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A189144 a(n) = lcm(n,n+1,n+2,n+3,n+4,n+5,n+6)/420.

%I A189144 #20 Oct 20 2024 00:56:04
%S A189144 0,1,2,6,6,66,66,858,858,429,572,9724,2652,50388,3876,3876,42636,
%T A189144 245157,28842,48070,32890,296010,296010,780390,33930,525915,841464,
%U A189144 712008,1344904,1344904,139128
%N A189144 a(n) = lcm(n,n+1,n+2,n+3,n+4,n+5,n+6)/420.
%C A189144 (n-1)!*12600*a(n)/(n+6)! produces a sequence of fractions (from offset 1).
%C A189144 The numerators have a period of 5, repeating [5,5,5,1,1]=4*(n^4 mod 5 +(n-1)^4 mod 5 -1)+1
%C A189144 The denominators have a period of 12, repeating [2,8,12,8,2,24,4,8,6,8,4,24]. This sequence factors to 2^p(n)*3^q(n) where p(n) is a sequence of period 4, repeating [1,3,2,3] and q(n) is a sequence of period 3, repeating [0,0,1]. p(n) = A131729(n+1)+2. q(n) = A022003(n-1).
%H A189144 Reinhard Zumkeller, <a href="/A189144/b189144.txt">Table of n, a(n) for n = 0..10000</a>
%H A189144 <a href="/index/Lc#lcm">Index entries for sequences related to lcm's</a>
%F A189144 a(n)= (n+6)!*f(n)/(12600*(n-1)!*2^p(n)*3^q(n)),n>0 where
%F A189144 f(n)= 4*(n^4 mod 5 +(n-1)^4 mod 5 -1)+1
%F A189144 p(n)=(9+2*cos((n-3)*Pi/2)+3*(-1)^n)/4
%F A189144 q(n)=((n-1)^5 -(n-1)^2) mod 3
%p A189144 seq(lcm(n,n+1,n+2,n+3,n+4,n+5,n+6)/420, n=0..30);
%p A189144 f:= n-> 4*(n^4 mod 5 +(n-1)^4 mod 5 -1)+1:p:= n->)=(9+2*cos((n-3)*Pi/2)+3*(-1)^n)/4:q:=n->)=((n-1)^5 -(n-1)^2) mod 3: seq((n+6)!*f(n)/(12600*(n-1)!*2^p(n)*3^q(n)),n=1..30);
%t A189144 Table[(LCM@@Range[n,n+6])/420,{n,0,30}] (* _Harvey P. Dale_, Jun 13 2015 *)
%o A189144 (Haskell)
%o A189144 a189144 n = (foldl1 lcm [n..n+6]) `div` 420
%o A189144 -- _Reinhard Zumkeller_, Apr 28 2011
%K A189144 nonn,look
%O A189144 0,3
%A A189144 _Gary Detlefs_, Apr 17 2011