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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A218011 Numbers n for which n’ = x’*y’, where x>0, y>0, n = x + y and n’, x’, y’ are the arithmetic derivatives of n, x, y.

Original entry on oeis.org

5, 7, 13, 19, 31, 43, 48, 55, 61, 73, 74, 87, 103, 106, 109, 117, 139, 146, 151, 159, 160, 178, 181, 193, 199, 202, 208, 212, 225, 229, 236, 241, 252, 267, 268, 271, 283, 285, 298, 313, 349, 357, 362, 386, 403, 411, 421, 433, 455, 463, 496, 511, 519, 523, 535
Offset: 1

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Author

Paolo P. Lava, Oct 18 2012

Keywords

Comments

The greatest prime in a twin primes couple is in the sequence. In fact if the twin primes are a and b, with a

Examples

			n= 612, x=85,  y=527; n’=1056, x’=22, y’=48 and 1056=22*48.
n= 752, x=361, y=391; n’=1520, x’=38, y’=40 and 1520=38*40.
n= 779, x=36,  y=743; n’=60,   x’=60, y’=1  and 60=60*1.
		

Crossrefs

Subsequences: A006512 (primes in this sequence), A370126 (k with a solution where both x and y are composite).

Programs

  • Maple
    with(numtheory);
    A218011:= proc(i)
    local a,b,c,n,p,pfs,q;
    for n from 1 to i do
    for q from 1 to trunc(n/2) do
      a:=q*add(op(2,p)/op(1,p),p= ifactors(q)[2]);
      b:=(n-q)*add(op(2,p)/op(1,p),p= ifactors(n-q)[2]);
      c:=n*add(op(2,p)/op(1,p),p= ifactors(n)[2]);
      if c=a*b then lprint(n,q,n-q); break; fi;
    od; od;
    end:
    A218011(1000000);
  • Mathematica
    dn[0] = 0; dn[1] = 0; dn[n_?Negative] := -dn[-n]; dn[n_] := Module[{f = Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Plus @@ (n*f[[2]]/f[[1]])]]; f[n_] := Select[Range[n/2], dn[#]*dn[n - #] == dn[n] &]; Select[Range[535], Length[f[#]] > 0 &] (* T. D. Noe, Oct 18 2012 *)
  • PARI
    up_to = 2^18;
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    v003415 = vector(up_to,n,A003415(n));
    isA218011(n) = { my(z=v003415[n]); for(x=2,ceil(n/2),if(!(z%v003415[x]), if(z==v003415[x]*v003415[n-x], return(1)))); (0); }; \\ Antti Karttunen, Feb 22 2024
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