Bernard Schott, Nov 19 2021

Proofs that all the bounds proposed by David Corneth in A349188 are indeed the corresponding a(n) for n = 6, 7, 8, 9.

1) Proof that A349188(6) = 1440

 David Corneth proposes A3491886(6) >= 1440, we prove here that indeed A3491886(6) = 1440. 
 
 First, we prove two short lemmas that will be used for the proof by induction.

  1.1) Lemma 1: every odd integer >= 483 is a 6-phile number.

     If m is a 5-phile number, then q = 1 + 2*m is a 6-phile odd number.
         Proof: if m = b_1 + b_2 + b_3 + b_4 +  b_5 is a 5-phile number,
         then, q = 1 + 2b_1 + 2b_2 +2b_3 +2b_4 +2b_5 satisfies the definition of the 6-phile numbers.

     Indeed, if m is a k-phile number, then q = 1 + 2*m is a (k+1)-phile odd number.

     Now, we know that the largest 5-phobe is A349188(5) = 240, so every integer >= 241 is 5-phile and 
         every odd integer >= 1 + 2*241 = 483 is 6-phile (QED).

  1.2) Lemma 2: If m is 6-phile, then, q = 2*m is also 6-phile.

        Proof: if m = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 is a 6-phile number, 
        then q = 2b_1 + 2b_2 +2b_3 +2b_4 +2b_5 + 2b_6 is also a 6-phile number, according to the definition.

     Indeed, if m is k-phile, then, q = 2*m is also k-phile, for k >= 2.


  1.3) Proof:

      From the computer results from David Corneth, we know that in the interval [1441, 2048], all terms are 6-phile.

      Now, proof by induction, that all terms >= 2048 are 6-phile.

       -> Basis: from the calculations of David Corneth up to 50000, 
             we know that in the interval [2048, 4096] = [2^11, 2^12], all terms are 6-phile.

       -> Induction step:

          Suppose that all terms in the interval [2^k, 2^(k+1)] are 6-phile for k >= 11; then, 
           because the two lemmas, lemma 1 for odd numbers and lemma 2 for even numbers, 
           in the interval [2^(k+1), 2^(k+2)] all terms are also 6-phile.

          By induction, we know now that all terms >= 2048 are 6-phile, hence all terms >= 1441 are 6-phile.

       Conclusion: 1440 is 6-phobe and every integer >= 1441 is 6-phile, hence A349188(6) = 1440.


2) Proof for A349188(7) = 2400 

 David Corneth proposes A3491886(7) >= 2400, we prove here that indeed A3491886(7) = 2400. 

   Lemma 1 becomes: every odd integer >= 2883 is a 7-phile number, because 2883 = 1 + 2*1441 when 1440 = A3491886(6).
   Lemma 2 becomes: If m is 7-phile, then, q = 2*m is also 7-phile.

   Proof: 
     From the calculations of David Corneth, we know that in the interval [2401, 4096], all terms are 7-phile.
     From the calculations of David Corneth up to 50000, 
       we know that in the interval [4096, 8192] = [2^12, 2^13], all terms are 7-phile.
     Suppose that all terms in the interval [2^k, 2^(k+1)] are 7-phile for k >= 12; 
       then, because the two lemmas, in the interval [2^(k+1), 2^(k+2)] all terms are also 7-phile.
     By induction, we know that all terms >= 4096 are 7-phile, hence all terms >= 2401 are 7-phile.

    Conclusion: 2400 is 7-phobe and every integer >= 2401 is 7-phile, hence A349188(7) = 2400.


3) Proof for A349188(8) = 7440 

 David Corneth proposes A3491886(8) >= 7440, we prove here that indeed A3491886(8) = 7440. 

    Lemma 1 becomes: every odd integer >= 4803 is a 8-phile number, because 4803 = 1 + 2*2401 when 2400 = A3491886(7).
    Lemma 2 becomes: If m is 8-phile, then, q = 2*m is also 8-phile.

    Proof: 
      From the calculations of David Corneth, we know that in the interval [7441, 8192], all terms are 8-phile.
      From the calculations of David Corneth up to 100000, 
         we know that in the interval [8192, 16384] = [2^13, 2^14], all terms are 8-phile.
      Suppose that all terms in the interval [2^k, 2^(k+1)] are 8-phile for k >= 13; 
        then, because the two lemmas, in the interval [2^(k+1), 2^(k+2)] all terms are also 8-phile.
      By induction, we know that all terms >= 8192 are 8-phile, hence all terms >= 7441 are 8-phile.

    Conclusion: 7440 is 8-phobe and every integer >= 7441 is 8-phile, hence A349188(8) = 7440.


4) For A349188(9) = 25920 

 David Corneth proposes A3491886(9) >= 25920, we prove here that indeed A3491886(9) = 25920. 

    Lemma 1 becomes: every odd integer >= 14883 is a 9-phile number, because 14883 = 1 + 2*7441 when 7440 = A3491886(8).
    Lemma 2 becomes: If m is 9-phile, then, q = 2*m is also 9-phile.

    Proof: 
       From the calculations of David Corneth, we know that in the interval [25921, 32768], all terms are 9-phile.
       From the calculations of David Corneth up to 100000, 
          we know that in the interval [32768, 65536] = [2^15, 2^16], all terms are 9-phile.
       Suppose that all terms in the interval [2^k, 2^(k+1)] are 9-phile for k >= 15; 
          then, because the two lemmas, in the interval [2^(k+1), 2^(k+2)] all terms are also 9-phile.
       By induction, we know that all terms >= 32768 are 9-phile, hence all terms >= 25921 are 9-phile.

    Conclusion: 25920 is 9-phobe and every integer >= 25921 is 9-phile, hence A349188(9) = 25920.


5) Consequence

   Bounds found by David for A349189(6), A349189(7), A349189(8) and A349189(9) are indeed the exact values for these terms.


6) Practical algorithm 

  If we know that A349188(n-1) = t, then all odd integers >= 1 + 2 * (t+1) = 2*t+3 are n-phile.
  If we get by programme A349188(n) >= q and we suspect A349188(n) = q.
    Let 2^r the smallest power of 2 that is > q. 
    If q > 2*t+3, and,
    If all numbers in [q+1, 2^(r+1)] are n-phile, 
       then A349188(n) = q.