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.

Showing 1-3 of 3 results.

A007850 Giuga numbers: composite numbers n such that p divides n/p - 1 for every prime divisor p of n.

Original entry on oeis.org

30, 858, 1722, 66198, 2214408306, 24423128562, 432749205173838, 14737133470010574, 550843391309130318, 244197000982499715087866346, 554079914617070801288578559178, 1910667181420507984555759916338506
Offset: 1

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Author

D. Borwein, J. M. Borwein, P. B. Borwein and R. Girgensohn

Keywords

Comments

There are no other Giuga numbers with 8 or fewer prime factors. I did an exhaustive search using a PARI script which implemented Borwein and Girgensohn's method for finding n factor solutions given n - 2 factors. - Fred Schneider, Jul 04 2006
One further Giuga number is known with 10 prime factors, namely:
420001794970774706203871150967065663240419575375163060922876441614\
2557211582098432545190323474818 =
2 * 3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519 * 8491659218261819498490029296021 * 58254480569119734123541298976556403 but this may not be the next term. (See the Butske et al. paper.)
Conjecture: Giuga numbers are the solution of the differential equation n' = n + 1, where n' is the arithmetic derivative of n. - Paolo P. Lava, Nov 16 2009
n is a Giuga number if and only if n' = a*n + 1 for some integer a > 0 (see our preprint in arXiv:1103.2298). - José María Grau Ribas, Mar 19 2011
A composite number n is a Giuga number if and only if Sum_{i = 1..n-1} i^phi(n) == -1 (mod n), where phi(n) = A000010(n). - Jonathan Sondow, Jan 03 2014
A composite number n is a Giuga number if and only if Sum_{prime p|n} 1/p = 1/n + an integer. (In fact, all known Giuga numbers n satisfy Sum_{prime p|n} 1/p = 1/n + 1.) - Jonathan Sondow, Jan 08 2014
The prime factors of a(n) are listed as n-th row of A236434. - M. F. Hasler, Jul 13 2015
Conjecture: let k = a(n) and k be the product of x(n) distinct prime factors where x(n) <= x(n+1). Then, for any even n, n/2 + 2 <= x(n) <= n/2 + 3 and, for any odd n, (n+1)/2 + 2 <= x(n) <= (n+1)/2 + 3. For any n > 1, there are y "old" distinct prime factors o(1)...o(y) such that o(1) = 2, o(2) = 3, and z "new" distinct prime factors n(1)...n(z) such that none of them - unlike the "old" ones - can be a divisor of a(q) while q < n; n(1) > o(y), y = x(n) - z >= 2, 2 <= z <= b where b is either 4, or 1/2*n. - Sergey Pavlov, Feb 24 2017
Conjecture: a composite n is a Giuga number if and only if Sum_{k=1..n-1} k^lambda(n) == -1 (mod n), where lambda(n) = A002322(n). - Thomas Ordowski and Giovanni Resta, Jul 25 2018
A composite number n is a Giuga number if and only if A326690(n) = 1. - Jonathan Sondow, Jul 19 2019
A composite n is a Giuga number if and only if n * A027641(phi(n)) == - A027642(phi(n)) (mod n^2). Note: Euler's phi function A000010 can be replaced by the Carmichael lambda function A002322. - Thomas Ordowski, Jun 07 2020
By von Staudt and Clausen theorem, a composite n is a Giuga number if and only if n * A027759(phi(n)) == A027760(phi(n)) (mod n^2). Note: Euler's phi function can be replaced by the Carmichael lambda function. - Thomas Ordowski, Aug 01 2020

Examples

			From _M. F. Hasler_, Jul 13 2015: (Start)
The prime divisors of 30 are {2, 3, 5}, and 2 divides 30/2-1 = 14, 3 divides 30/3-1 = 9, and 5 divides 30/5-1 = 5.
The prime divisors of 858 are {2, 3, 11, 13} and 858/2-1 = 428 is even, 858/3-1 = 285 is divisible by 3, 858/11-1 = 77 is a multiple of 11, and 858/13-1 = 65 = 13*5.
(End)
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 30, pp 11, Ellipses, Paris 2008.

