A073734 GCD of consecutive members of the EKG sequence A064413.
1, 2, 2, 3, 3, 3, 4, 2, 5, 5, 3, 2, 7, 7, 3, 8, 4, 2, 11, 11, 3, 3, 5, 5, 7, 2, 13, 13, 3, 4, 2, 17, 17, 3, 2, 19, 19, 3, 5, 4, 2, 23, 23, 3, 2, 2, 2, 2, 7, 7, 3, 5, 5, 5, 2, 29, 29, 3, 2, 31, 31, 3, 8, 4, 2, 37, 37, 3, 3, 2, 4, 2, 41, 41, 3, 3, 7, 11, 2, 43, 43, 3, 5, 5, 5, 4, 2, 47, 47, 3, 2, 7
Offset: 2
Examples
a(8) = 4 because gcd(A064413(7), A064413(8)) = gcd(12, 8) = 4. From _Michael De Vlieger_, Sep 27 2023: (Start) Let b(n) = A064413(n): a(11068) = 12 since gcd(b(11067), b(11068)) = gcd(11484, 11472) = 12, a(58836) = 18 since gcd(b(58835), b(58836)) = gcd(60786, 60678) = 18. (End)
Links
- Jianing Song, Table of n, a(n) for n = 2..10000 (based on the data of A064413; terms n = 2..1000 from T. D. Noe)
- J. C. Lagarias, E. M. Rains and N. J. A. Sloane, The EKG sequence, Exper. Math. 11 (2002), 437-446; arXiv:math/0204011 [math.NT], 2002.
- Index entries for sequences related to EKG sequence
Programs
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Haskell
a073734 n = a073734_list !! (n-2) a073734_list = zipWith gcd a064413_list $ tail a064413_list -- Reinhard Zumkeller, Sep 17 2001
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Mathematica
t = {1, 2}; Join[{1}, Table[k = 3; While[MemberQ[t, k] || (y = GCD[Last[t], k]) == 1, k++];AppendTo[t, k]; y, {91}]] (* Jayanta Basu, Jul 09 2013 *)
Comments