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A064224 Numbers having more than one representation as the product of consecutive integers > 1. 5


%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 any member of A045619 we can construct a member of this sequence by equating n(n+1)(n+2)...(x-1) to (n+2)(n+3)...(x-1)x. Also, as demonstrated in my examples, 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

%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]

%Y Cf. A003015 (numbers occurring 5 or more times in Pascal's triangle).

%Y Cf. A002378, A045619.

%Y Cf. A163263 (non-overlapping case). [_T. D. Noe_, Jul 29 2009]

%K nonn

%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_

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Last modified March 19 13:51 EDT 2018. Contains 300855 sequences. (Running on oeis4.)