A326691 -id:A326691 - 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

A326715 Values of n for which the denominator of (Sum_{prime p | n} 1/p - 1/n) is 1.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241

Offset: 1

Jonathan Sondow, Jul 20 2019

nonn

n is in the sequence iff either n = 1 or n is a prime or n is a Giuga number, by one definition of Giuga numbers A007850.

			a(30) = denominator(Sum_{prime p | 30} 1/p - 1/30) = denominator(1/2 + 1/3 + 1/5 - 1/30) = denominator(1/1) = 1, and 30 is a Giuga number.

Robert Israel, Table of n, a(n) for n = 1..10000
Wikipedia, Giuga number

Cf. A007850, A326689, A326690, A326691, A326692.

filter:= proc(n) local p;
   denom(add(1/p, p = numtheory:-factorset(n))-1/n)=1
end proc:
select(filter, [$1..300]); # Robert Israel, Dec 15 2020

PrimeFactors[n_] := Select[Divisors[n], PrimeQ];
f[n_] := Denominator[Sum[1/p, {p, PrimeFactors[n]}] - 1/n];
Select[Range[148], f[#] == 1 &]

n such that A326690(n) = 1.

A326692 Values of k for which the denominator of (Sum_{prime p | k} 1/p - 1/k) is k.

Original entry on oeis.org

1, 4, 8, 9, 15, 16, 20, 24, 25, 27, 28, 32, 33, 35, 36, 40, 44, 45, 49, 51, 52, 60, 63, 64, 65, 68, 69, 72, 76, 77, 81, 85, 87, 88, 91, 92, 95, 96, 99, 100, 104, 108, 112, 115, 116, 117, 119, 121, 123, 124, 125, 128, 133, 135, 136, 140, 141, 143, 144, 145, 148

Offset: 1

Jonathan Sondow, Jul 20 2019

nonn

Any prime power p^k with k > 1 is a term, as 1/p - 1/p^k = (p^(k-1) - 1)/p^k which is in reduced form and has denominator p^k.

Are there infinitely many Carmichael numbers A002997 in the sequence?

			1/3 + 1/5 - 1/15 = 7/15 has denominator 15, so 15 is a term.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A002997, A326689, A326690, A326691, A326715.

PrimeFactors[n_] := Select[Divisors[n], PrimeQ];
f[n_] := Denominator[Sum[1/p, {p, PrimeFactors[n]}] - 1/n];
Select[Range[148], f[#] == # &]

is(k) = {my(p = factor(k)[,1]); denominator(sum(i = 1, #p, 1/p[i]) - 1/k) == k;} \\ Amiram Eldar, Apr 26 2024

Solutions of A326690(x) = x. That is, fixed points of A326690.

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

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A326715 Values of n for which the denominator of (Sum_{prime p | n} 1/p - 1/n) is 1.

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Maple

Mathematica

Formula

A326692 Values of k for which the denominator of (Sum_{prime p | k} 1/p - 1/k) is k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula