A160642 -id:A160642 - cp's OEIS frontend

A064224 Numbers having more than one representation as the product of consecutive integers > 1.

120, 210, 720, 5040, 175560, 17297280, 19958400, 259459200, 20274183401472000, 25852016738884976640000, 368406749739154248105984000000

Offset: 1

Jon Perry, Sep 22 2001

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
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]
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]

			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.

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. [_T. D. Noe_, Jul 29 2009]
Robert Munafo, Page dealing with this sequence

Cf. A003015 (numbers occurring 5 or more times in Pascal's triangle).
Cf. A002378, A160642.
Subsequence of A045619, A100934.
Cf. A163263 (non-overlapping case). [T. D. Noe, Jul 29 2009]

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]

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

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

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A064224 Numbers having more than one representation as the product of consecutive integers > 1.

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