Crossrefs

Programs

  • Mathematica
    fQ[n_] := AllTrue[First /@ FactorInteger@ n, Divisible[n/# - 1, #] &]; Select[Range@ 100000, CompositeQ@ # && fQ@ # &] (* Michael De Vlieger, Oct 05 2015 *)
  • PARI
    is(n)=if(isprime(n), return(0)); my(f=factor(n)[,1]); for(i=1,#f, if((n/f[i])%f[i]!=1, return(0))); n>1 \\ Charles R Greathouse IV, Apr 28 2015
    
  • Python
    from itertools import count, islice
    from sympy import isprime, primefactors
    def A007850_gen(startvalue=2): # generator of terms >= startvalue
        return filter(lambda x: not isprime(x) and all((x//p-1) % p == 0 for p in primefactors(x)), count(max(startvalue,2)))
    A007850_list = list(islice(A007850_gen(),4)) # Chai Wah Wu, Feb 19 2022

Formula

Sum_{i = 1..a(n)-1} i^phi(a(n)) == -1 (mod a(n)). - Jonathan Sondow, Jan 03 2014

Extensions

a(12) from Fred Schneider, Jul 04 2006
Further references from Fred Schneider, Aug 19 2006
Definition corrected by Jonathan Sondow, Sep 16 2012

A054377 Primary pseudoperfect numbers: numbers n > 1 such that 1/n + sum 1/p = 1, where the sum is over the primes p | n.

Original entry on oeis.org

2, 6, 42, 1806, 47058, 2214502422, 52495396602, 8490421583559688410706771261086
Offset: 1

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Author

Keywords

Comments

Primary pseudoperfect numbers are the solutions of the "differential equation" n' = n-1, where n' is the arithmetic derivative of n. - Paolo P. Lava, Nov 16 2009
Same as n > 1 such that 1 + sum n/p = n (and the only known numbers n > 1 satisfying the weaker condition that 1 + sum n/p is divisible by n). Hence a(n) is squarefree, and is pseudoperfect if n > 1. Remarkably, a(n) has exactly n (distinct) prime factors for n < 9. - Jonathan Sondow, Apr 21 2013
From the Wikipedia article: it is unknown whether there are infinitely many primary pseudoperfect numbers, or whether there are any odd primary pseudoperfect numbers. - Daniel Forgues, May 27 2013
Since the arithmetic derivative of a prime p is p' = 1, 2 is obviously the only prime in the sequence. - Daniel Forgues, May 29 2013
Just as 1 is not a prime number, 1 is also not a primary pseudoperfect number, according to the original definition by Butske, Jaje, and Mayernik, as well as Wikipedia and MathWorld. - Jonathan Sondow, Dec 01 2013
Is it always true that if a primary pseudoperfect number N > 2 is adjacent to a prime N-1 or N+1, then in fact N lies between twin primes N-1, N+1? See A235139. - Jonathan Sondow, Jan 05 2014
Also, integers n > 1 such that A069359(n) = n - 1. - Jonathan Sondow, Apr 16 2014

