (exp), A320962 (log(x+1)), A320956 (sec+tan), A320958 (arcsin), this sequence (arctanh).

Programs

  • Maple
    # The function ExpLim is defined in A320956.
    L := [ExpLim(28, arctanh)]: seq(L[2*n], n=1..13);
  • Mathematica
    m = 13; CoefficientList[ArcTanh[x] + O[x]^(2 m + 1), x]*Range[0, 2 m - 1]!*BellB[Range[0, 2 m - 1]] // DeleteCases[#, 0]& (* Jean-François Alcover, Jul 23 2019 *)

Formula

For n >= 3 and odd, -a(m)*Zeta(m) = g(n), where g denotes the exponential limit of log(Gamma(x + 1)) and m = (n-1)/2.

A162445 A sequence related to the Beta function.

Original entry on oeis.org

1, 8, 384, 46080, 2064384, 3715891200, 392398110720, 1428329123020800, 274239191619993600, 1678343852714360832000, 102043306245033138585600, 4714400748520531002654720000, 160144566965128191597871104000
Offset: 0

Views

Author

Johannes W. Meijer, Jul 06 2009

Keywords

Comments

We define F(z) = Beta(1/2-z/2,1/2+z/2)/Beta(1/2,1/2) = 1/sin(Pi*(1+z)/2) with Beta(z,w) the Beta function. See A008956 for a closely related function.
For the Taylor series expansion of F(z) we can write F(z) = sum(b(n)*(Pi*z)^(2*n)/a(n), n=0..infinity) with b(n) = A046976(n) and a(n) the sequence given above.
We can also write F(z) = sum(c(n)*(Pi*z)^(2*n)/d(n), n=0..infinity) with c(n) = A000364(n) and d(n) = A067624(n).
If p(n) is the exponent of the prime factor 2 in a(n) than p(n) = A120738(n) and 2^p(n) = A061549(n) = abs((4*n)!!/A117972(n)).

Crossrefs

Bisection of A050971
Equals 2^(2*n)*A046977(n)

Programs

  • Mathematica
    Denominator[Table[EulerE[2n]/(4n)!!,{n,0,20}]] (* Harvey P. Dale, Jun 23 2013 *)

Formula

a(n) = denom(euler(2*n)/(4*n)!!)

A334366 Decimal expansion of Sum_{k>=0} 1/(4*k)!!.

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Apr 24 2020

Keywords

Comments

This constant is transcendental.

Examples

			1/(2^0*0!) + 1/(2^2*2!) + 1/(2^4*4!) + ... = 1.1276259652063807852...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Cosh[1/2], 10, 110] [[1]]
  • PARI
    cosh(1/2) \\ Michel Marcus, Apr 25 2020

Formula

Equals cosh(1/2).
Equals Product_{k>=0} 1 + 1/((2*k+1)*Pi)^2. - Amiram Eldar, Jul 16 2020
Showing 1-6 of 6 results.