OFFSET
0,2
COMMENTS
The standard offset is changed to zero to simplify formulas related to the a(n).
The sequence of fractions A145911(k)/(k+1) is 0, 1/2, 1/3, 1/2, 2, 5/6, 1, 7/2, 4/9, 1/2, 5, 11/6, 2, 13/2, 7/3, 5/2, 8, 17/18, 1, 19/2, 10/3, 7/2, 11, 23/6, 4, 25/2, 13/9, 3/2, 14, 29/6, ....
Its numerators are A106619. Integer values appear at indices of the form 6*n and 4+6*n.
The sequence of denominators of the fractions appears to have a period of length 18.
a(n+18)-a(n) = 3*(a(n+6)-a(n)) = 3, 27, 9, 27, 9, 27, 3, 27, 9, ,... are multiples of 3, apparently with a period of length 6.
The recurrence a(n) = 2a(n-6)-a(n-12) shows that the sequence consists of 6 interleaved first-order polynomials: a(6*n)=n. a(1+6*n) = 2+9*n. a(2+6*n) = 1+3*n = A016777(n). a(3+6*n) = 5+9*n. a(4+6*n) = 2+3*n = A016789(n). a(5+6*n) = 8+9*n. - Paul Curtz, Feb 23 2011
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
FORMULA
a(1+2*n) = 2+3*n = A106619(4+6*n).
a(6*n) = n.
From R. J. Mathar, Feb 10 2011: (Start)
a(n)= +2*a(n-6) -a(n-12).
G.f.: x*(2+x +5*x^2 +2*x^3 +8*x^4 +x^5 +7*x ^6 +2*x^7 +4*x^8 +x^9 +x^10) / ( (x-1)^2*(1+x)^2*(1+x+x^2)^2*(x^2-x+1)^2 ). (End)
a(n) = A014682(n) if n is not a multiple of 6. - Paul Curtz, Feb 23 2011
MAPLE
MATHEMATICA
CoefficientList[Series[x*(2 + x + 5*x^2 + 2*x^3 + 8*x^4 + x^5 + 7*x^6 + 2*x^7 + 4*x^8 + x^9 + x^10)/((x - 1)^2*(1 + x)^2*(1 + x + x^2)^2*(x^2 - x + 1)^2), {x, 0, 50}], x] (* G. C. Greubel, Jul 11 2017 *)
PROG
(PARI) x='x+O('x^50); concat([0], Vec(x*(2 + x + 5*x^2 + 2*x^3 + 8*x^4 + x^5 + 7*x^6 + 2*x^7 + 4*x^8 + x^9 + x^10)/((x - 1)^2*(1 + x)^2*(1 + x + x^2)^2*(x^2 - x + 1)^2))) \\ G. C. Greubel, Jul 11 2017
CROSSREFS
KEYWORD
nonn,easy,less
AUTHOR
Paul Curtz, Feb 05 2011
STATUS
approved