cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Jonathan Sondow, Jul 18 2019

Keywords

Comments

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

Examples

			-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.
		

Crossrefs

Numerators are A326689. Quotients n/a(n) are A326691.
Cf. A069359, A007947 (denominator of Sum_{prime p | n} 1/p).

Programs

  • Magma
    [1] cat [Denominator(&+[1/p:p in PrimeDivisors(k)]-1/k):k in [2..72]]; // Marius A. Burtea, Jul 27 2019
  • Maple
    A326690 := n -> denom((A069359(n)-1)/n):
    seq(A326690(n), n=1..72); # Peter Luschny, Jul 22 2019
  • Mathematica
    PrimeFactors[n_] := Select[Divisors[n], PrimeQ];
    f[n_] := Denominator[Sum[1/p, {p, PrimeFactors[n]}] - 1/n];
    Table[ f[n], {n, 100}]
  • PARI
    a(n) = denominator(sumdiv(n, d, isprime(d)/d) - 1/n); \\ Michel Marcus, Jul 19 2019
    
  • SageMath
    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
    

Formula

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