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.

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

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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)).