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.

User: Joseph Abate

Joseph Abate's wiki page.

Joseph Abate has authored 3 sequences.

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

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)).

A280442 Numerators of coefficients in the Taylor series expansion of Sum_{n>=0} exp((-1)^n*euler(2*n)*x^n/(2*n)).

Original entry on oeis.org

1, 1, 11, 173, 22931, 1319183, 233526463, 29412432709, 39959591850371, 8797116290975003, 4872532317019728133, 1657631603843299234219, 2718086236621937756966743, 1321397724505770800453750299, 1503342018433974345747514544039
Offset: 0

Author

Johannes W. Meijer and Joseph Abate, Jan 03 2017

Keywords

Comments

This sequence is related in a peculiar way to A223067, a sequence related to the period T of a simple gravity pendulum for arbitrary amplitudes. See A280443 for more information.

Crossrefs

Cf. A046161 (denominators).
Cf. A000364 (Euler numbers), A223067, A255881, A280443.

Programs

  • Maple
    nmax:=14: 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 a(n) := numer(coeff(f, x, n)) od: seq(a(n), n=0..nmax);
  • Sage
    def A280442_list(prec):
        P. = PowerSeriesRing(QQ, default_prec=2*prec)
        def g(x): return 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^(2*n - sum(n.digits(2)))
        return [d(n)*R[n] for n in (0..prec)]
    print(A280442_list(14)) # Peter Luschny, Jan 05 2017

Formula

a(n) = numerators of coefficients in the Taylor series expansion of Sum_{n>=0} exp((-1)^n * euler(2*n)*x^n/(2*n)).
Let S = Sum_{n>=0} (-1)^n*euler(2*n)*x^n/(2*n) and w(n) = A005187(n) then a(n) = 2^w(n) * [x^n] exp(S). - Peter Luschny, Jan 05 2017

A178792 Dot product of the rows of triangle A046899 with vector (1,2,4,8,...) (= A000079).

Original entry on oeis.org

1, 5, 31, 209, 1471, 10625, 78079, 580865, 4361215, 32978945, 250806271, 1916280833, 14698053631, 113104519169, 872801042431, 6751535300609, 52337071357951, 406468580343809, 3162019821780991, 24634626678980609, 192179216026959871
Offset: 0

Author

Joseph Abate, Jun 15 2010

Keywords

Comments

Hankel transform is A133460.

Examples

			a(3) = (1,4,10,20)dot(1,2,4,8) = 209.
		

Crossrefs

Row sums of A091811.

Programs

  • Maple
    a := n -> binomial(2*n+2,n+1)*hypergeom([-n, n + 1], [n + 2], -1)/2:
    seq(simplify(a(n)), n=0..20); # Peter Luschny, Feb 21 2017
  • Mathematica
    CoefficientList[Series[(3-Sqrt[1-8*x])/(2*(1+x)*Sqrt[1-8*x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)
    Table[Sum[2^k*Binomial[n + k, k], {k, 0, n}], {n, 0, 20}] (* Michael De Vlieger, Oct 28 2016 *)
    a[n_] := (-1)^(n + 1) - 2^(n + 1) (2n + 1) Binomial[2n, n] Hypergeometric2F1[1, 2n + 2, n + 2, 2]/(n + 1); Array[a, 22, 0] (* Robert G. Wilson v, Jul 21 2018 *)

Formula

a(n) = Sum_{k = 0..n} A046899(n,k)*2^k = Sum_{k = 0..n} 2^k * binomial(n+k,k).
G.f.: (1/3)*(4/sqrt(1 - 8*x) - 1/(1 - x*c(2*x))) with c(x) the g.f. of the Catalan numbers A000108.
a(n) = (1/3)*(4*2^n*A000984(n) - A064062(n)).
a(n) + a(n+1) = 6*2^n*A001700(n).
O.g.f.: (3 - sqrt(1 - 8*x))/(2*(1 + x)*sqrt(1 - 8*x)). - Peter Bala, Apr 10 2012
a(n) = 2^n *binomial(2+2*n,1+n)*2F1(1, 2+2*n; 2+n;-1). - Olivier Gérard, Aug 19 2012
D-finite with recurrence n*a(n) = (7*n - 4)*a(n-1) + 4*(2*n - 1)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(3n+2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 20 2012
a(n) = Sum_{k = 0..n} binomial(k+n,n) * binomial(2*n+1,n-k). - Vladimir Kruchinin, Oct 28 2016
a(n) = 1/2*(n + 1)*binomial(2*n+2,n+1)*Sum_{k = 0..n} binomial(n,k)/(n + k + 1). - Peter Bala, Feb 21 2017
a(n) = binomial(2*n+2,n+1)*hypergeom([-n, n+1], [n+2], -1)/2. - Peter Luschny, Feb 21 2017
a(n) = (-1)^(n+1) - 2^(n+1)*(2*n+1)*binomial(2*n,n)*hypergeom([1, 2*n+2], [n+2], 2)/(n+1). - John M. Campbell, Jul 14 2018
From Akiva Weinberger, Dec 06 2024: (Start)
a(n) = (2*n + 1)!/(n!^2) * Integral_{t=0..1} (t + t^2)^n dt.
a(n) = (Integral_{t=0..1} (t + t^2)^n dt) / (Integral_{t=0..1} (t - t^2)^n dt). (End)
a(n) = (-1)^n * Sum_{k=0..n} binomial(2*n+1,k) * (-2)^k. - André M. Timpanaro, Dec 15 2024
From Seiichi Manyama, Aug 04 2025: (Start)
a(n) = [x^n] (1+x)^(2*n+1)/(1-x)^(n+1).
a(n) = [x^n] 1/((1-x) * (1-2*x)^(n+1)). (End)