A267068 a(n) = (n+1) / A189733(n).

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, 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, 1, 61, 2, 3, 2, 65, 1, 67, 2, 3, 2

Offset: 0

Author

Paul Curtz, Jan 10 2016

Keywords

• nonn

Comments

A189733(n) is the denominator of an autosequence of the first kind (the main diagonal is A000004).

Crossrefs

Cf. A000004, A047238, A047245, A052901, A130196, A189733.

Programs

Mathematica

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 *)
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 *)

Formula

a(2n+1) = A130196(n+1).

A052901(n+2) = period 3: 2, 3, 2 is at rank A047245(n+1) = 1, 2, 3, 7, 8, 9, ... .

Conjectures from Colin Barker, Jan 10 2016: (Start)

a(n) = 2*a(n-6) - a(n-12) for n>11.

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).

(End)

a(3n) + a(3n+1) + a(3n+2) = A047238(n+3).