OFFSET
1,1
COMMENTS
Solutions to the equations: x(x+1)...(x+m)=y(y+1)...(y+n) with x>1, y>1.
Some patterns are impossible, e.g. x(x+1)(x+2)(x+3)=y(y+1) has been proved impossible.
The early terms in this sequence each have two representations. Is two the maximum possible? The sequence is infinite: for any n, the number n*(n+1)*...*(n^2+n-1) is in this sequence. The next number of this form is 20274183401472000, which is obtained when n=4. - T. D. Noe, Nov 22 2004
MacLeod and Barrodale prove that the equation x(x+1)...(x+m-1) = y(y+1)...(y+n-1) has no solutions x>1 and y>1 for the following pairs of (m,n): (2,4), (2,6), (2,8), (2,12), (4,8), (5,10). They also show that (2,3) has two solutions and (3,6) has one solution. They conjecture that (2,k) has no solution for k>3. - T. D. Noe, Jul 29 2009
The pair (m,n) = (2,3) produces only one term 210, while (m,n) = (3,4) produces only two terms: 120 and 175560. Terms from a(6) onward have max(m,n) >= 5. - Max Alekseyev, Sep 21 2025
For all k = A002378(n) > 2 we can construct a member of this sequence by equating n(n+1)(n+2)...(k-1) to (n+2)(n+3)...(k-1)k. In other words, n(n+1)(n+2)...(n^2+n-1) is a term for all n > 1. Also, as demonstrated in my examples below, 5040 is related to 720 as 259459200 is to 210. So we also know that 6244042313569035223343873483125151604764341428027427022254596874567680000000000000 is a term. - Robert Munafo, Aug 17 2007; edited by Peter Munn, Aug 20 2023 and Max Alekseyev, Sep 27 2025
LINKS
H. L. Abbott, P. Erdos and D. Hanson, On the numbers of times an integer occurs as a binomial coefficient, Amer. Math. Monthly, (March 1974), 256-261.
R. A. MacLeod and I. Barrodale, On equal products of consecutive integers, Canad. Math. Bull., 13 (1970) 255-259.
Robert Munafo, Page dealing with this sequence
EXAMPLE
120 is here because 120 = 2*3*4*5 = 4*5*6.
a(2)=210 because we can write 210=5*6*7 or 14*15. The term a(8) = 259459200 = 5*6*7*8*9*10*11*12*13 = 8*9*10*11*12*13*14*15 is related to 210 by adding the intervening integers (8 through 13) to both products.
MATHEMATICA
nn=10^10; t3={}; Do[m=0; p=n; While[m++; p=p(n+m); p<=nn, t3={t3, p}], {n, 2, Sqrt[nn]}]; t3=Sort[Flatten[t3]]; lst={}; Do[If[t3[[i]]==t3[[i+1]], AppendTo[lst, t3[[i]]]], {i, Length[t3]-1}]; Union[lst]
PROG
(Python)
import heapq
def aupton(terms, verbose=False):
p = 2*3; h = [(p, 2, 3)]; nextcount = 4; alst = []; oldv = None
while len(alst) < terms:
(v, s, l) = heapq.heappop(h)
if v == oldv and v not in alst:
alst.append(v)
if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")
if v >= p:
p *= nextcount
heapq.heappush(h, (p, 2, nextcount))
nextcount += 1
oldv, olds, oldl = v, s, l
v //= s; s += 1; l += 1; v *= l
heapq.heappush(h, (v, s, l))
return alst
print(aupton(8, verbose=True)) # Michael S. Branicky, Jun 24 2021
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Jon Perry, Sep 22 2001
EXTENSIONS
a(1), a(7) and a(8) from T. D. Noe, Nov 22 2004
a(9) and a(10) from Robert Munafo, Aug 13 2007
a(11) from Robert Munafo, Aug 17 2007
Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar
a(13) from Robert Munafo confirmed and a(12) added by Max Alekseyev, Sep 27 2025
STATUS
approved