Search: id:A011784
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%I A011784
%S A011784 1,2,2,3,4,7,14,42,213,2837,175450,139759600,6837625106787,
%T A011784 266437144916648607844,508009471379488821444261986503540,
%U A011784 37745517525533091954736701257541238885239740313139682,5347426383812697233786139576220450142250373277499130252\
554080838158299886992660750432
%N A011784 Levine's sequence. First construct a triangle as follows. Row 1 is {1,
1}; if row n is {r_1, ..., r_k} then row n+1 consists of {r_k 1's,
r_{k-1} 2's, r_{k-2} 3's, etc.}; sequence consists of the final elements
in each row.
%H A011784 Johan Claes, Table of n, a(n) for n = 1..18
%H A011784 N. J. A. Sloane,
My favorite integer sequences, in Sequences and their Applications
(Proceedings of SETA '98).
%F A011784 Additional remarks: The sequence is generated by this array, the final
term in each row forming the sequence:
%F A011784 1 1
%F A011784 1 2
%F A011784 1 1 2
%F A011784 1 1 2 3
%F A011784 1 1 1 2 2 3 4
%F A011784 1 1 1 1 2 2 2 3 3 4 4 5 6 7
%F A011784 1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 7 7 7 8 8 9
9 10 10 11 12 13 14
%F A011784 ...
%F A011784 where we start with the first row {1 1} and produce the rest of the array
recursively as follows:
%F A011784 Suppose line n is {a_1, ..., a_k}; then line n+1 contains a_k 1's, a_{k-1}
2's, etc.
%F A011784 So the fifth line contains three 1's, two 2's, one 3 and one 4.
%F A011784 The sequence is 1,2,2,3,4,7,14,42,213,2837,175450,...,
%F A011784 where the n-th term a(n) is the sum of the elements in row n-2
%F A011784 = the number of elements in row n-1
%F A011784 = the last element in row n
%F A011784 = the number of 1's in row n+1
%F A011784 = ...
%F A011784 If the n-th row is r_{n,i} then
%F A011784 Sum_{i=1..f(n+1)} (a(n+1) - i + 1)*r_{n,i} ) = a(n+3)
%F A011784 Let {a( )} be the sequence; s(i,j) = j-th partial sum of the i-th row,
%F A011784 L(i) is the length of that row and S(i) = its sum. Then
%F A011784 L(i+1) = a(i+2) = S(i) = s(i,a(i+1));
%F A011784 L(i+2) = SUM(s(i,j));
%F A011784 L(i+3) = SUM(s(i,j)*(1+s(i,j))/2) (Allan Wilks).
%F A011784 Eric Rains and Bjorn Poonen have shown (6/97) that the log of the n-th
term is asymptotic to constant times phi^n, where phi = golden number.
%F A011784 This follows from the inequalities S(n) <= a(n)L(n) and S(n+1) >= ([L(n+1)/
a(n)]+1) choose 2)*a(n).
%F A011784 The n-th term is approximately exp(a*phi^n)/I, where phi = golden number,
a = .05427 (last digit perhaps 6 or 8), I = .277 (last digit perhaps
6 or 8) (Colin Mallows).
%F A011784 a(n+2) = n-th row sum of A012257; e.g. 5-th row of A012257 is {1, 1,
1, 2, 2, 3, 4} and the sum of elements is 1+1+1+2+2+3+4=14=a(7) -
Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 06 2003
%e A011784 {1,1}, {1,2}, {1,1,2}, {1,1,2,3}, {1,1,1,2,2,3,4}, {1,1,1,1,2,2,2,3,3,
4,4,5,6,7}
%Y A011784 Cf. A012257.
%Y A011784 Sequence in context: A053638 A051920 A023105 this_sequence A032252 A112708
A147558
%Y A011784 Adjacent sequences: A011781 A011782 A011783 this_sequence A011785 A011786
A011787
%K A011784 nonn,nice
%O A011784 1,2
%A A011784 Lionel Levine (levine(AT)ultranet.com)
%E A011784 a(12) from C. L. Mallows (colinm(AT)research.avayalabs.com), a(13) from
N. J. A. Sloane (njas(AT)research.att.com), a(14) and a(15) from
Allan Wilks.
%E A011784 a(16) from Johan Claes (Johan.Claes(AT)luc.ac.be), Jun 09 2004.
%E A011784 a(17) (an 85-digit number) from Johan Claes (Johan.Claes(AT)luc.ac.be),
Jun 18 2004.
%E A011784 Edited by N. J. A. Sloane (njas(AT)research.att.com), Mar 08 2006
%E A011784 a(18) (a 137-digit number) from Johan Claes (Johan.Claes(AT)luc.ac.be),
Aug 19 2008
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