OFFSET
1,2
COMMENTS
For other primes after a few exceptions it seems that all denominators of harmonic numbers are divisible by that prime. For 11 there are many more exceptions. Maybe infinitely many?
EXAMPLE
33 is in the sequence since H(33) = p/q and 11 is not a divisor of q. Here H(n) = Sum_{i=1..n} 1/i.
Of course if H(33) has no denominator with a factor 11 the same is true for 34, 35, ..., 43.
PROG
(Sage)
n = 10000
sum11 = 0
resu = [0]
for i in range(11, n, 11):
D = (1 / i).partial_fraction_decomposition()[1]
sum11 += sum(v for v in D if 11.divides(v.denominator()))
if sum11 >= 1:
sum11 -= 1
if sum11 == 0:
resu.append(i)
resu
(PARI) lista(nn) = {forstep (n=0, nn, 11, if (denominator(sum(k=2, n, 1/k)) % 11, print1(n, ", ")); ); } \\ Michel Marcus, Oct 19 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Matthijs Coster, Oct 18 2014
STATUS
approved