A127727 -id:A127727 - cp's OEIS frontend

A127724 k-imperfect numbers for some k >= 1.

1, 2, 6, 12, 40, 120, 126, 252, 880, 2520, 2640, 10880, 30240, 32640, 37800, 37926, 55440, 75852, 685440, 758520, 831600, 2600640, 5533920, 6917400, 9102240, 10281600, 11377800, 16687440, 152182800, 206317440, 250311600, 475917120, 715816960, 866829600

Offset: 1

T. D. Noe, Jan 25 2007

For prime powers p^e, define a multiplicative function rho(p^e) = p^e - p^(e-1) + p^(e-2) - ... + (-1)^e. A number n is called k-imperfect if there is an integer k such that n = k*rho(n). Sequence A061020 gives a signed version of the rho function. As with multiperfect numbers (A007691), 2-imperfect numbers are also called imperfect numbers. As shown by Iannucci, when rho(n) is prime, there is sometimes a technique for generating larger imperfect numbers.
Zhou and Zhu find 5 more terms, which are in the b-file. - T. D. Noe, Mar 31 2009
Does this sequence follow Benford's law? - David A. Corneth, Oct 30 2017
If a term t has a prime factor p from A065508 with exponent 1 and does not have the corresponding prime factor q from A074268, then t*p*q is also a term. - Michel Marcus, Nov 22 2017
For n >= 1, the least n-imperfect numbers are 1, 2, 6, 993803899780063855042560. - Michel Marcus, Feb 13 2018
For any m > 0, if n*p^(2m-1) is k-imperfect, q = rho(p^(2m)) is prime and gcd(pq,n) = 1, then n*p^(2m)*q is also k-imperfect. - M. F. Hasler, Feb 13 2020

			126 = 2*3^2*7, rho(126) = (2-1)*(9-3+1)*(7-1) = 42.  3*42 = 126, so 126 is 3-imperfect. - _Jud McCranie_ Sep 07 2019

R. K. Guy, Unsolved Problems in Theory of Numbers, Springer, 1994, B1.

Giovanni Resta, Table of n, a(n) for n = 1..50 (terms < 10^13, a(1)-a(39) from T. D. Noe (from Iannucci, Zhou, and Zhu), a(40)-a(44) from Donovan Johnson)
David A. Corneth, Conjectured to be the terms up to 10^28
Douglas E. Iannucci, On a variation of perfect numbers, INTEGERS: Electronic Journal of Combinatorial Number Theory, 6 (2006), #A41.
Donovan Johnson, 43 terms > 2*10^11
Andrew Lelechenko, 4-imperfect numbers, Apr 19 2014.
Michel Marcus, More 4-imperfect numbers, Nov 07 2017.
Michel Marcus, Least known integers with small denominator-fractional k's, Feb 13 2018.
Allan Wechsler, Some progress in k-imperfect numbers (A127724), Seqfan, Feb 13 2020. Gives a first instance of a 5-imperfect number.
Weiyi Zhou and Long Zhu, On k-imperfect numbers, INTEGERS: Electronic Journal of Combinatorial Number Theory, 9 (2009), #A01.

Cf. A127725 (2-imperfect numbers), A127726 (3-imperfect numbers), A127727 (related primes), A309806 (the k values).
Cf. A061020 (signed version of rho function), A206369 (the rho function).
Cf. A065508, A074268.

f[p_,e_]:=Sum[(-1)^(e-k) p^k, {k,0,e}]; rho[n_]:=Times@@(f@@@FactorInteger[n]); Select[Range[10^6], Mod[ #,rho[ # ]]==0&]

isok(n) = denominator(n/sumdiv(n, d, d*(-1)^bigomega(n/d))) == 1; \\ Michel Marcus, Oct 28 2017

upto(ulim) = {res = List([1]); rhomap = Map(); forprime(p = 2, 3, for(i = 1, logint(ulim, p), mapput(rhomap, p^i, rho(p^i)); iterate(p^i, mapget(rhomap, p^i), ulim))); listsort(res, 1); res}
iterate(m, rhoo, ulim) = {my(c); if(m / rhoo == m \ rhoo, listput(res, m); my(frho = factor(rhoo)); for(i = 1, #frho~, if(m%frho[i, 1] != 0, for(e = 1, logint(ulim \ m, frho[i, 1]), if(mapisdefined(rhomap, frho[i, 1]^e) == 0, mapput(rhomap, frho[i, 1]^e, rho(frho[i, 1]^e))); iterate(m * frho[i, 1]^e, rhoo * mapget(rhomap, frho[i, 1]^e), ulim)); next(2))))}
rho(n) = {my(f = factor(n), res = q = 1); for(i=1, #f~, q = 1; for(j = 1, f[i, 2], q = -q + f[i, 1]^j); res * =q); res} \\ David A. Corneth, Nov 02 2017

A127724_vec=concat(1, select( {is_A127724(n)=!(n%A206369(n))}, [1..10^5]*2))
  /* It is known that the least odd term > 1 is > 10^49. This code defines an efficient function is_A127724, but A127724_vec is better computed with upto(.) */
  A127724(n)=A127724_vec[n] \\ Used in other sequences. - M. F. Hasler, Feb 13 2020

Small correction in name from Michel Marcus, Feb 13 2018

A383649 Numbers k such that A206369(k) is prime.

Original entry on oeis.org

3, 4, 6, 8, 9, 16, 18, 49, 64, 81, 98, 162, 169, 338, 625, 729, 1024, 1250, 1458, 4096, 4489, 6241, 8978, 12482, 14641, 19321, 22801, 26569, 29282, 37249, 38642, 45602, 53138, 65536, 74498, 113569, 121801, 143641, 208849, 227138, 243602, 262144, 287282, 292681, 375769, 413449, 417698

Offset: 1

Shreyansh Jaiswal, May 04 2025

nonn

The corresponding primes are 2, 3, 2, 5, 7, 11, 7, 43, 43, 61, 43, 61, 157, 157, 521... Sorting and removing duplicates from this sequence of primes appears to give A127727.

All the terms are either prime powers or twice prime powers since A206369 is multiplicative and A206369(n) = 1 only for n = 1 and 2. 3 is the only prime term since A206369(p) = p-1 for a prime p. - Amiram Eldar, May 04 2025

			3 is a term since A206369(3) = 2, and 2 is a prime.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..122 from Shreyansh Jaiswal)

Cf. A127727, A206369.

seq[max_] := Module[{s = {3, 6}, e}, Do[e = Select[Range[2, Floor[Log[p, max]]], PrimeQ[Sum[(-1)^(# - k)*p^k, {k, 0, #}]] &]; s = Join[s, p^e]; If[p > 2, e = Select[e, # <= Floor[Log[p, max/2]] &]; s = Join[s, 2*p^e]], {p, Prime[Range[Floor[Sqrt[max]]]]}]; Sort[s]]; seq[500000] (* Amiram Eldar, May 04 2025 *)

isok(k) = ispseudoprime(sumdiv(k, d, eulerphi(k/d) * issquare(d))); \\ Michel Marcus, May 04 2025

from math import prod; from sympy import *
def ok(n): return isprime(prod((lambda x:x[0]+int((x[1]<<1)>=p+1))(divmod(p**(e+1), p+1)) for p, e in factorint(n).items())) # using code from Chai Wah Wu in A206369
print([k for k in range(1,10**5) if ok(k)])

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A127724 k-imperfect numbers for some k >= 1.

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A383649 Numbers k such that A206369(k) is prime.

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