A124162 Number of quadruples [i,j,k,l] with all entries between 1 and n such that gcd(i,j) = gcd(k,l).

1, 10, 51, 132, 373, 590, 1287, 1992, 3209, 4402, 7323, 9004, 13949, 17558, 22159, 27696, 38897, 45018, 60931, 70100, 83653, 98142, 126391, 139928, 170489, 195170, 228139, 254428, 311661, 334886, 405087, 451008, 507329, 563530, 635795, 680772, 799861

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

Sequence in context: A370737 A264044 A143855 * A077044 A069038 A213563

Adjacent sequences: A124159 A124160 A124161 * A124163 A124164 A124165

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