|
|
A074313
|
|
a(n) = the maximal length of a sequence of primes {s_1 = prime(n), s_2 = f(s1), s_3 = f(s_2), ....} formed by repeated application of f(m) = Floor(m/2) on prime(n).
|
|
2
|
|
|
1, 1, 2, 2, 3, 1, 1, 1, 4, 1, 1, 1, 1, 1, 5, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
The smallest value of n such that a(n) = 6 is n = 417.
|
|
LINKS
|
|
|
EXAMPLE
|
To compute a(9): prime(9) = 23, f(23) = 11, f(11) = 5, f(5) = 2, f(2) = 1, where f(m) = Floor(m/2). Hence the sequence (of length 4) 23, 11, 5, 2 is the sequence of primes of maximal length formed by repeated application of f to prime(9) = 23. Therefore a(9) = 4.
|
|
MATHEMATICA
|
f[n_] := Module[{i}, i = 0; m = n; While[PrimeQ[m], m = Floor[m/2]; i++ ]; i]; Table[f[Prime[i]], {i, 1, 100}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|