A182471 If n = Product (p_j^k_j), j=1 to m, then a(n) = Floor( Sum (p_j + k_j) / 2m ) (a(1) = 0 by convention).

0, 1, 2, 2, 3, 1, 4, 2, 2, 2, 6, 2, 7, 2, 2, 3, 9, 2, 10, 2, 3, 3, 12, 2, 3, 4, 3, 3, 15, 2, 16, 3, 4, 5, 3, 2, 19, 5, 4, 2, 21, 2, 22, 4, 2, 6, 24, 2, 4, 2, 5, 4, 27, 2, 4, 3, 6, 8, 30, 2, 31, 8, 3, 4, 5, 3, 34, 5, 7, 2, 36, 2, 37, 10, 2, 6, 5, 3, 40, 3, 3

Offset

1,3

Comments

a(n) = k (a selected constant) for only a finite number of values of n. The maximal value of n for a desired value of a(n) grows rapidly.

Thus for a(n)=2, n=2^1*3^1*5^5 is maximal.

For a(n)=3, n=2^1*3^1*5^1*7^11 is maximal.

Links

Table of n, a(n) for n=1..81.

Example

21 = (3^1)(7^1), with 2 distinct prime factors. Floor(((3+1)+(7+1))/(2*2)) = 3, thus a(21) = 3.

Prog

(PARI) {for(k=2, 175,
M=factor(k); smt = 0; h = 0;
for(i=1, matsize(M)[1], for(j=1, matsize(M)[2], smt=smt+M[i, j]; h++));
print1( floor(smt/h), ", "))} \\\ Douglas Latimer, Apr 30 2012

Crossrefs

A008475

Sequence in context: A072078 A322316 A260439 * A078378 A141197 A035207

Adjacent sequences: A182468 A182469 A182470 * A182472 A182473 A182474

Keyword

nonn

Author

Douglas Latimer, May 01 2012

Status

approved