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A341358
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Deficient numbers k > 1 such that k*p is abundant for all primes p dividing k.
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1
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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