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Revision History for A145384

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The number of terms of A050791 bracketed by successive terms of A141326
(history; published version)
#4 by N. J. A. Sloane at Thu Nov 11 07:34:06 EST 2010
LINKS

Lewis Mammel, <a href="/A145384/b145384.txt">Table of n, a(n) for n = 1..122</a>

KEYWORD

nonn,new

nonn

#3 by N. J. A. Sloane at Sun Jul 11 03:00:00 EDT 2010
EXAMPLE

0 = number of terms of A050791 preceding the 1st first term of A141326

1 = number of terms of A050791 between the 1st first and 2nd terms of A141326

KEYWORD

nonn,new

nonn

#2 by N. J. A. Sloane at Fri Feb 27 03:00:00 EST 2009
COMMENTS

A141326 is a simply generated subsequence of A050791, and by observation it forms a natural measure of the parent sequence. The first several hundred terms of the parent sequence not belonging to A141326 are bracketed into groups with a small integral number of terms ( including 0 ) by the successive terms of the subsequence, A141326.

a(107),a(108) are the first occurrence of 2 consecutive 0's, and a(119),a(120),a(121) are the first occurrence of 3 consecutive 0's. This leads to the following conjecture:

LINKS

Lewis Mammel, <a href="http://www.research.att.com/~njas/sequences/b145384.txt">Table of n, a(n) for n = 1..122</a>

KEYWORD

nonn,new

nonn

#1 by N. J. A. Sloane at Fri Jan 09 03:00:00 EST 2009
NAME

The number of terms of A050791 bracketed by successive terms of A141326

DATA

0, 1, 2, 3, 2, 3, 2, 2, 6, 6, 0, 3, 1, 3, 2, 3, 2, 4, 4, 3, 0, 3, 5, 0, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 5, 1, 1, 4, 2, 0, 1, 3, 1, 3, 3, 2, 2, 2, 4, 2, 1, 2, 4, 2, 0, 1, 2, 3, 1, 1, 1, 3, 0, 3, 1, 0, 3, 1, 1, 4, 2, 2, 1, 3, 3, 1, 2, 0, 3, 2, 5, 1, 1, 3, 6, 2, 4, 1, 0, 5, 2, 2, 2, 2, 3, 2, 3, 3, 0, 1

OFFSET

1,3

COMMENTS

A141326 is a simply generated subsequence of A050791, and by observation it forms a natural measure of the parent sequence. The first several hundred terms of the parent sequence not belonging to A141326 are bracketed into groups with a small integral number of terms ( including 0 ) by the successive terms of the subsequence, A141326.

a(107),a(108) are the first occurrence of 2 consecutive 0's, and a(119),a(120),a(121) are the first occurrence of 3 consecutive 0's. This leads to the following conjecture:

<a(n)> -> 0 as n ->inf

where <a(n)> = ( sum m=1,n of a(m) )/n

LINKS

Lewis Mammel, <a href="http://www.research.att.com/~njas/sequences/b145384.txt">Table of n, a(n) for n = 1..122</a>

FORMULA

a(1) = A145383(1) - 1

a(n) = A145383(n) - A145383(n-1) - 1 ; n>1

EXAMPLE

0 = number of terms of A050791 preceding the 1st term of A141326

1 = number of terms of A050791 between the 1st and 2nd terms of A141326

2 = number of terms of A050791 between the 2nd and 3rd terms of A141326

CROSSREFS
KEYWORD

nonn

AUTHOR

Lewis Mammel (l_mammel(AT)att.net), Oct 10 2008

STATUS

approved