A309268 -id:A309268 - cp's OEIS frontend

A326690 Denominator of the fraction (Sum_{prime p | n} 1/p - 1/n).

1, 1, 1, 4, 1, 3, 1, 8, 9, 5, 1, 4, 1, 7, 15, 16, 1, 9, 1, 20, 7, 11, 1, 24, 25, 13, 27, 28, 1, 1, 1, 32, 33, 17, 35, 36, 1, 19, 13, 40, 1, 21, 1, 44, 45, 23, 1, 16, 49, 25, 51, 52, 1, 27, 11, 8, 19, 29, 1, 60, 1, 31, 63, 64, 65, 11, 1, 68, 69, 35, 1, 72

Offset: 1

Jonathan Sondow, Jul 18 2019

Theorem. If n is a prime or a Carmichael number, then a(n) = A309132(n) = denominator of (N(n-1)/n + D(n-1)/n^2), where B(k) = N(k)/D(k) is the k-th Bernoulli number. This is a generalization of Theorem 1 in A309132 that A309132(p) = 1 if p is a prime. The proof generalizes that in A309132. As an application of Theorem, for n a prime or a Carmichael number one can compute A309132(n) without calculating Bernoulli numbers; see A309268.
A composite number n is a Giuga number A007850 if and only if a(n) = 1. (In fact, Sum_{prime p | n} 1/p - 1/n = 1 for all known Giuga numbers n.)
Semiprimes m = pq such that 1/p + 1/q - 1/m = p/q are exactly A190275. - Amiram Eldar and Thomas Ordowski, Jul 22 2019
The preceding comment may be rephrased as "Semiprimes m = pq such that A326689(m) = p and a(m) = q are exactly A190275." - Jonathan Sondow, Jul 22 2019
More generally, semiprimes m = pq such that 1/p + 1/q - 1/m = P/Q are exactly A190273, where P <> Q are primes. In other words, semiprimes m such that A326689(m) is prime and a(m) is prime are exactly A190273. - Amiram Eldar and Thomas Ordowski, Jul 25 2019

			-1/1, 0/1, 0/1, 1/4, 0/1, 2/3, 0/1, 3/8, 2/9, 3/5, 0/1, 3/4, 0/1, 4/7, 7/15, 7/16, 0/1, 7/9, 0/1, 13/20, 3/7, 6/11, 0/1, 19/24, 4/25, 7/13, 8/27, 17/28, 0/1, 1/1
a(12) = denominator of (Sum_{prime p | 12} 1/p - 1/12) = denominator of (1/2 + 1/3 - 1/12) = denominator of 3/4 = 4.
Computing A309132(561) involves numerator(B(560)) which has 865 digits. But 561 is a Carmichael number, so Theorem implies A309132(561) = a(561) = denominator(1/3 + 1/11 + 1/17 - 1/561) = denominator(90/187) = 187.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Wikipedia, Bernoulli number
Wikipedia, Carmichael number
Wikipedia, Giuga number

Numerators are A326689. Quotients n/a(n) are A326691.
Cf. A069359, A007947 (denominator of Sum_{prime p | n} 1/p).
Cf. A002997, A007850, A190275, A309132, A309235, A309268, A309378, A326692.

[1] cat [Denominator(&+[1/p:p in PrimeDivisors(k)]-1/k):k in [2..72]]; // Marius A. Burtea, Jul 27 2019

A326690 := n -> denom((A069359(n)-1)/n):
seq(A326690(n), n=1..72); # Peter Luschny, Jul 22 2019

PrimeFactors[n_] := Select[Divisors[n], PrimeQ];
f[n_] := Denominator[Sum[1/p, {p, PrimeFactors[n]}] - 1/n];
Table[ f[n], {n, 100}]

a(n) = denominator(sumdiv(n, d, isprime(d)/d) - 1/n); \\ Michel Marcus, Jul 19 2019

p = lambda n: [n//f[0] for f in factor(n)]
A326690 = lambda n: ((sum(p(n)) - 1)/n).denominator()
[A326690(n) for n in (1..72)] # Peter Luschny, Jul 22 2019

a(n) = 1 if n is a prime or a Giuga number A007850.
a(n) = denominator of (N(n-1)/n + D(n-1)/n^2) if n is a Carmichael number A002997.
a(n) = denominator((A069359(n) - 1)/n). - Peter Luschny, Jul 22 2019

A309132 a(n) is the denominator of F(n) = A027641(n-1)/n + A027642(n-1)/n^2.

