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%I #14 Apr 29 2023 20:46:56
%S 1,3,3,3,19,51,115,115,115,627,627,2675,2675,2675,2675,35443,35443,
%T 166515,166515,166515,1215091,3312243,3312243,3312243,3312243,36866675
%N 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).
%C a(n+1) - a(n) = 2^n or 0; see A127015.
%C In the 2-adic integers, lim_{n->oo} a(n) = 11001110010100010100110001...; see A127015.
%D N. Koblitz, p-adic Numbers, p-adic Analysis and Zeta-Functions, 2nd ed., Springer, New York, 1996.
%D 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.
%H J. Sondow and K. Schalm, <a href="http://arxiv.org/abs/0709.0671">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</a>.
%F A(a(n)) = A138761(n) = Sum_{k=0..a(n)} a(n)!/k! for n > 0. - _Jonathan Sondow_, Jun 12 2009
%e 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.
%t a522[n_] := E Gamma[n + 1, 1];
%t 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]]];
%t Table[a[n], {n, 1, 17}] (* _Jean-François Alcover_, Feb 20 2019 *)
%Y Cf. A000522, A127015, A138761.
%K nonn
%O 1,2
%A Kyle Schalm (kschalm(AT)math.utexas.edu), Jan 07 2007
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