A127014 a(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).

1, 3, 3, 3, 19, 51, 115, 115, 115, 627, 627, 2675, 2675, 2675, 2675, 35443, 35443, 166515, 166515, 166515, 1215091, 3312243, 3312243, 3312243, 3312243, 36866675

Offset

1,2

Comments

a(n+1) - a(n) = 2^n or 0; see A127015.

In the 2-adic integers, lim_{n->oo} a(n) = 11001110010100010100110001...; see A127015.

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.

Links

Table of n, a(n) for n=1..26.

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), II.

Formula

A(a(n)) = A138761(n) = Sum_{k=0..a(n)} a(n)!/k! for n > 0. - Jonathan Sondow, Jun 12 2009

Example

A(0) = 1, A(1) = 2, A(2) = 5 and A(3) = 16 = 2^4, so a(1) = 1 and a(2) = a(3) = a(4) = 3. Also, A(19) = 330665665962404000 is the first A(k) divisible by 2^5, so a(5) = 19.

Mathematica

a522[n_] := E Gamma[n + 1, 1];
a[1] = 1; a[n_] := a[n] = For[k = a[n - 1], True, k++, If[Mod[a522[k], 2^n] == 0, Print[n, " ", k]; Return[k]]];
Table[a[n], {n, 1, 17}] (* Jean-François Alcover, Feb 20 2019 *)

Crossrefs

Cf. A000522, A127015, A138761.

Sequence in context: A239125 A341752 A325892 * A377278 A073748 A289118

Adjacent sequences: A127011 A127012 A127013 * A127015 A127016 A127017

Keyword

nonn

Author

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

Status

approved