cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A360328 Numbers k such that A360327(k) > 2*k.

Original entry on oeis.org

7425, 8415, 22275, 25245, 37125, 42075, 46035, 66825, 75735, 76725, 81675, 92565, 101475, 111375, 126225, 138105, 143055, 182655, 185625, 200475, 210375, 227205, 230175, 245025, 260865, 277695, 304425, 334125, 345015, 355725, 378675, 383625, 408375, 414315, 429165
Offset: 1

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Author

Amiram Eldar, Feb 03 2023

Keywords

Comments

Analogous to abundant numbers (A005101) with divisors that are restricted to numbers that have only prime-indexed prime factors.
The abundancy index of numbers in A076610 (i.e., numbers whose prime factors are only prime-indexed primes) is bounded by P = Product_{p in A006450} p/(p-1) which seems to be less than 4 (see A076610). Therefore, there are no terms k of A076610 with sigma(k) >= 4*k, or equivalently, no even terms in this sequence, and all the terms of this sequence are in A076610. Also, assuming that P < 15/4 = 3.75, there are no terms in this sequence that are coprime to 15.
Since P > 3 there are terms that are not divisible by 3. The least of them must be larger than Product_{k=1..21826870} A006450(k) = 3 * 5 * 11 * ... * 8958801613 > 10^206662375, because Product_{k=2..m} A006450(k)/(A006450(k)-1) > 2 only for m >= 21826870.
The least term that is not divisible by 5 is 789909738655399955305165431.
The least term that is not divisible by 11 is a(30) = 355725.
The least squarefree term is 14093057715.

Crossrefs

Intersection of A005101 (or A005231) and A076610.

Programs

  • Mathematica
    f[p_, e_] := If[PrimeQ[PrimePi[p]], (p^(e + 1) - 1)/(p - 1), 1]; s[n_] := Times @@ (f @@@ FactorInteger[n]); Select[Range[10^6], s[#] > 2*# &]
  • PARI
    is(n) = {my(f = factor(n), p = f[,1], e = f[,2]); prod(i = 1, #p, if(isprime(primepi(p[i])), (p[i]^(e[i]+1)-1)/(p[i]-1), 1)) > 2*n;}