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%I #23 Dec 10 2020 01:31:32
%S 53,159,1735,4508,3222,18238,31499,16965,78013,114722,54348,225124,
%T 303425,133515,519187,662408,277794,1035370,1272023,515697,1864393,
%U 2228174,880920,3112528,3642317,1412343,4901599,5641460,2154030,7368982,8368163,3155229,10667605,11980538
%N Smallest positive integer m such that m, m+1, m+2, m+3 are divisible by 2n+1, 2n+3, 2n+5, 2n+7 respectively.
%H Joseph Myers, <a href="/A279259/b279259.txt">Table of n, a(n) for n = 0..1000</a>
%H United Kingdom Mathematics Trust, <a href="https://bmos.ukmt.org.uk/home/bmo1-2017.pdf">2016/17 British Mathematical Olympiad Round 1</a>, Problem 6.
%H <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,5,0,0,-10,0,0,10,0,0,-5,0,0,1).
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical competitions</a>.
%F a(n) = (2*n+1 + lcm(2*n+1, 2*n+3, 2*n+5, 2*n+7))/2.
%F G.f.: (8*x^14 +10*x^12 -89*x^11 -153*x^10 -1777*x^9 -4173*x^8 -2445*x^7 -9489*x^6 -9563*x^5 -2427*x^4 -4243*x^3 -1735*x^2 -159*x-53) / ((x-1)^5*(x^2+x+1)^5). - _Alois P. Heinz_, Dec 08 2016
%F From _Bernard Schott_, Dec 08 2020: (Start)
%F If n == 1 (mod 3), a(n) = (2*n+1)* ((2*n+3)*(2*n+5)*(2*n+7)/3 + 1)/2.
%F If n == 0, 2 (mod 3), a(n) = (2*n+1)* ((2*n+3)*(2*n+5)*(2*n+7) + 1)/2. (End)
%e 53 is the smallest positive integer such that 53, 54, 55, 56 are divisible by 1, 3, 5, 7 respectively, hence a(0) = 53. - _Bernard Schott_, Dec 08 2020
%t LinearRecurrence[{0,0,5,0,0,-10,0,0,10,0,0,-5,0,0,1},{53,159,1735,4508,3222,18238,31499,16965,78013,114722,54348,225124,303425,133515,519187},40] (* _Harvey P. Dale_, Dec 29 2017 *)
%K nonn,easy
%O 0,1
%A _Joseph Myers_, Dec 08 2016
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