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 A124162 Number of quadruples [i,j,k,l] with all entries between 1 and n such that gcd(i,j) = gcd(k,l). 0
 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 (list; graph; refs; listen; history; text; internal format)
 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 FORMULA Let c(n) = A018805(n). Then a(n) = Sum_{ 1 <= d <= n} c(floor(n/d))^2. Comment from Gareth McCaughan, Dec 04 2006: 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 over d of # { p,q,r,s in [1,n] : (p,q) = (r,s) = d } = (1/n^4) sum over d of (# { p,q in [1,n] : (p,q) = d })^2 = (1/n^4) sum over d of (# { p,q in [1,n/d] : (p,q) = 1 })^2 ~ (1/n^4) sum over d of (6/pi^2 (n/d)^2)^2 = (36/pi^4) sum over d of 1/d^4 = (36/pi^4) (pi^4/90) = 2/5. Comment from Eugene Salamin (gene_salamin(AT)yahoo.com), Dec 04 2006: 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). CROSSREFS Sequence in context: A106041 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

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Last modified October 13 16:34 EDT 2019. Contains 327966 sequences. (Running on oeis4.)