|
| |
|
|
A168066
|
|
If n = Product p(k)^e(k) then a(n) = {Product (p(k)+1)^e(k) - Product (p(k)-1)^e(k)}/2, a(1) = 0.
|
|
4
|
|
|
|
0, 1, 1, 4, 1, 5, 1, 13, 6, 7, 1, 17, 1, 9, 8, 40, 1, 22, 1, 25, 10, 13, 1, 53, 10, 15, 28, 33, 1, 32, 1, 121, 14, 19, 12, 70, 1, 21, 16, 79, 1, 42, 1, 49, 40, 25, 1, 161, 14, 46, 20, 57, 1, 92, 16, 105, 22, 31, 1, 104, 1, 33, 52, 364, 18, 62, 1, 73, 26, 60, 1, 214, 1, 39, 56, 81, 18, 72
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,4
|
|
|
COMMENTS
|
a(n) = 0 iff n is 1;
a(n) = 1 iff n prime p;
a(n) = p+q iff n is biprime, i.e. n = pq, p <= q primes;
a(n) = (pq+pr+qr)+1 iff n is triprime, i.e. n = pqr, p <= q <= r primes;
a(n) = (pqr+pqs+prs+qrs)+(p+q+r+s) iff n is quadprime, i.e. n = pqrs, p <= q <= r <= s primes;
...
|
|
|
LINKS
|
Daniel Forgues, Table of n, a(n) for n=1..100000
|
|
|
FORMULA
|
a(n) = {A003959(n) - A003958(n)}/2
|
|
|
PROG
|
(PARI) a(n) = {f = factor(n); return ((prod(k=1, #f~, (f[k, 1]+1)^f[k, 2]) - prod(k=1, #f~, (f[k, 1]-1)^f[k, 2]))/2); } \\ Michel Marcus, Jun 13 2013
|
|
|
CROSSREFS
|
Cf. A003958, A003959, A168065.
Sequence in context: A302055 A086300 A028271 * A029666 A269593 A194512
Adjacent sequences: A168063 A168064 A168065 * A168067 A168068 A168069
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Daniel Forgues, Nov 18 2009
|
|
|
STATUS
|
approved
|
| |
|
|
