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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

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Last modified August 18 04:50 EDT 2019. Contains 326072 sequences. (Running on oeis4.)