%I #46 Jul 18 2024 08:52:00
%S 7,8,5,5,3,4,4,6,9,7,8,8,7,8,5,5,3,4,4,6,9,7,8,8,7,8,5,5,3,4,4,6,9,7,
%T 8,8,7,8,5,5,3,4,4,6,9,7,8,8,7,8,5,5,3,4,4,6,9,7,8,8,7,8,5,5,3,4,4,6,
%U 9,7,8,8,7,8,5,5,3,4,4,6,9,7,8,8
%N Number of letters in English words for months of year.
%C Period 12: repeat [7, 8, 5, 5, 3, 4, 4, 6, 9, 7, 8, 8].
%C According to the definition this should strictly speaking be finite: there is no 13th month of the year. But for several reasons we prefer to see this as an infinite periodic sequence. - _M. F. Hasler_, Mar 05 2018
%D GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 60.
%H Time and Date, <a href="http://www.timeanddate.com/calendar/">A Calendar website</a>.
%H <a href="/index/Ca#calendar">Index entries for sequences related to calendars</a>.
%H <a href="/index/Periodic#12">Index to 12-periodic sequences</a>.
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,0,0,1).
%F From _Elmo R. Oliveira_, Jul 18 2024: (Start)
%F G.f.: x*(7 + 8*x + 5*x^2 + 5*x^3 + 3*x^4 + 4*x^5 + 4*x^6 + 6*x^7 + 9*x^8 + 7*x^9 + 8*x^10 + 8*x^11)/(1 - x^12).
%F a(n) = a(n-12) for n > 12. (End)
%e a(1) = 7 because January has 7 letters.
%t PadRight[{}, 72, {7, 8, 5, 5, 3, 4, 4, 6, 9, 7, 8, 8}] (* or *)
%t Array[StringLength@ DateString[DateObject[{0, Mod[#, 12] + 1, 1, 0, 0, 0}, "Month"], {"MonthName"}] &, 72, 0] (* _Michael De Vlieger_, Feb 25 2018 *)
%o (PARI) A031139(n)=digits(879644355878)[12-n%12] \\ _M. F. Hasler_, Mar 05 2018
%Y Cf. A089746, A161376, A182495.
%K nonn,word,easy
%O 1,1
%A Dmitri Papichev (Dmitri.Papichev(AT)iname.com)