|
| |
|
|
A164878
|
|
Maximum number of copies of prime divisors of n, with repetition, required to express n as a sum; a(1) = 0 by convention.
|
|
3
|
|
|
|
0, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 3, 3, 2, 1, 3, 3, 2, 3, 4, 1, 3, 1, 4, 3, 2, 5, 4, 1, 2, 3, 4, 1, 4, 1, 4, 5, 2, 1, 5, 4, 5, 3, 4, 1, 6, 5, 5, 3, 2, 1, 5, 1, 2, 6, 6, 5, 5, 1, 4, 3, 5, 1, 6, 1, 2, 6, 4, 7, 5, 1, 7, 7, 2, 1, 6, 5, 2, 3, 6, 1, 8, 7, 4, 3, 2, 5, 8, 1, 7, 6, 8, 1, 5, 1, 6, 7
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,6
|
|
|
COMMENTS
|
For p prime, a(p^k) = ceiling(p^(k-1)/k); in particular, a(p) = 1. For p and q distinct primes, a(pq) = min(p,q).
a(n) >= n/sopfr(n), where sopfr is A001414; when the right hand side is an integer, this is an equality.
|
|
|
LINKS
|
Antti Karttunen, Table of n, a(n) for n = 1..12024
|
|
|
EXAMPLE
|
For n = 12, we have the representation 2+2+2+3+3. Since there are two 2's available, this can be done with 2 copies of the prime factors: 2+2'+3 from the first copy, and 2+3 from the second. Thus a(12) = 2.
|
|
|
PROG
|
(PARI) a(n)=local(fm, p); if(n<=1, return(0)); fm=factor(n); p=prod(i=1, matsize(fm)[1], (1+x^fm[i, 1])^fm[i, 2]); for(k=0, n, if(polcoeff(p^k, n)!=0, return(k)))
|
|
|
CROSSREFS
|
Cf. A164879, A164880, A027746.
Sequence in context: A263569 A105220 A083654 * A319695 A029428 A167866
Adjacent sequences: A164875 A164876 A164877 * A164879 A164880 A164881
|
|
|
KEYWORD
|
nonn,look
|
|
|
AUTHOR
|
Franklin T. Adams-Watters, Aug 29 2009
|
|
|
STATUS
|
approved
|
| |
|
|
