A244656 Least product of consecutive positive integers which is divisible by each of 1, 2, ..., n.

2, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 55440, 55440, 360360, 360360, 360360, 2162160, 85765680, 85765680, 33522128640, 33522128640, 33522128640, 33522128640, 19275223968000, 19275223968000, 19275223968000

Offset: 1

Author

Rick L. Shepherd, Jul 03 2014, Sep 14 2014

Keywords

• nonn

Comments

For n > 1, clearly a(n) is bounded below by lcm(1,2,...,n) and bounded above by n!. Further, a(n) is a positive multiple of lcm(1,2,...,n). Any product of two or more consecutive positive integers may be expressed as m!/k!, where 0 <= k <= m-2. For this sequence, the m corresponding to a(n) may or may not be a multiple of n. Whenever a(n) can be expressed as the product of exactly two consecutive integers, it is a term of A002378. See the a-file link for further comments.

Examples

			a(7) = 20*21 = 21!/19! = 420 because 420 is divisible by 1, 2, 3, 4, 5, 6, and 7, and no positive integer less than 420 is divisible by each of these. Here, 420 = lcm(1,2,3,4,5,6,7). 420 is an oblong (or promic) number (A002378).
a(11) = 7*8*9*10*11 = 11!/6! = 55440. Here, 27720 = lcm(1,2,3,4,5,6,7,8,9,10,11), but 27720 cannot be represented as a product of consecutive positive integers.
a(31) = 6081487775*6081487776 = 36984493563555938400, also a promic number.
		

Links

• Rick L. Shepherd, Table of n, a(n) for n = 1..36
• Rick L. Shepherd, Sample program output and calculation notes

Crossrefs

Cf. A003418, A000142, A025527, A002378.

Programs

PARI

{a(n) =
local(L, M, i, k = 0, s = 0, ret = 0, d, divs2,
   st, pr, prt = 1); /* Use prt = 0 to suppress printing. */
if(n < 1, return, if(n < 3, ret = 2,
L = lcm(vector(n, i, i));
M = n!/L;
while(k < M,
   k++;
   s += L; d = divisors(s); divs2 = #d \ 2;
   st = 2; pr = d[st];
   i = 0;
   while(st + i <= divs2,
      if(d[st + i + 1] == d[st + i] + 1,
         pr *= d[st + i + 1]; i++;
         if(pr == s,
            if(prt,
               print1("k*L = ", k, "*", L, " = ",
                 s, " = ", d[st], "*");
               if(d[st + i] > d[st] + 2, print1("...*"),
                 if(d[st + i] == d[st] + 2,
                   print1(d[st] + 1, "*")));
               print(d[st + i], " = ", d[st + i], "!/",
                 d[st] - 1, "!"));
            ret = s; break(2),
            if(pr > s, st++; pr = d[st]; i = 0)),
         if(pr < s, st += i + 1, st++); pr = d[st]; i = 0)))));
return(ret)}