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.

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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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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

A236433 List of primes generated by factoring successive primary pseudoperfect numbers (A054377).

Original entry on oeis.org

2, 2, 3, 2, 3, 7, 2, 3, 7, 43, 2, 3, 11, 23, 31, 2, 3, 11, 23, 31, 47059, 2, 3, 11, 17, 101, 149, 3109, 2, 3, 11, 23, 31, 47059, 2217342227, 1729101023519
Offset: 1

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Author

Jonathan Sondow, Jan 25 2014

Keywords

Comments

It is unknown whether there are infinitely many primary pseudoperfect numbers or any odd ones.
See A054377 for other comments, references, links, etc.
For prime factors of Giuga numbers, see A236434; of terms in Sylvester's sequence, see A126263.

Examples

			2 = 2.
6 = 2 * 3.
42 = 2 * 3 * 7.
1806 = 2 * 3 * 7 * 43.
47058 = 2 *  3 * 11 * 23 * 31.
2214502422 =  2 *  3 * 11 * 23 * 31 * 47059.
52495396602 =  2 * 3 * 11 * 17 * 101 * 149 * 3109.
8490421583559688410706771261086 = 2 *  3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519.
		

Crossrefs

Programs

  • Mathematica
    A054377 = Cases[Import["https://oeis.org/A054377/b054377.txt", "Table"], {, }][[All, 2]];
    First /@ Flatten[FactorInteger[A054377], 1] (* Robert Price, Mar 14 2020 *)

A216826 Largest prime factor of the n-th primary pseudoperfect number A054377(n).

Original entry on oeis.org

2, 3, 7, 43, 31, 47059, 3109, 1729101023519
Offset: 1

Views

Author

Jonathan Sondow, Sep 20 2012

Keywords

Comments

See A054377 for comments, references, and links.

Examples

			42 = 2*3*7 is the third primary pseudoperfect number, so a(3) = 7.
		

Crossrefs

Programs

Showing 1-3 of 3 results.