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