A291545 a(n) is the smallest integer k > n such that (k+1)(k+2)...(2k-n)/(k(k-1)...(k-n+1)) is an integer.

6, 6, 9, 10, 16, 16, 27, 27, 28, 28, 95, 95, 96, 96, 121, 121, 122, 122, 123, 123, 124, 124, 125, 125, 126, 126, 537, 537, 538, 538, 539, 539, 540, 540, 905, 905, 906, 906, 1149, 1149, 1150, 1150, 1349, 1349, 1350, 1350, 1351, 1351, 1352, 1352

Offset: 1

Author

Bernard Schott, Aug 26 2017

Keywords

• nonn

Comments

This sequence comes from an exercise proposed by Paul Erdős for Crux Mathematicorum (see link). In the solution, it's proved that for n >= 4, the fraction is always an integer for k = (n+1)! - 2. Be careful, n and k are swapped between Crux Mathematicorum and this sequence. A stronger statement is proposed in A290791.

Erdős also proves that lim a(n)/n is infinite. That is, for any constant C, a(n) > C*n for all large enough n. - Charles R Greathouse IV, Aug 26 2017

From Jon E. Schoenfield, Aug 29 2017: (Start)

Observations up through a(294)=2010880 (and a lower bound on a(295)):

- for even n (except at n=4), a(n) = a(n-1);

- for odd n > 1, a(n) = a(n-1) + 1 except at n = 3, 5, 7, 11, 15, 27, 35, 39, 43, 67, 71, 87, 103, 143, 171, 191, 223, 227, 235, 263, 295, ...

A lower bound is given by a(n) >= A104138(j) + j where j = floor((n+1)/2) and A104138(j) is the smallest prime that is followed by j or more nonprimes. Conjecture: this bound is tight for all n > 6. (End)

Examples

			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.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..200
• Paul Erdös, Problem 556, Crux Mathematicorum, Vol. 7 (Nov. 81), p. 282.

Crossrefs

Cf. A104138, A290791.

Programs

PARI

a(n)=my(k=n+1,t=(n+2)/k!); while(denominator(t)>1, k++; t*=(2*k-n)*(2*k-n-1)*(k-n)/k^2); k \\ Charles R Greathouse IV, Aug 26 2017

Formula

a(n) = min_{k > n} : (k!/(k-n)!) | ((2k-n)!/k!). - Jon E. Schoenfield, Aug 28 2017

Extensions

Corrected and extended by Charles R Greathouse IV, Aug 26 2017