|
|
A093879
|
|
First differences of A004001.
|
|
16
|
|
|
0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
All the terms are 0 or 1: it is easy to show that if {b(n)} = A004001, b(n)>=b(n-1) and b(n)<n, therefore the first differences form an infinite binary word. - Benoit Cloitre, Jun 05 2004
|
|
LINKS
|
|
|
FORMULA
|
Other identities. For all n >= 1:
a(n) = 1 if and only if A051135(n+1) = 1.
(End)
|
|
MATHEMATICA
|
a[1] = a[2] = 1; a[n_] := a[n] = a[a[n - 1]] + a[n - a[n - 1]]; t = Table[a[n], {n, 110}]; Drop[t, 1] - Drop[t, -1] (* Robert G. Wilson v, May 28 2004 *)
|
|
PROG
|
(PARI) {m=106; v=vector(m, j, 1); for(n=3, m, a=v[v[n-1]]+v[n-v[n-1]]; v[n]=a); for(n=2, m, print1(v[n]-v[n-1], ", "))}
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|