

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; arXiv:math/1812.06566 [math.NT], 2018.
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



