A203618 -id:A203618 - cp's OEIS frontend

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

2, 6, 42, 1806, 47058, 2214502422, 52495396602, 8490421583559688410706771261086

Offset: 1

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

			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

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]
W. Butske, L. M. Jaje, and D. R. Mayernik, On the Equation Sum_{p|N} 1/p + 1/N = 1, Pseudoperfect numbers and partially weighted graphs, Math. Comput., 69 (1999), 407-420. [Title corrected by _Jonathan Sondow_, Apr 11 2012]
J. M. Grau, A. M. Oller-Marcen, and J. Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv:1309.7941 [math.NT], 2013.
J. M. Grau, A. M. Oller-Marcén, and D. Sadornil, On µ-Sondow Numbers, arXiv:2111.14211 [math.NT], 2021.
John Machacek, Egyptian Fractions and Prime Power Divisors, arXiv:1706.01008 [math.NT], 2017.
J. Sondow and K. MacMillan, Reducing the Erdos-Moser 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 Erdos-Moser equation, Amer. Math. Monthly, 124 (2017) 232-240; arXiv:math/1812.06566 [math.NT], 2018.
J. Sondow and E. Tsukerman, The p-adic order of power sums, the Erdos-Moser 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.

Cf. A005835, A007850, A069359, A168036, A190272, A191975, A203618, A216825, A216826, A230311, A235137, A235138, A235139, A236433.

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 *)

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

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

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

A203617 Numbers m such that (m'-1)' = m+1, where m' denotes the arithmetic derivative of m.

Original entry on oeis.org

30, 210, 246, 858, 1722, 66198, 235290, 282342, 1929378, 1976394, 2214408306

Offset: 1

Paolo P. Lava, Jan 20 2012

The differential equation whose solutions are the Giuga numbers is m' = k*m+1, with k a positive integer. Let us rewrite the equation as m'-1 = k*m and then take the derivative: (m'-1)' = (k*m)' = k'*m + k*m' = k'*m + k*(k*m+1) = (k'+k^2)*m+k.
Let k=1: (m'-1)' = m+1. The solutions of this equation are the Giuga numbers plus pairs of numbers (x,y) for which x' = y+1 and y' = x+1.
A007850 is a subsequence of this sequence.
a(11) > 10^9. - Michel Marcus, Nov 05 2014
a(12) > 10^10. - Giovanni Resta, Jun 04 2016

			235290' = 282343; (282343 - 1)' = 282342' = 235291 = 235290 + 1, so 235290 is a term.
282342' = 235291; (235291 - 1)' = 235290' = 282343 = 282342 + 1, so 282342 is a term.

Cf. A007850, A185222, A203618.

with(numtheory);
P:=proc(i)
local a,n,p,pfs;
for n from 1 to i do
  pfs:=ifactors(n)[2]; a:=n*add(op(2,p)/op(1,p),p=pfs) ;
  pfs:=ifactors(a-1)[2]; a:=(a-1)*add(op(2,p)/op(1,p),p=pfs) ;
  if a=n+1 then print(n); fi;
od;
end:
P(10000000);

ad(n) = sum(i=1, #f=factor(n)~, n/f[1, i]*f[2, i]);
isok(n) = my(m = ad(n)-1); (m) && ad(m) == n+1; \\ Michel Marcus, Nov 05 2014

a(11) from Giovanni Resta, Jun 04 2016

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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.

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A203617 Numbers m such that (m'-1)' = m+1, where m' denotes the arithmetic derivative of m.

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A185222 Composite numbers n such that (n'+1)' = n', where n' is the arithmetic derivative of n.

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