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A054377
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Primary pseudoperfect numbers: numbers n > 1 such that 1/n + sum 1/p = 1, where the sum is over the primes p | n.
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39
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. 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]
J. M. Grau, A. M. Oller-Marcén, and D. Sadornil, On µ-Sondow Numbers, arXiv:2111.14211 [math.NT], 2021.
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FORMULA
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a(n) == 36*(n-2) + 6 (mod 288) for n = 2,3,..,8. - Kieren MacMillan and Jonathan Sondow, Sep 20 2017
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EXAMPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A005835, A007850, A069359, A168036, A190272, A191975, A203618, A216825, A216826, A230311, A235137, A235138, A235139, A236433.
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KEYWORD
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nonn,more,hard
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AUTHOR
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STATUS
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approved
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