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.

A240978 The largest prime divisor of A246053(n).

Original entry on oeis.org

2, 2, 7, 31, 127, 73, 691, 8191, 3617, 131071, 524287, 593, 2294797, 657931, 362903, 1001259881, 2147483647, 151628697551, 26315271553053477373, 154210205991661, 1897170067619, 1520097643918070802691, 1798482437, 67568238839737, 153289748932447906241
Offset: 0

Views

Author

Peter Luschny, Aug 12 2014

Keywords

Comments

According to theorem 2 of the Milnor paper a(2) and a(4) through a(8) are lower bounds for the number of distinct differentiable structures on spheres S^(4*k-1) for k = 2 and 4,..,8. Better bounds are given in A242032.

Crossrefs

Programs

  • Sage
    h = lambda x: zeta(2*x)*(4^x-2)
    A246053 = lambda n: Integer((h((n+1)//2)*h(n//2)/h(n)).denominator())
    A240978 = lambda n: max(prime_divisors(A246053(n)))
    [A240978(n) for n in range(25)]

Formula

a(n) = A006530(A246053(n)). - Michel Marcus, Aug 18 2014

A242050 Denominator of A242035(n) / A246053(floor(n/2)).

Original entry on oeis.org

2, 2, 2, 2, 1, 7, 1, 2, 1, 1, 1, 5, 1, 1, 7, 10, 1, 7, 1, 1, 1, 1, 3, 7, 1, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 5, 1, 1, 3, 1, 1, 1, 1, 62, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Peter Luschny, Aug 17 2014

Keywords

Comments

Let h(x) = zeta(2*x)*(4^x-2) and r(n) = h([(n+1)/2])*h([n/2])/ h(n). Then the denominator of r([n/2]) almost divides the numerator of r(n), where 'a almost divides b' means that a divides b or the denominator of the quotient b/a (reduced to lowest terms) is very small compared to the size of b.

Examples

			a(34) = 1 because A246053(17) divides A242035(34).
a(35) = 7 because A242035(35)/A246053(17) = 8416843291197378595121525029762095062482279733/7.
		

Crossrefs

Programs

  • Sage
    h = lambda x: zeta(2*x)*(4^x-2)
    r = lambda n: h((n+1)//2)*h(n//2)/h(n)
    A242050 = lambda n: (r(n).numerator()/r(n//2).denominator()).denominator()
    [A242050(n) for n in range(1000)]

A246052 Triangle read by rows: denominator of h(n-k)*h(k)/h(n) where h(x) = zeta(2*x)*(4^x-2), 0<=k<=n.

Original entry on oeis.org

2, 2, 2, 2, 7, 2, 2, 62, 62, 2, 2, 381, 381, 381, 2, 2, 5110, 365, 365, 5110, 2, 2, 1414477, 2828954, 1414477, 2828954, 1414477, 2, 2, 1720110, 49146, 573370, 573370, 49146, 1720110, 2, 2, 16931177, 50793531, 1638501, 118518239, 1638501, 50793531, 16931177, 2
Offset: 0

Views

Author

Peter Luschny, Aug 11 2014

Keywords

Comments

Conjecture: A240978(n) divides T(n,k) for k in (1..n-1) and n>=2.

Examples

			             2
            2, 2
          2, 7, 2
        2, 62, 62, 2
    2, 381, 381, 381, 2
2, 5110, 365, 365, 5110, 2
		

Crossrefs

Cf. A246051 (numerators), A240978, A246053.

Programs

  • Maple
    h := x -> Zeta(2*x)*(4^x-2);
    A246052 := (n, k) -> denom(h(n-k)*h(k)/h(n));
    seq(print(seq(A246052(n, k), k=0..n)), n=0..8);
  • Sage
    h = lambda n: zeta(2*n)*(4^n-2)
    A246052 = lambda n, k: (h(n-k)*h(k)/h(n)).denominator()
    for n in range(8): [A246052(n, k) for k in (0..n)]

A242035 The numerator of h(n-k)*h(k)/h(n) where h(x) = zeta(2*x)*(4^x-2) and k = floor(n/2).

Original entry on oeis.org

1, 1, 5, 49, 343, 341, 1374230, 562991, 117628797, 5722552563, 274111769750, 767094923209, 29727071936873882, 860722536439030, 65045120396044500, 1850097086237495825037, 16555136396811464938269, 962684710425111932621, 29167062964422333027973288250
Offset: 0

Views

Author

Peter Luschny, Aug 12 2014

Keywords

Crossrefs

Cf. A246053 (denominator), A242050, A246051, A246052.

Programs

  • Sage
    h = lambda x: zeta(2*x)*(4^x-2)
    A242035 = lambda n: Integer((h((n+1)//2)*h(n//2)/h(n)).numerator())
    [A242035(n) for n in range(19)]

A242032 A sequence related to lower bounds for the number of distinct differentiable structures on spheres of the form S^(4*k-1).

Original entry on oeis.org

2, 2, 7, 31, 127, 73, 1414477, 8191, 16931177, 5749691557, 91546277357, 3324754717, 1982765468311237, 22076500342261, 65053034220152267, 925118910976041358111, 16555640865486520478399, 8089941578146657681, 29167285342563717499865628061
Offset: 0

Views

Author

Peter Luschny, Aug 12 2014

Keywords

Comments

Definition: Divide the prime factorization of an integer M into two parts: L(k, M) = product{p^v(M) the highest power of a prime p which divides M and p < k} and H(k, M) = M / L(k, M). Then a(n) = H(2*floor((n+1)/2), A246053(n)).
For some n the a(n) are lower bounds for the number of distinct differentiable structures on spheres. Compare theorem 2 of the Milnor 1959 paper which asserts that a(2) and a(4) through a(8) are lower bounds for the spheres S^(4*k-1) for k = 2 and 4,..,8.
Let b(n) = numerator(B(2*n)/(2*n)!*(4^n-2)*(-1)^(n-1)), B(n) Bernoulli number. Apart from n=0 and n=1 the first n such that a(n) != b(n) is n = 1437. Thus in the range [2..1436] the a(n) are the numerators in the Taylor series for x*cosec(x), A036280.

Examples

			a( 0) = 2
a( 1) = 2
a( 2) = 7
a( 3) = 31
a( 4) = 127
a( 5) = 73
a( 6) = 23 * 89 * 691
a( 7) = 8191
a( 8) = 31 * 151 * 3617
a( 9) = 43867 * 131071
a(10) = 283 * 617 * 524287
a(11) = 127 * 131 * 337 * 593
a(12) = 47 * 103 * 178481 * 2294797
a(13) = 31 * 601 * 1801 * 657931
		

Crossrefs

Programs

  • Mathematica
    h[x_] := Zeta[2x] (4^x-2);
    a[n_] := Module[{M, k, p}, M = Denominator[h[Quotient[n+1, 2]] h[Quotient[ n, 2]]/h[n]]; k = 2 Quotient[n+1, 2]; p = 2; While[p < k, While[ Divisible[M, p], M = M/p]; p = NextPrime[p]]; M];
    Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 12 2019, from Sage code *)
  • Sage
    def A242032(n):
        h = lambda x: zeta(2*x)*(4^x-2)
        M = Integer((h((n+1)//2)*h(n//2)/h(n)).denominator())
        k = 2*((n+1)//2)
        P = Primes()
        p = P.first()
        while p < k:
            while p.divides(M):
                M /= p
            p = P.next(p)
        return M
    [A242032(n) for n in (0..30)]
Showing 1-5 of 5 results.