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%I #12 Sep 14 2017 03:57:40
%S 2,3,6,3,3,5,3,5,4,6,4,5,4,3,8,2,3,6,3,8,6,3,8,6,2,6,8,3,4,4,4,6,4,3,
%T 8,8,2,3,7,3,8,12,3,4,8,2,8,8,3,4,8,4,8,9,4,10,6,2,5,8,2,8,6,4,6,5,3,
%U 8,4,3,8,9,4,6,7,3,8,4,4,8,5,4,8,6,5,6
%N a(n) = number of terms in row n of A214850.
%C Number of multiplicative finite groups G(p) with elements {T(2n+1,k)/pZ} where T(2n+1,k) is the reduced trajectory of the Collatz problem whose elements are all odd and p <= A075684(n) + 1.
%H Michel Lagneau, <a href="/A224504/b224504.txt">Table of n, a(n) for n = 1..500</a>
%e a(18) = 6 because there exist 6 finite groups given by row 18 of A214850 where p = 2, 4, 6, 8, 12 and 18. The Collatz trajectory of the number 2*18 + 1 = 37 with odd numbers is T(37,k) = {37, 7, 11, 17, 13, 5, 1}, and the 6 groups G(p) are:
%e G(2) = {T(37,k)/2Z} = {1}
%e G(4) = {T(37,k)/4Z} = {1, 3}
%e G(6) = {T(37,k)/6Z} = {1, 5}
%e G(8) = {T(37,k)/8Z} = {1, 3, 5, 7}
%e G(12) = {T(37,k)/12Z} = {1, 5, 7, 11}
%e G(18) = {T(37,k)/18Z} = {1, 5, 7, 11, 13, 17}
%e G(18) is a cyclic group because the element 5 (or 11) generates the group:
%e 5^1 == 5, 5^2 == 7, 5^3 == 17, 5^4 == 13, 5^5 == 11, 5^6 == 1 (mod 18).
%e G(8) is not a cyclic group.
%e a(170) = 32 because there exist 32 finite groups with two elements given by row 170 of A214850 where p = 2, 4, 6, 8, 10, 12, 18, 20, 24, 30, 34, 36, 38, 40, 60, 68, 72, 76, 90, 102, 114, 120, 136, 152, 170, 180, 190, 204, 228, 306, 340, 342. The Collatz trajectory of the number 2*170 + 1 = 341 with odd numbers is T(341,k) = {1, 341}.
%Y Cf. A075680, A075684, A214850.
%K nonn
%O 1,1
%A _Michel Lagneau_, Apr 08 2013
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