Of Different Prime-Factorization Exponents

[Dan Dima]

 n = p_1^{u_1} * p_2^{u_2} * ... * p_k^{u_k}
 Consider for each (2^k-2) subsets A_i of {1,2,...,k}, && 1 \leq {card A_i}
\leq k-1:
- all (2^{k-{card A_i}}-1) subsets B_i,j of {1,2,...,k} \ A_i, && 1 \leq
{card B_i,j} \leq k-{card A_i}
- A_i && B_i,j -disjoint and their reunion is a subset of {1,2,...,k}
- for all (3^k - 2^{k+1} + 1) such possible pairs (A_i,B_i,j) we have to
count a floor term:

a(n) = (p_1^{u_1} - 1) * (p_2^{u_2} - 1) * ... * (p_k^{u_k} - 1) +
+ \sum_{all pairs (A_i,B_i,j)}  (-1)^{k-{card A_i}-{card B_i,j}}  floor(
(\prod_{all i in A_i} p_i^{u_i} - 1) / (\prod_{all j in B_i,j} p_j) ).

for ex.if k=3, n=p_1^{u_1} * p_2^{u_2} * p_3^{u_3}:
a(n) = (p_1^{u_1} - 1) * (p_2^{u_2} - 1) * (p_3^{u_3} - 1) -
+ floor( (p_1^{u_1} * p_2^{u_2} - 1) / p_3 )
+ floor( (p_2^{u_2} * p_3^{u_3} - 1) / p_1 )
+ floor( (p_3^{u_3} * p_1^{u_1} - 1) / p_2 )
- floor( (p_1^{u_1} - 1) / p_2 ) - floor( (p_1^{u_1} - 1) / p_3 ) + floor(
(p_1^{u_1} - 1) / (p_2 * p_3) )
- floor( (p_2^{u_2} - 1) / p_3 ) - floor( (p_2^{u_2} - 1) / p_1 ) + floor(
(p_2^{u_2} - 1) / (p_3 * p_1) )
- floor( (p_3^{u_3} - 1) / p_1 ) - floor( (p_3^{u_3} - 1) / p_2 ) + floor(
(p_3^{u_3} - 1) / (p_1 * p_2) ).

I hope it's OK now...

Regards,
 Dan Dima
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20060319/f6efbaa4/attachment-0001.htm>