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A256496 a(n) = 15(n mod 2) + 10(n mod 3) + 6(n mod 5).

%I A256496 #45 Oct 31 2016 17:11:47
%S A256496 31,32,33,34,35,6,37,38,39,10,41,12,43,44,15,16,47,18,49,20,21,22,53,
%T A256496 24,25,26,27,28,59,0,31,32,33,34,35,6,37,38,39,10,41,12,43,44,15,16,
%U A256496 47,18,49,20,21,22,53,24,25,26,27,28,59,0,31,32,33,34,35,6
%N A256496 a(n) = 15(n mod 2) + 10(n mod 3) + 6(n mod 5).
%C A256496 After 0 it cycles again from 31.
%C A256496 This is the simplest example of a(n) = b*c*(n mod a) + a*c*(n mod b) + a*b*(n mod c) with a=2, b=3, c=5.
%C A256496 a(n) mod abc = n mod abc.
%C A256496 Other values for a,b,c require modification so that bc + ac + ab = abc + 1. For example, for a=3, b=5, c=7, a(n)= 2b*c*(n mod a) + a*c*(n mod b) + a*b*(n mod c) so that 2*5*7 + 3*7 + 3*5 = 3*5*7 = 70 + 21 + 15 = 106.
%C A256496 This expression (with a=3, b=5, c=7) (A255818) is found in the Kol Bo (Hebrew: כלבו), a book of religious Jewish law from the 13th to 14th centuries. It is given there as a method for calculating a person's age without anyone saying it explicitly.
%C A256496 a(n) = n for n = 6, 10, 12, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 47, 49, 53, 59.
%H A256496 Colin Barker, <a href="/A256496/b256496.txt">Table of n, a(n) for n = 1..1000</a>
%H A256496 Aaron Hacohen from Lunil?, <a href="http://www.hebrewbooks.org/pdfpager.aspx?req=15217&amp;st=&amp;pgnum=370">כלבו</a> (in Hebrew).
%H A256496 Menachem Mendel Schneerson, <a href="http://www.hebrewbooks.org/pdfpager.aspx?req=15860&amp;st=&amp;pgnum=280&amp;hilite=">אגרות קודש</a> (in Hebrew).
%H A256496 Wikipedia, <a href="http://en.wikipedia.org/wiki/Kol_Bo">Kol Bo</a>
%H A256496 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (-2,-2,-1,0,1,2,2,1).
%F A256496 a(n) = 15(n mod 2) + 10(n mod 3) + 6(n mod 5).
%F A256496 G.f.: -x*(59*x^6+146*x^5+201*x^4+195*x^3+159*x^2+94*x+31) / ((x-1)*(x+1)*(x^2+x+1)*(x^4+x^3+x^2+x+1)). - _Colin Barker_, Apr 07 2015
%F A256496 a(n) = -2*a(n-1) - 2*a(n-2) - a(n-3) + a(n-5) + 2*a(n-6) + 2*a(n-7) + a(n-8), for n>=9. - _Vaclav Kotesovec_, Apr 07 2015
%t A256496 Table[15 Mod[n, 2] + 10 Mod[n, 3] + 6 Mod[n, 5], {n, 60}] (* _Michael De Vlieger_, Mar 31 2015 *)
%t A256496 LinearRecurrence[{-2,-2,-1,0,1,2,2,1},{31,32,33,34,35,6,37,38},70] (* or *) PadRight[ {},70,{31,32,33,34,35,6,37,38,39,10,41,12,43,44,15,16,47,18,49,20,21,22,53,24,25,26,27,28,59,0}] (* _Harvey P. Dale_, Oct 31 2016 *)
%o A256496 (PARI) vector(30, n, 15*(n%2) + 10*(n%3) + 6*(n%5)) \\ _Michel Marcus_, Mar 31 2015
%Y A256496 Cf. A255818 for an example with 1 modification, A256643 for 2 modifications and A256668 for 3 modifications.
%K A256496 nonn,easy
%O A256496 1,1
%A A256496 _Aaron Kastel_, Mar 31 2015