Examples

			From _Daniel Forgues_, May 24 2013: (Start)
With a(1) = 2, we have 1/2 + 1/2 = (1 + 1)/2 = 1;
with a(2) = 6 = 2 * 3, we have
  1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = (1*3 + 3)/(2*3) = (1 + 1)/2 = 1;
with a(3) = 42 = 6 * 7, we have
  1/2 + 1/3 + 1/7 + 1/42 = (21 + 14 + 6 + 1)/42 =
  (3*7 + 2*7 + 7)/(6*7) = (3 + 2 + 1)/6 = 1;
with a(4) = 1806 = 42 * 43, we have
  1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = (903 + 602 + 258 + 42 + 1)/1806 =
  (21*43 + 14*43 + 6*43 + 43)/(42*43) = (21 + 14 + 6 + 1)/42 = 1;
with a(5) = 47058 (not oblong number), we have
  1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/47058 =
  (23529 + 15686 + 4278 + 2046 + 1518 + 1)/47058 = 1.
For n = 1 to 8, a(n) has n prime factors:
  a(1) = 2
  a(2) = 2 * 3
  a(3) = 2 * 3 *  7
  a(4) = 2 * 3 *  7 * 43
  a(5) = 2 * 3 * 11 * 23 *  31
  a(6) = 2 * 3 * 11 * 23 *  31 * 47059
  a(7) = 2 * 3 * 11 * 17 * 101 *   149 *       3109
  a(8) = 2 * 3 * 11 * 23 *  31 * 47059 * 2217342227 * 1729101023519
If a(n)+1 is prime, then a(n)*[a(n)+1] is also primary pseudoperfect. We have the chains: a(1) -> a(2) -> a(3) -> a(4); a(5) -> a(6). (End)
A primary pseudoperfect number (greater than 2) is oblong if and only if it is not the initial member of a chain. - _Daniel Forgues_, May 29 2013
If a(n)-1 is prime, then a(n)*(a(n)-1) is a Giuga number (A007850). This occurs for a(2), a(3), and a(5). See A235139 and the link "The p-adic order . . .", Theorem 8 and Example 1. - _Jonathan Sondow_, Jan 06 2014
		

Crossrefs

Programs

  • Mathematica
    pQ[n_] := (f = FactorInteger[n]; 1/n + Sum[1/f[[i]][[1]], {i, Length[f]}] == 1)
    Select[Range[2, 10^6], pQ[#] &] (* Robert Price, Mar 14 2020 *)
  • PARI
    isok(n) = if (n > 1, my(f=factor(n)[,1]); 1/n + sum(k=1, #f, 1/f[k]) == 1); \\ Michel Marcus, Oct 05 2017
  • Python
    from sympy import primefactors
    A054377 = [n for n in range(2,10**5) if sum([n/p for p in primefactors(n)]) +1 == n] # Chai Wah Wu, Aug 20 2014
    

Formula

A031971(a(n)) (mod a(n)) = A233045(n). - Jonathan Sondow, Dec 11 2013
A069359(a(n)) = a(n) - 1. - Jonathan Sondow, Apr 16 2014
a(n) == 36*(n-2) + 6 (mod 288) for n = 2,3,..,8. - Kieren MacMillan and Jonathan Sondow, Sep 20 2017

A235137 a(n) = Sum_{k = 1..n} k^phi(n), where phi(n) = A000010(n).

Original entry on oeis.org

1, 3, 14, 30, 979, 91, 184820, 8772, 978405, 25333, 40851766526, 60710, 36720042483591, 19092295, 5666482312, 9961449608, 76762718946972480009, 105409929, 164309788542828686799730, 70540730666, 15909231318568907, 67403375450475, 1433191209985108404653810959324, 351625763020, 15975648280734359596251725645
Offset: 1

Views

Author

Jonathan Sondow and Emmanuel Tsukerman, Jan 03 2014

Keywords

Comments

a(n) == -1 (mod n) if and only if n is prime or is a Giuga number A007850.
a(n) == 1 (mod n) if (and probably only if) n is a primary pseudoperfect number A054377.

Examples

			a(4) = 30 since 1^(phi(4)) + 2^(phi(4)) + 3^(phi(4)) + 4^(phi(4))= 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30.
a(5) = 979, since phi(5) = 4 and 1^4 + 2^4 + 3^4 + 4^4 + 5^4 = 1 + 16 + 81 + 256 + 625 = 979.
a(6) = 91, since phi(6) = 2 and 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[PowerMod[i, EulerPhi@n, n], {i, n}]
  • PARI
    a(n) = sum(k=1, n , k^eulerphi(n)); \\ Michel Marcus, Oct 21 2015

Formula

a(n) (mod n) = A235138(n).
Showing 1-3 of 3 results.