A329061 Greatest k such that A002805(k) is not divisible by n, or a(n) = 0 if there's no such k.

0, 1, 68, 3, 124, 68, 719102, 7, 206, 124, 11130347490407364042652446389727, 68, 2196, 719102, 124, 15, 4912, 206, 16128612858, 124, 719102, 11130347490407364042652446389727, 12166, 68, 624, 2196, 620, 719102, 20171036, 124, 27488495831, 31, 11130347490407364042652446389727

Offset

1,3

Comments

There are two cases where a(n) = 0: (a) n divides A002805(k) for all k, which only happens for n = 1; (b) there are infinitely many k such that n does not divide A002805(k), which may happen for some primes p and their multiples.

For k > a(n) > 0, A002805(k) is always divisible by n.

For prime p and k >= p, A002805(k) = (the denominator of s + (Sum_{i=1..floor(k/p)} 1/i)/p) is not divisible by p if and only if p divides A001008(floor(k/p)) = (the numerator of Sum_{i=1..floor(k/p)} 1/i), because the denominator of s = Sum_{1 <= i <= k, i is not divisible by p} 1/i can never be divisible by p.

If k == -1 or 0 (mod p), then p divides A001008(k) iff p^2 divides A001008(floor(k/p)), otherwise p divides A001008(k) iff p divides the numerator of (Sum_{i=floor(k/p)*p+1..k} 1/i) + (Sum_{i=1..floor(k/p)} 1/i)/p, where p is an odd prime and k >= p. (Since Sum_{i=1..p-1} (p-1)!/i = (-1)^((p-1)/2)*((p-1)/2)!*(Sum_{i=1..(p-1)/2} ((p-1)/2)!/i) + ((p-1)/2)!*(Sum_{i=1..(p-1)/2} (-1)^((p-1)/2)*((p-1)/2)!/(-i)) == 0 (mod p), odd prime p divides the numerator of Sum_{1 <= i <= floor(k/p)*p, i is not divisible by p} 1/i.)

Links

Table of n, a(n) for n=1..33.

Jinyuan Wang, PARI program and examples for n <= 10

Eric Weisstein's World of Mathematics, Harmonic Number

Formula

If n = Product_{j=1..i} p_j^e_j, p_1 < ... < p_i are primes and a(p_j^e_j) > 0, then a(n) = Max_{j=1..i} a(p_j^e_j).

a(p^e) = p^(e-1)*(a(p)+1) - 1 for prime p and a(p) > 0. Proof: A001008(k)/A002805(k) = (Sum_{1 <= i <= k, i is not divisible by p^e} 1/i) + (Sum_{i=1..floor(k/p^e)} 1/i)/p^e), hence A002805(k) is not divisible by p^e if and only if p divides A001008(floor(k/p^e)). From the comment, we know that (a(p)+1)/p - 1 is the greatest m such that p divides A001008(m). Therefore, a(p^e) = p^e*((a(p)+1)/p-1) + p^e - 1 = p^(e-1)*(a(p)+1) - 1.

a(prime(i)) = (A177734(i)+1)*prime(i) - 1, where prime(i) = A000040(i). - Jinyuan Wang, Feb 06 2020

Example

For p = 3, 3 divides numerator(1+1/2), so 2*3, 2*3 + 1 and 2*3 + 2 are such k that A002805(k) can't be divisible by 3. Similarly, 7*3, 7*3 + 1 and 7*3 + 2 are such k. Mod(A001008(7), 3) > 0 and Mod(numerator(1/22 + (Sum_{i=1..7} 1/i)/3), 3) = 0, hence 3 divides A001008(22), which means 22*3, 22*3 + 1 and 22*3 + 2 are also such k. a(3) = 68 because A001008(k) can never be divisible by 3 for k = 66, 67 and 68.

Crossrefs

Cf. A001008, A002805, A177734, A329293.

Sequence in context: A281888 A282338 A198210 * A033388 A196109 A196106

Adjacent sequences: A329058 A329059 A329060 * A329062 A329063 A329064

Keyword

nonn

Author

Jinyuan Wang, Dec 07 2019

Extensions

More terms from Jinyuan Wang, Feb 06 2020

Status

approved