%I A064224
%S A064224 120,210,720,5040,175560,17297280,19958400,259459200,20274183401472000,
%T A064224 25852016738884976640000,368406749739154248105984000000
%N A064224 Numbers having more than one representation as the product of consecutive 
               integers > 1.
%C A064224 Solutions to the equations: x(x+1)...(x+m)=y(y+1)...(y+n) with x>1, y>
               1.
%C A064224 Some patterns are impossible, e.g. x(x+1)(x+2)(x+3)=y(y+1) has been proved 
               impossible.
%C A064224 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 
               (noe(AT)sspectra.com), Nov 22 2004
%C A064224 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 (mrob27(AT)gmail.com), Aug 17 2007
%C A064224 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. [From T. D. Noe (noe(AT)sspectra.com), 
               Jul 29 2009]
%D A064224 H. L. Abbott, P. Erdos and D Hanson, On the number of times an integer 
               occurs as a binomial coefficient, Amer. Math. Monthly, Vol. 81, No. 
               3 (Mar., 1974), 256-261.
%D A064224 R. A. MacLeod and I. Barrodale, On equal products of consecutive integers, 
               Canad. Math. Bull., 13 (1970) 255-259. [From T. D. Noe (noe(AT)sspectra.com), 
               Jul 29 2009]
%H A064224 Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-a100933.html">
               Page dealing with this sequence</a>
%e A064224 120 is here because 120 = 2*3*4*5 = 4*5*6.
%e A064224 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 A064224 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 A064224 Cf. A003015 (numbers occurring 5 or more times in Pascal's triangle).
%Y A064224 Cf. A002378, A045619.
%Y A064224 A163263 (non-overlapping case) [From T. D. Noe (noe(AT)sspectra.com), 
               Jul 29 2009]
%Y A064224 Sequence in context: A056994 A114823 A069790 this_sequence A069674 A003015 
               A098565
%Y A064224 Adjacent sequences: A064221 A064222 A064223 this_sequence A064225 A064226 
               A064227
%K A064224 nonn
%O A064224 1,1
%A A064224 Jon Perry (perry(AT)globalnet.co.uk), Sep 22 2001
%E A064224 a(1), a(7) and a(8) from T. D. Noe (noe(AT)sspectra.com), Nov 22 2004
%E A064224 a(9) and a(10) from Robert Munafo (mrob27(AT)gmail.com), Aug 13 2007
%E A064224 a(11) from Robert Munafo (mrob27(AT)gmail.com), Aug 17 2007
%E A064224 Edited by N. J. A. Sloane (njas(AT)research.att.com), Sep 14 2008 at 
               the suggestion of R. J. Mathar