OFFSET
1,2
COMMENTS
This is based on a naive multiplication of A033172 with a fixed number of seconds per year, 24*3600 = 86400. It ignores that leap years are not regularly occurring after 4 years (but after 400 years, note the formula that relates a(n+4) to a(n) and also the simple Mma implementation), ignores leap seconds, and any other influences that align the slowing down of the Earth rotation in an astronomical fixed coordinate system measured relative to atomic clocks. In summary, the use of "year" in the definition is not commensurate with years in standard astronomical or earth observational terms. - R. J. Mathar, Aug 21 2012
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Wikipedia, Unix time
Wikipedia, Leap second
Wikipedia, Revised Julian Calendar
FORMULA
From Alexander R. Povolotsky, Aug 20 2012: (Start)
a(n) = 10800*(2922*n + (-1)^n + (1+i)*(-i)^n + (1-i)*i^n - 2923).
a(n+4) = a(n) + 126230400.
G.f.: 86400*(365*x +365*x^2 +366*x^3 +365*x^4)/((1-x)^2*(1+x+x^2+x^3)). (End)
MATHEMATICA
lst = {}; t = 86400; Do[e = t*(365*(n - 1) + Ceiling[n/4]); If[! Mod[n, 4] == 0, e = e - t]; AppendTo[lst, e], {n, 25}]; lst (* Arkadiusz Wesolowski, Aug 20 2012 *)
CoefficientList[Series[86400*(365*x + 365*x^2 + 366*x^3 + 365*x^4)/((x - 1)^2*(1 +x +x^2 +x^3)), {x, 0, 50}], x] (* G. C. Greubel, Feb 26 2017 *)
PROG
(PARI) x='x+O('x^50); Vec(86400*(365*x +365*x^2 +366*x^3 +365*x^4)/((1-x)^2*(1+x+x^2+x^3))) \\ G. C. Greubel, Feb 26 2017
CROSSREFS
KEYWORD
easy,nonn,less
AUTHOR
Kyle Stern, Aug 09 2012
EXTENSIONS
a(11)-a(25) from Arkadiusz Wesolowski, Aug 20 2012
STATUS
approved