Search: id:A064224 Results 1-1 of 1 results found. %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, Page dealing with this sequence %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 Search completed in 0.002 seconds