OFFSET
1,1
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, ...
LINKS
Michel Marcus, Table of n, a(n) for n = 1..200
Paul Erdös, Problem 556, Crux Mathematicorum, Vol. 7 (Nov. 81), p. 282.
FORMULA
a(n) = min_{k > n} : (k!/(k-n)!) | ((2k-n)!/k!). - Jon E. Schoenfield, Aug 28 2017
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Aug 26 2017
EXTENSIONS
Corrected and extended by Charles R Greathouse IV, Aug 26 2017
STATUS
approved