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A092935 a(1) = 1; a(n) = floor {(n+1)(n+2)(n+3)...(n+k)}/{(n-1)(n-2)(n-3)...(n-k)} for the least value of k. 1

%I

%S 1,3,2,5,14,42,6,429,1430,4862,16796,58786,2261,742900,2674440,66861,

%T 35357670,129644790,2274470,1767263190,25246617,24466267020,

%U 91482563640,343059613650,1289904147324,4861946401452,18367353072152,69533550916004

%N a(1) = 1; a(n) = floor {(n+1)(n+2)(n+3)...(n+k)}/{(n-1)(n-2)(n-3)...(n-k)} for the least value of k.

%C The first occurrence of an integer in the sequence (n+1)/(n-1), (n+1)(n+2)/{(n-1)(n-2)}, (n+1)(n+2)(n+3)/{(n-1)(n-2)(n-3)},... The triangle of these numbers with initial value of n = 2, for k = 1 to n-1 is:

%C 3

%C 2 10

%C 5/3 5 35

%C 3/2 7/2 14 126

%C 7/5 14/5 42/5 42 462

%C 4/3 12/5 6 22 132 1716

%C ...

%C Sequence contains the first integer in each row.

%C The leading diagonal of the triangle is given by A001700 = C(2n+1,n+1). i.e. eventually an integer occurs for k < n-1.

%e a(6)= 42: the relevant numbers are 7/5, 7*8/(5*4), 7*8*9/(5*4*3), (7*8*9*10)/(5*4*3*2),...or 1.4, 2.8,8.4,42,...

%o (PARI) { a(n) = local(p, q, r); p=1; q=1; for(k=1,n, p*=n+k; q*=n-k; r=gcd(p,q); p\=r; q\=r; if(q==1,break)); return(p) } (Alekseyev)

%Y Cf. A001700.

%Y Sequence of corresponding values of k is A103634

%K nonn

%O 1,2

%A _Amarnath Murthy_, Mar 22 2004

%E More terms from _Max Alekseyev_, Feb 11 2005

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Last modified June 1 20:49 EDT 2020. Contains 334765 sequences. (Running on oeis4.)