|
%I #11 Jan 19 2016 10:58:01
%S 1,2,3,2,5,1,7,2,3,2,11,1,13,2,3,2,17,1,19,2,3,2,23,1,25,2,3,2,29,1,
%T 31,2,3,2,35,1,37,2,3,2,41,1,43,2,3,2,47,1,49,2,3,2,53,1,55,2,3,2,59,
%U 1,61,2,3,2,65,1,67,2,3,2
%N a(n) = (n+1) / A189733(n).
%C A189733(n) is the denominator of an autosequence of the first kind (the main diagonal is A000004).
%F a(2n+1) = A130196(n+1).
%F A052901(n+2) = period 3: 2, 3, 2 is at rank A047245(n+1) = 1, 2, 3, 7, 8, 9, ... .
%F Conjectures from _Colin Barker_, Jan 10 2016: (Start)
%F a(n) = 2*a(n-6) - a(n-12) for n>11.
%F G.f.: (1+2*x+3*x^2+2*x^3+5*x^4+x^5+5*x^6-2*x^7-3*x^8-2*x^9+x^10-x^11) / ((1-x)^2*(1+x)^2*(1-x+x^2)^2*(1+x+x^2)^2).
%F (End)
%F a(3n) + a(3n+1) + a(3n+2) = A047238(n+3).
%t CoefficientList[Series[(1 + 2 x + 3 x^2 + 2 x^3 + 5 x^4 + x^5 + 5 x^6 - 2 x^7 - 3 x^8 - 2 x^9 + x^10 - x^11)/((1 - x)^2 (1 + x)^2 (1 - x + x^2)^2 (1 + x + x^2)^2), {x, 0, 69}], x] (* or *)
%t b[m_, n_] := b[m, n] = Which[m == n, 0, n == m + 1, (-1)^(n + 1)/n, n > m, b[m, n - 1] + b[m + 1, n - 1], n < m, b[m - 1, n + 1] - b[m - 1, n]]; Table[(n + 1)/Denominator@ b[0, n], {n, 0, 69}] (* _Michael De Vlieger_, Jan 15 2016, _Jean-François Alcover_ at A189733 *)
%Y Cf. A000004, A047238, A047245, A052901, A130196, A189733.
%K nonn
%O 0,2
%A _Paul Curtz_, Jan 10 2016
|
