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 1-2 of 2 results.

A127015 Digits of the 2-adic integer lim_{n->oo} A127014(n).

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1
Offset: 1

Views

Author

Kyle Schalm (kschalm(AT)math.utexas.edu), Jan 07 2007

Keywords

Comments

A127014(n) = smallest k such that A(k) == 0 (mod 2^n), where A(0) = 1 and A(k) = k*A(k-1) + 1 = A000522(k).

Examples

			In 2-adic notation (aka reverse binary) A127014(26) = 11001110010100010100110001.
		

References

  • N. Koblitz, p-adic Numbers, p-adic Analysis and Zeta-Functions, 2nd ed., Springer, New York, 1996.
  • J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.

Crossrefs

A138761 a(n) is the smallest member of A000522 divisible by 2^n, where A000522(m) = total number of arrangements of a set with m elements.

Original entry on oeis.org

1, 2, 16, 16, 16, 330665665962404000, 4216377920843140187197325631474390438452208808916276571342090223552
Offset: 0

Views

Author

Jonathan Sondow, Apr 01 2008

Keywords

Comments

a(n) < A000522(2^n) for n > 0; see Sondow and Schalm, Proposition A.13 part (ii).

Examples

			a(5) = A000522(19) = 330665665962404000 because that is the smallest member of A000522 divisible by 2^5.
		

References

  • J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.

Crossrefs

Programs

  • Mathematica
    a522[n_] := E Gamma[n + 1, 1];
    (* b = A127014 *)
    b[1] = 1; b[n_] := b[n] = For[k = b[n - 1], True, k++, If[Mod[a522[k], 2^n] == 0, Return[k]]];
    a[0] = 1; a[n_] := a522[b[n]];
    Table[a[n], {n, 0, 6}] (* Jean-François Alcover, Feb 20 2019 *)

Formula

a(n) = A000522(A127014(n)) = Sum_{k=0..A127014(n)} A127014(n)!/k! for n > 0.
Showing 1-2 of 2 results.