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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). 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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

In the 2-adic integers, lim_{n->infty} 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. [From 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-Fran├žois Alcover, Feb 20 2019 *)

CROSSREFS

Cf. A000522, A127015, A138761.

Sequence in context: A229934 A239125 A325892 * A073748 A289118 A131445

Adjacent sequences:  A127011 A127012 A127013 * A127015 A127016 A127017

KEYWORD

nonn

AUTHOR

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

STATUS

approved

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Last modified November 22 13:47 EST 2019. Contains 329393 sequences. (Running on oeis4.)