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.

A334245 Let S = A061228, T = A070229, and F = S o T (where o denotes composition). Then a(n) is the least number in the intersection of the sequences F^k(n) and F^k(S(n)), k>=0, or the intersection of the sequences T(F^k(n)) and T(F^k(S(n))), k>=0, or 0 if these intersections are empty.

Original entry on oeis.org

12, 12, 12, 15, 12, 30, 15, 21, 15, 77, 21, 77, 30, 77, 30, 77, 30, 57, 77, 30, 77, 91, 77, 77, 77, 77, 35, 105, 77, 133, 77, 91, 77, 60, 77, 111, 57, 77, 77, 133, 77, 165, 91, 65, 91, 147, 60, 105, 65, 70, 65, 165, 77, 105, 105, 91, 105, 180, 133, 231, 133, 91, 105, 105, 105, 234, 105, 147, 91
Offset: 2

Views

Author

Ali Sada and Robert Israel, Apr 19 2020

Keywords

Comments

60420 >= a(n) > 0 for n <= 6042, but either a(6043) = 0 or a(6043) > 10^30.
If p is prime, a(p) = a(2*p).
From n = 5, 55, 2, 24, 245, ... begin successive strings of exactly 1, 2, 3, 4, 5, ... identical merging points that are: 15, 105, 12, 77, 713, ... . - Bernard Schott, Jul 04 2020

Examples

			For n=5, 15 = T(F(5)) = T(S(5)) so a(5) = 15.
For n=6, 12 = F(6) = F(S(6)) so a(6) = 12.
For n=10, 15 = T(10) = T(F(10)) so a(10) = 15.
		

Crossrefs

Programs

  • Maple
    S:= x -> x + min(numtheory:-factorset(x)):
    T:= x -> x + max(numtheory:-factorset(x)):
    f:= proc(n) local s,t,i,j,ct;
      s:= S(n); i:= 0; t:= T(n); j:= 1;
      for ct from 1 to 1000 do
        if s = t and i=j then return s fi;
        if s <= t then
          if i = 0 then s:= T(s); i:= 1
          else s:= S(s); i:= 0
          fi
        elif j = 0 then t:= T(t); j:= 1
        else t:= S(t); j:= 0
        fi
      od;
      0
    end proc:
    map(f, [$2..100]);
  • Mathematica
    S[x_] := x + FactorInteger[x][[1, 1]];
    T[x_] := x + FactorInteger[x][[-1, 1]];
    f[n_] := Module[{s = S[n], t = T[n], i = 0, j = 1, ct}, For[ct = 1, ct <= 1000, ct++, If[s == t && i == j, Return[s]]; If[s <= t, If[i == 0, s = T[s]; i = 1, s = S[s]; i = 0], If[j == 0, t = T[t]; j = 1, t = S[t]; j = 0]]]; 0];
    f /@ Range[2, 100] (* Jean-François Alcover, Aug 15 2020, after Maple *)

Formula

a(2^m+2) = a(2^m). - Bernard Schott, Jun 27 2020