

A054377


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


26




OFFSET

1,1


COMMENTS

Primary pseudoperfect numbers are the solutions of the "differential equation" n' = n1, 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 N1 or N+1, then in fact N lies between twin primes N1, N+1? See A235139.  Jonathan Sondow, Jan 05 2014
Also, integers n > 1 such that A069359(n) = n  1.  Jonathan Sondow, Apr 16 2014


LINKS

Table of n, a(n) for n=1..8.
M. A. Alekseyev, J. M. Grau, A. M. OllerMarcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT]
W. Butske, L. M. Jaje, and D. R. Mayernik, On the Equation Sum_{pN} 1/p + 1/N = 1, Pseudoperfect numbers and partially weighted graphs, Math. Comput., 69 (1999), 407420. [Title corrected by Jonathan Sondow, Apr 11 2012]
J. M. Grau, A. M. OllerMarcen, and J. Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with nm, arXiv:1309.7941 [math.NT], 2013.
John Machacek, Egyptian Fractions and Prime Power Divisors, arXiv:1706.01008 [math.NT], 2017.
J. Sondow and K. MacMillan, Reducing the ErdosMoser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2, Integers 11 (2011), #A34.
J. Sondow and K. MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the ErdosMoser equation, Amer. Math. Monthly, 124 (2017) 232240.
J. Sondow and E. Tsukerman, The padic order of power sums, the ErdosMoser equation, and Bernoulli numbers, arXiv:1401.0322 [math.NT], 2014; see section 4.
Eric Weisstein's World of Mathematics, Primary pseudoperfect number.
Wikipedia, Primary pseudoperfect number.
OEIS Wiki, Primary pseudoperfect numbers.


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*(n2) + 6 (mod 288) for n = 2,3,..,8.  Kieren MacMillan and Jonathan Sondow, Sep 20 2017


EXAMPLE

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 padic order . . .", Theorem 8 and Example 1.  Jonathan Sondow, Jan 06 2014


PROG

(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
(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


CROSSREFS

Cf. A005835, A007850, A069359, A168036, A190272, A191975, A203618, A216825, A216826, A230311, A235137, A235138, A235139, A236433.
Sequence in context: A123137 A014117 A242927 * A230311 A276416 A007018
Adjacent sequences: A054374 A054375 A054376 * A054378 A054379 A054380


KEYWORD

nonn,more,hard


AUTHOR

Eric W. Weisstein


STATUS

approved



