OFFSET
1,2
COMMENTS
Based on a posting from Dan Asimov, Dec 03 2006, challenging people to find the limit a(n)/n^4.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
FORMULA
Let c(n) = A018805(n). Then a(n) = Sum_{d=1..n} c(floor(n/d))^2.
Comment from Gareth McCaughan, Dec 04 2006: (Start)
To find a(n)/n^2, note that (1/n^4) # { p,q,r,s in [1,n] : (p,q) = (r,s) }
= (1/n^4) Sum_{d} # { p,q,r,s in [1,n] : (p,q) = (r,s) = d }
= (1/n^4) Sum_{d} (# { p,q in [1,n] : (p,q) = d })^2
= (1/n^4) Sum_{d} (# { p,q in [1,n/d] : (p,q) = 1 })^2
~ (1/n^4) Sum_{d} ((6/Pi^2)*(n/d)^2)^2
= (36/Pi^4) Sum_{d} 1/d^4 = (36/Pi^4)*(Pi^4/90) = 2/5. (End)
Comment from Eugene Salamin (gene_salamin(AT)yahoo.com), Dec 04 2006: (Start)
More generally:
(i) The probability that gcd(i[1],...,i[n]) = gcd(j[1],...,j[n]) is zeta(2n)/zeta(n)^2.
(ii) The probability that k r-tuples of random integers all have the same gcd is zeta(kr)/zeta(r)^k.
(iii) The probability that the gcd of an r-tuple of random integers divides the gcd of an n-tuple of random integers is zeta(n+r)/zeta(n). (End)
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Dec 03 2006
EXTENSIONS
More terms from N. J. A. Sloane and several other people, Dec 04 2006
STATUS
approved