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%I #18 Jun 22 2022 14:45:15
%S 1,1,1,1,2,2,3,3,4,5,6,6,7,8,9,10,11,11,12,13,14,15,16,17,18,19,20,21,
%T 22,23,24,26,27,28,29,30,31,32,33,35,36,37,38,39,40,42,43,44,45,47,48,
%U 49,50,52,53,54,55,57,58,59,60,62,63,64,66,67,68,70,71,72,74,75,76,78
%N Number of digits of Bell number A000110(n).
%C The positive integers which are in the complement to this sequence are: 25, 34, 41, 46, 51, 56, 61, 65, 69, 73, 77, 80, 84, 88, 91, 94, 98, 101, ... because there is no Bell number with 25 digits (B(30) = 846749014511809332450147 has 24 digits, B(31) = 10293358946226376485095653 has 26 digits).
%C Since a(n) >> n log n, there are infinitely many numbers (indeed, almost all positive integers) in the complement of this sequence. [_Charles R Greathouse IV_, Aug 10 2011]
%H Charles R Greathouse IV, <a href="/A113865/b113865.txt">Table of n, a(n) for n = 0..10000</a>
%H John Sokol, <a href="http://www.dnull.com/bells/bell1000.html">The First 1000 Bell Numbers</a>.
%F a(n) = ceiling(log_10 A000110(n)).
%F a(n) ~ nk log n with k = 1/log 10. More specifically, a(n) = (n log n + n log log n - n + n/W(n) + log n - 0.5 log W(n) - 1)/log 10 + o(1), where W is Lambert's W function W(x)*exp(W(x)) = x. [_Charles R Greathouse IV_, Aug 11 2011]
%e a(0) = 1 because Bell(0) = 1, which has one digit.
%e a(1) = 1 because Bell(1) = 1, which has one digit.
%e a(2) = 1 because Bell(2) = 2, which has one digit.
%e a(3) = 1 because Bell(3) = 5, which has one digit.
%e a(4) = 2 because Bell(4) = 15, which has two digits.
%p seq(length(bell(n)), n = 0 .. 73); # _Zerinvary Lajos_, Aug 07 2007
%o (Python)
%o from sympy import bell
%o def A113865(n): return len(str(bell(n))) # _Chai Wah Wu_, Jun 22 2022
%Y A113015(n) = a(10^n).
%Y Cf. A000110.
%K base,nonn
%O 0,5
%A _Jonathan Vos Post_, Jan 25 2006
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