login
Let b(0)=0. For n >= 1, b(n) is the least k > b(n-1)+1 such that k divides (k-1)!/b(n-1)!, and a(n) = (b(n)-1)!/(b(n-1)!*b(n)).
6

%I #25 Mar 25 2019 10:08:13

%S 1,20,4620,12697776,159845400,941432800,158800433792,1895312483064000,

%T 3438271897004237230080,933561026438040,2562849175892544,

%U 640904462719404383808000,1528364130975,2352733350786,959393282698730880000,6142080926952

%N Let b(0)=0. For n >= 1, b(n) is the least k > b(n-1)+1 such that k divides (k-1)!/b(n-1)!, and a(n) = (b(n)-1)!/(b(n-1)!*b(n)).

%C Group the natural numbers so that every 2n-th group product is divisible by the single number in the next group. (1), (2,3,4,5), (6), (7,8,9,10,11), (12), (13,14,15,16,17,18,19),(20), (21,22,23,24,25,26,27),(28),...Sequence contains the ratio of the product of terms in 2n-th group and the (2n+1)-th group.

%H Robert Israel, <a href="/A079759/b079759.txt">Table of n, a(n) for n = 1..10001</a> (corrected by Robert Israel, Jan 20 2019)

%e a(1) = 1*2*3*4*5/6 = 20, a(2) = 7*8*9*10*11/12 = 4620, a(3) = 13*14*15*16*17*18*19/20 = 12697776, a(4) = 159845400 = 21*22*...*27/28.

%p t:= 0:

%p for n from 1 to 30 do

%p p:= t+1;

%p for j from t+2 while not (p/j)::integer do p:= p*j od;

%p A[n]:= p/j;

%p t:= j;

%p od:

%p seq(A[i],i=1..30); # _Robert Israel_, Jul 16 2018

%t a[1] = 1; t = 0; nmax = 16; For[n = 1, n <= nmax, n++, p = t+1; For[j = t+2, Not[IntegerQ[p/j]], j++, p = p*j]; a[n+1] = p/j; t = j];

%t Table[a[n], {n, 1, nmax}] (* _Jean-François Alcover_, Mar 25 2019, after _Robert Israel_ *)

%Y Cf. A079759, A079760, A109895, A109896, A109897.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Jan 10 2003

%E More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com) and _Sascha Kurz_, Jan 12 2003

%E Edited by _N. J. A. Sloane_, Nov 04 2018 at the suggestion of _Georg Fischer_. This entry now contains the merger of two identical sequences submitted by the same author.