A290791 a(n) is the smallest integer k > n such that (k+1)(k+2)...(2k-2n+1)/(k(k-1)...(k-n+1)) is an integer.

6, 9, 16, 27, 28, 95, 96, 121, 122, 123, 124, 125, 126, 537, 538, 539, 540, 905, 906, 1149, 1150, 1349, 1350, 1351, 1352, 1353, 1354, 1355, 1356, 1357, 1358, 1359, 1360, 9585, 9586, 15719, 15720, 15721, 15722, 15723, 15724, 15725, 15726, 19653, 19654, 19655

Offset

1,1

Comments

This sequence comes from a small change of an exercise proposed by Paul Erdős for Crux Mathematicorum (see link). In the solution, they show that for n >= 3, the fraction is always an integer for k = (n+1)! - 2. Be careful, n and k are swapped between Crux Mathematicorum and this sequence.

Links

Chai Wah Wu, Table of n, a(n) for n = 1..100

Wojcich Komornicki, Problem 556, Crux Mathematicorum, page 49, Vol. 8, Feb. 82.

Example

If n = 1, for k = 2, 3, 4, 5, the fraction is respectively equal to 3/2, (4*5)/3, (5*6*7)/4, (6*7*8*9)/5 but for k = 6, the quotient is (7*8*9*10*11)/6 = 9240 and so a(1) = 6.

Mathematica

a[n_] := Block[{k = n+1}, While[! IntegerQ[(1 + 2*k - 2*n)! (k-n)! / (k!)^2], k++]; k]; Array[a, 30] (* Giovanni Resta, Aug 11 2017 *)

Crossrefs

Sequence in context: A316066 A282498 A031326 * A132107 A007262 A129317

Adjacent sequences: A290788 A290789 A290790 * A290792 A290793 A290794

Keyword

nonn

Author

Bernard Schott, Aug 10 2017

Extensions

a(6)-a(46) from Giovanni Resta, Aug 11 2017

Status

approved