%I #24 Aug 20 2023 16:56:45
%S 120,210,720,5040,175560,17297280,19958400,259459200,
%T 20274183401472000,25852016738884976640000,
%U 368406749739154248105984000000
%N Numbers having more than one representation as the product of consecutive integers > 1.
%C Solutions to the equations: x(x+1)...(x+m)=y(y+1)...(y+n) with x>1, y>1.
%C Some patterns are impossible, e.g. x(x+1)(x+2)(x+3)=y(y+1) has been proved impossible.
%C 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
%C Using an improved algorithm I have performed an exhaustive search up to 2.15 * 10^33 and can confirm the terms shown above are all that exist up to that point. 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. Also, as demonstrated in my examples below, 5040 is related to 720 as 259459200 is to 210. So we also know that 36055954861352887137197787308347629783163600896000000000 and 6244042313569035223343873483125151604764341428027427022254596874567680000000000000 are terms. - _Robert Munafo_, Aug 17 2007 [edited by _Peter Munn_, Aug 20 2023]
%C 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]
%H H. L. Abbott, P. Erdos and D. Hanson, <a href="http://www.jstor.org/stable/2319526">On the numbers of times an integer occurs as a binomial coefficient</a>, Amer. Math. Monthly, (March 1974), 256-261.
%H R. A. MacLeod and I. Barrodale, <a href="http://dx.doi.org/10.4153/CMB-1970-052-8">On equal products of consecutive integers</a>, Canad. Math. Bull., 13 (1970) 255-259. [_T. D. Noe_, Jul 29 2009]
%H Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-a100933.html">Page dealing with this sequence</a>
%e 120 is here because 120 = 2*3*4*5 = 4*5*6.
%e 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.
%t 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]
%o (Python)
%o import heapq
%o def aupton(terms, verbose=False):
%o p = 2*3; h = [(p, 2, 3)]; nextcount = 4; alst = []; oldv = None
%o while len(alst) < terms:
%o (v, s, l) = heapq.heappop(h)
%o if v == oldv and v not in alst:
%o alst.append(v)
%o if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")
%o if v >= p:
%o p *= nextcount
%o heapq.heappush(h, (p, 2, nextcount))
%o nextcount += 1
%o oldv, olds, oldl = v, s, l
%o v //= s; s += 1; l += 1; v *= l
%o heapq.heappush(h, (v, s, l))
%o return alst
%o print(aupton(8, verbose=True)) # _Michael S. Branicky_, Jun 24 2021
%Y Cf. A003015 (numbers occurring 5 or more times in Pascal's triangle).
%Y Cf. A002378, A160642.
%Y Subsequence of A045619, A100934.
%Y Cf. A163263 (non-overlapping case). [_T. D. Noe_, Jul 29 2009]
%K nonn,more
%O 1,1
%A _Jon Perry_, Sep 22 2001
%E a(1), a(7) and a(8) from _T. D. Noe_, Nov 22 2004
%E a(9) and a(10) from _Robert Munafo_, Aug 13 2007
%E a(11) from _Robert Munafo_, Aug 17 2007
%E Edited by _N. J. A. Sloane_, Sep 14 2008 at the suggestion of _R. J. Mathar_