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

0, 1, 2, 6, 6, 66, 66, 858, 858, 429, 572, 9724, 2652, 50388, 3876, 3876, 42636, 245157, 28842, 48070, 32890, 296010, 296010, 780390, 33930, 525915, 841464, 712008, 1344904, 1344904, 139128

Offset: 0

Author

Gary Detlefs, Apr 17 2011

Keywords

• nonn
• look

Comments

(n-1)!*12600*a(n)/(n+6)! produces a sequence of fractions (from offset 1).

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

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).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to lcm's

Programs

Haskell

a189144 n = (foldl1 lcm [n..n+6]) `div` 420
-- Reinhard Zumkeller, Apr 28 2011

Maple

seq(lcm(n,n+1,n+2,n+3,n+4,n+5,n+6)/420, n=0..30);
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);

Mathematica

Table[(LCM@@Range[n,n+6])/420,{n,0,30}] (* Harvey P. Dale, Jun 13 2015 *)

Formula

a(n)= (n+6)!*f(n)/(12600*(n-1)!*2^p(n)*3^q(n)),n>0 where

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