

A244656


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


2



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
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OFFSET

1,1


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 <= m2. 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 afile link for further comments.


LINKS

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


EXAMPLE

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.


PROG

(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)}


CROSSREFS

Cf. A003418, A000142, A025527, A002378.
Sequence in context: A216641 A191970 A175516 * A159322 A049313 A049954
Adjacent sequences: A244653 A244654 A244655 * A244657 A244658 A244659


KEYWORD

nonn


AUTHOR

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


STATUS

approved



