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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.

%I A291545 #31 Nov 23 2024 09:16:49
%S A291545 6,6,9,10,16,16,27,27,28,28,95,95,96,96,121,121,122,122,123,123,124,
%T A291545 124,125,125,126,126,537,537,538,538,539,539,540,540,905,905,906,906,
%U A291545 1149,1149,1150,1150,1349,1349,1350,1350,1351,1351,1352,1352
%N 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.
%C A291545 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.
%C A291545 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
%C A291545 From _Jon E. Schoenfield_, Aug 29 2017: (Start)
%C A291545 Observations up through a(294)=2010880 (and a lower bound on a(295)):
%C A291545 - for even n (except at n=4), a(n) = a(n-1);
%C A291545 - 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, ...
%C A291545 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)
%H A291545 Michel Marcus, <a href="/A291545/b291545.txt">Table of n, a(n) for n = 1..200</a>
%H A291545 Paul Erdös, <a href="https://cms.math.ca/publications/crux/issue/?volume=7&amp;issue=9">Problem 556</a>, Crux Mathematicorum, Vol. 7 (Nov. 81), p. 282.
%F A291545 a(n) = min_{k > n} : (k!/(k-n)!) | ((2k-n)!/k!). - _Jon E. Schoenfield_, Aug 28 2017
%e A291545 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.
%o A291545 (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
%Y A291545 Cf. A104138, A290791.
%K A291545 nonn
%O A291545 1,1
%A A291545 _Bernard Schott_, Aug 26 2017
%E A291545 Corrected and extended by _Charles R Greathouse IV_, Aug 26 2017