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.

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A309769 Even numbers m having at least one odd prime divisor p for which there exists a positive integer k < p-1 such that p-k|m-k.

Original entry on oeis.org

20, 28, 42, 44, 50, 52, 66, 68, 70, 76, 78, 80, 88, 92, 102, 104, 110, 112, 114, 116, 124, 130, 136, 138, 140, 148, 152, 154, 156, 164, 170, 172, 174, 176, 182, 184, 186, 188, 190, 196, 200, 204, 208, 212, 222, 228, 230, 232, 236, 238, 242, 244, 246, 248, 252
Offset: 1

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Author

David James Sycamore, Aug 16 2019

Keywords

Comments

Complement in A005843 of A309239. Every odd number > 1 has the property mentioned in Name, but these are the only even numbers with this property. No term is either a power of 2 or a semiprime. A number m is a term if and only if m = 2rp, where r >= 2, and p is a prime > q, the smallest prime divisor of 2r-1 (k=p-q). For any given r, 2rz is the smallest multiple of 2r in this sequence, where z=nextprime(q). If m = 2rp is a term and 2r-1 is prime, then p is the greatest prime divisor of m (the converse is not true; e.g., m=70=10*7).

Examples

			20 = 4*5 is a term because with k=2, 5-k|20-k.
66 = 6*11 is a term (k=6), although when expressed as 66=22*3 no k exists.
110 = 10*11 = 22*5 is a term for two reasons, since with both of its odd prime factors it has the required property; 5-2|110-2 and 11-8|110-8. This is the smallest term having two distinct odd prime factors, both of which have the above property (see A309780, A309781).
		

Crossrefs

Programs

  • Mathematica
    kQ[n_, p_] := Module[{ans = False}, Do[If[Divisible[n - k, p - k], ans = True; Break[]], {k, 1, p - 2}]; ans]; aQ[n_] := EvenQ[n] && Length[(p = FactorInteger[ n][[2 ;; -1, 1]])] > 0 && AnyTrue[p, kQ[n, #] &]; Select[Range[252], aQ] (* Amiram Eldar, Aug 17 2019 *)
  • PARI
    getk(p, m) = {for (k=1, p-2, if (((m-k) % (p-k)) == 0, return(k)););}
    isok(m) = {if ((m % 2) == 0, my(f = factor(m)[,1]~); if (#f == 1, return (0)); for (i=2, #f, if (getk(f[i], m), return(1));););} \\ Michel Marcus, Aug 26 2019
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