Original entry on oeis.org

1, 1, 1, 16, 1, 36, 1, 64, 27, 100, 1, 144, 1, 196, 75, 256, 1, 324, 1, 400, 49, 484, 1, 576, 125, 676, 243, 784, 1, 900, 1, 1024, 363, 1156, 1225, 1296, 1, 1444, 169, 1600, 1, 1764, 1, 1936, 135, 2116, 1, 2304, 343, 2500, 867, 2704, 1, 2916, 3025, 3136, 361, 3364, 1, 3600, 1, 3844, 1323, 4096, 845, 4356, 1

Offset: 1

Thomas Ordowski, Jul 14 2019

It seems that the numerator of F(n) is the numerator of (B(n-1) + 1/n), where B(k) is the k-th Bernoulli number; if so, for n > 2, the numerator of F(n) is A174341(n-1). How to prove it?
Conjecture: for n > 1, a(n) = 1 if and only if n is prime.
Is this conjecture equivalent to the Agoh-Giuga conjecture?
Theorem 1. If p is prime, then a(p) = 1. Proof. a(2) = 1, so let p be an odd prime. By the von Staudt-Clausen theorem, if k is even, then B(k) = A(k) - Sum_{prime q, q-1 | k} 1/q, where A(k) is an integer and the sum is over all primes q such that q-1 divides k. Thus B(k) = N(k)/D(k) with D(k) = Product_{prime q, q-1 | k} q. Now let k = p-1. Then N(p-1)/D(p-1) = B(p-1) = A(p-1) - 1/p - Sum_{prime q < p, q-1 | p-1} 1/q (*). Add 1/p to both sides of (*) and multiply by p*D(p-1) to get p*N(p-1) + D(p-1) = p*D(p-1)*(A(p-1) - Sum_{prime q < p, q-1 | p-1} 1/q) (**). Now p | D(p-1), so p^2 | p*D(p-1) in (**). The denominators on the right side of (**) are all of the form q < p. Therefore, p^2 divides both sides of (**). Hence F(p) = N(p-1)/p + D(p-1)/p^2 is an integer, so a(p) = 1. - Jonathan Sondow, Jul 14 2019
Conjecture: composite numbers n such that a(n) is squarefree are only the Carmichael numbers A002997. Cf. A309235. - Thomas Ordowski, Jul 15 2019
Conjecture checked up to n = 101101. - Amiram Eldar, Jul 16 2019
Theorem 2. If n is a prime or a Carmichael number, then a(n) = A326690(n) = denominator of (Sum_{prime p | n} 1/p - 1/n). The proof is a generalization of that of Theorem 1. (Note that Theorem 2 implies Theorem 1, since if n is prime, then (Sum_{prime p | n} 1/p - 1/n) = 1/n - 1/n = 0/1, so a(p) = A326690(n) = 1.) For n a prime or a Carmichael number, an application of Theorem 2 is computing a(n) without calculating Bernoulli(n-1) which may be huge; see A309268 and A326690. - Jonathan Sondow, Jul 19 2019
The values of F(n) when n is prime are A327033. - Jonathan Sondow, Aug 16 2019

			F(n) = 2/1, 0/1, 1/1, 1/16, 1/1, 1/36, 1/1, 1/64, 7/27, 1/100, 1/1, 1/144, -37/1, 1/196, 37/75, 1/256, -211/1, 1/324, 2311/1, 1/400, -407389/49, ...

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem
Wikipedia, Agoh-Giuga conjecture
Wikipedia, Bernoulli number: Related sequences

Cf. A000040, A000146, A002997, A027641, A027642, A110936, A166062, A174341, A174342, A309235, A326690, A327033.

[Denominator(Numerator(Bernoulli(n-1))/n + Denominator(Bernoulli(n-1))/n^2): n in [1..70]]; // Vincenzo Librandi, Jul 14 2019

Table[Denominator[Numerator[BernoulliB[n - 1]] / n + Denominator[ BernoulliB[ n - 1]] / n^2], {n, 70}] (* Vincenzo Librandi, Jul 14 2019 *)

a(n) = denominator(numerator(bernfrac(n-1))/n + denominator(bernfrac(n-1))/n^2); \\ Michel Marcus, Jul 14 2019

a(p) = 1 for prime p.
a(2k) = (2k)^2 for k > 1.
Conjecture: for k > 0, a(2k+1) = (2k+1)^2 iff 2k+1 is in A121707.
Denominator(F(p)/p) = 1 for the primes p = 2 and p = 1277 but for no other prime p < 1.5 * 10^4. Does denominator(F(p)/p) = 1 for any prime p > 1.5 * 10^4? - Jonathan Sondow, Jul 14 2019
Similarly, Sum_{k=1..p-1} k^(p-1) == -1 (mod p^2) for the prime p = 1277. - Thomas Ordowski, Jul 15 2019
a(n) = denominator(Sum_{prime p | n} 1/p - 1/n) if n is a prime or a Carmichael number. - Jonathan Sondow, Jul 19 2019

cp's OEIS Frontend

A326690 Denominator of the fraction (Sum_{prime p | n} 1/p - 1/n).

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