A341358 Deficient numbers k > 1 such that k*p is abundant for all primes p dividing k.

315, 884, 1155, 26325, 27027, 40365, 44650, 55335, 63248, 70564, 72675, 85936, 100804, 106425, 116624, 130815, 145222, 224750, 244036, 318250, 321470, 421515, 489645, 526688, 531310, 569625, 629145, 702405, 730125, 820808, 829521, 852776, 885928, 933224, 969969

Offset: 1

Author

Jianing Song, Feb 09 2021

Keywords

• nonn

Comments

These numbers are quite close to being abundant.

Note that if k is a perfect number, then k*m is abundant for all m > 1. For deficient k we cannot expect this to happen since k*p must be deficient for all primes p that are large enough. However, it seems that the smallest p making k*p deficient is big compared with k. For example, the smallest prime p such that 315*p is deficient is p = 107, and the smallest prime p such that 884*p is deficient is p = 443.

It's actually possible that a multiple of a term (other than the term itself) is again in this sequence. However, it seems that this rarely happens. For example, both 1329548 and 1329548 * 16619 are terms.

Examples

			The sum of divisors of 315 = 3^2 * 5 * 7 is 624 < 315 * 2, so 315 is deficient; however, all of 315 * 3, 315 * 5 and 315 * 7 are abundant, thus 315 is a term.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..1000 (terms 1..303 from Jianing Song)

Crossrefs

Cf. A005101 (abundant numbers), A005100 (deficient numbers).

Programs

Mathematica

f[p_, e_] := (p^(e + 1) - 1)/(p - 1); ff[p_, e_] := (p^(e + 2) - 1)/(p^(e + 2) - p); q[n_] := Module[{fct = FactorInteger[n], s}, s = (Times @@ f @@@ fct)/n; s < 2 && AllTrue[ff @@@ fct, # > 2/s &]]; Select[Range[2, 10^5], q] (* Amiram Eldar, Mar 14 2024 *)

PARI

isA341358(n) = if(sigma(n) < 2*n && n > 1, my(d=factor(n), k=omega(n)); for(i=1, k, if(!isA005101(n*d[i,1]), return(0))); return(1), 0) \\ see A005101 for its program

PARI

is(n) = {my(f = factor(n), p = f[, 1], e = f[, 2], s = 2/sigma(f,-1)); if(n == 1 || s <= 1, return(0)); for(i = 1, #p, if((p[i]^(e[i]+2) - 1)/(p[i]^(e[i]+2) - p[i]) < s, return(0))); 1;} \\ Amiram Eldar, Mar 14 2024