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Smallest k > 1 such that n*(n+1)*...*(n+k-1) / (n+(n+1)+...+(n+k-1)) is an integer.
3

%I #52 May 25 2025 23:30:19

%S 2,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,

%T 3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,9,3,3,5,3,3,5,3,3,5,3,3,5,3,3,

%U 7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3,3,5,3,3,5,3,3,5,3,3,5,3,3,7,3

%N Smallest k > 1 such that n*(n+1)*...*(n+k-1) / (n+(n+1)+...+(n+k-1)) is an integer.

%C a(n) = 7 for n == 8 (mod 15) (provided n != 53 (mod 105)).

%C a(n) = 5 for n == 2 (mod 3) (provided n != 8 (mod 15)).

%C a(n) = 9 for n == 53 (mod 105). - _Jon E. Schoenfield_, Mar 14 2014

%C a(n) = 3 for n == {0,1} (mod 3). - _Zak Seidov_, Mar 14 2014

%H Antti Karttunen, <a href="/A239278/b239278.txt">Table of n, a(n) for n = 0..20000</a>

%e 1*2/(1+2) = 2/3 is not an integer. 1*2*3/(1+2+3) = 1 is an integer. Thus a(1) = 3.

%e 2*3/(2+3) = 6/5 is not an integer. 2*3*4/(2+3+4) = 24/9 is not an integer. 2*3*4*5/(2+3+4+5) = 120/14 is not an integer. 2*3*4*5*6/(2+3+4+5+6) = 720/20 = 36 is an integer. Thus a(2) = 5.

%e a(0) = 2 as 0*(0+(2-1)) / 0+(0+(2-1)) = 0/1 = 0 is an integer. - _Antti Karttunen_, Jan 18 2025

%o (Python)

%o def A239278(n):

%o (m, s, t) = (n, n, n+1)

%o while 1:

%o m *= t

%o s += t

%o if 0 == (m%s): return (1+t-n)

%o else: t += 1

%o # _Antti Karttunen_, Jan 18 2025

%o (PARI) a(n) = {k = 2; while ( prod(i=0, k-1, n+i) % sum(i=0, k-1, n+i), k++); k;} \\ _Michel Marcus_, Mar 14 2014

%o (PARI) A239278(n) = { my(m=n, s=n); for(k=2, oo, m *= (n+(k-1)); s += (n+(k-1)); if(!(m%s), return(k))); }; \\ _Antti Karttunen_, Jan 18 2025

%Y A284721 has the same start.

%K nonn

%O 0,1

%A _Derek Orr_, Mar 13 2014

%E Term a(0)=2 prepended, original Python program replaced with one that works. - _Antti Karttunen_, Jan 18 2025