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A265100 a(n) = 9*A005836(n) + 5, n >= 1.

%I A265100 #31 Dec 23 2024 14:53:44
%S A265100 5,14,32,41,86,95,113,122,248,257,275,284,329,338,356,365,734,743,761,
%T A265100 770,815,824,842,851,977,986,1004,1013,1058,1067,1085,1094,2192,2201,
%U A265100 2219,2228,2273,2282,2300,2309,2435,2444,2462,2471,2516,2525
%N A265100 a(n) = 9*A005836(n) + 5, n >= 1.
%C A265100 Let C(m) denote the m-th Catalan number (A000108). Let == denote congruence and =!= its negation. _Vladimir Reshetnikov_ asked (see link) how many n exist such that C(n) == 1 (mod 6). It was pointed out by _Robert Israel_ that the only known n are in {1,3,31,255}. Since C(n) is odd if and only if n = 2^m - 1, for some m, _Emmanuel Vantieghem_ (see links) stated the stronger conjecture that C(2^n-1) == 0 (mod 3), for all n>8. This is the motivation for the following.
%C A265100 If n is an integer such that the congruences C(n) == 0 (mod 3) and C(n-1) =!= 0 (mod 3) hold simultaneously, then we call n a "block number." A sequence {n, n+1, ..., n+k-1} of consecutive numbers is called a "block" (of order k), if C(n+i) == 0 (mod 3), for all i such that 0 <= i < k, and if C(n-1) =!= 0 (mod 3) (i.e., if n is a block number) and C(n+k) =!= 0 (mod 3).
%C A265100 If m is an integer such that the congruences C(m) =!= 0 (mod 3) and C(m-1) == 0 (mod 3) hold simultaneously, then we call m a "gap number." A sequence {m, m+1, ..., m+j-1} of consecutive numbers is called a "gap" (of order j), if C(m+i) =!= 0 (mod 3), for all i such that 0 <= i < j, and if C(m-1) == 0 (mod 3) (i.e., if m is a gap number) and C(m+j) == 0 (mod 3). (The sequence A265104 is conjectured to contain all possible gap numbers.) If C(n) == 0 (mod 3), then we say that n is "gap-avoiding."
%C A265100 It follows that if {n, n+1, ..., n+k-1} is a block with block number n, then n+k is a gap number, and if {m, m+1, ..., m+j-1} is a gap with gap number m, then m+j is a block number.
%C A265100 Conjecture 1: The sequence contains all possible block numbers.
%C A265100 Conjecture 2: If m is a block number, then 3*m - 1 is a block number.
%C A265100 Conjecture 3: If C(n) == 0 (mod 3), then C(3*n-1) == 0 (mod 3) or, what is the same thing, if n lies in a block, then 3*n - 1 lies in a block.
%C A265100 Conjecture 4: Assuming that A265104 contains all possible gap numbers, let B(n) denote the block with block number a(n), n >= 1, so that B(n) = {a(n), a(n)+1, ..., A265104(n)-1}. The (flattened) sequence {B(1), B(2), ...} of blocks contains all numbers m such that the base 3 representations of m and m+1 both contain at least one 2 and is identical to A111018.
%C A265100 Conjecture 5: C(n) == 0 (mod 3) if and only if the base 3 representations of n and n + 1 both contain at least one 2. [This conjecture has been proved by _Robert Israel_ (see link for the proof)].
%C A265100 Theorem 1: The following statements are equivalent to Vantieghem's conjecture stated above: (i) For all m>8, 2^m-1 is gap-avoiding; (ii) C(2^n-1) == 0 (mod 3) if and only if the base 3 representations of 2^n - 1 and 2^n both contain at least one 2.
%C A265100 Proof: For (i), the statement obviously follows from the definitions, and (ii) follows from the proof of Conjecture 5.
%H A265100 Robert Israel, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-December/015749.html">A000108(n) == 1 (mod 6)</a>, Sequence Fans Mailing List, December 2015.
%H A265100 Vladimir Reshetnikov, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-November/015578.html">A000108(n) == 1 (mod 6)</a>, Sequence Fans Mailing List, November 2015.
%H A265100 Emmanuel Vantieghem, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-November/015629.html">A000108(n) == 1 (mod 6)</a>, Sequence Fans Mailing List, November 2015.
%H A265100 Emmanuel Vantieghem, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-November/015648.html">A000108(n) == 1 (mod 6)</a>, Sequence Fans Mailing List, November 2015.
%F A265100 Conjecture: a(n) = A265104(n) - A085296(n).
%t A265100 a005836[1] := 0; a005836[n_] := If[OddQ[n], 3*a005836[Floor[(n + 1)/2]], a005836[n - 1] + 1]; a265100[n_] := 9*a005836[n] + 5; Table[a265100[n], {n, 46}]
%t A265100 5 + 9 Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* _Vincenzo Librandi_, Dec 03 2015 *)
%Y A265100 A000108 (Catalan numbers).
%Y A265100 Cf. A085296, A111018, A265104.
%K A265100 nonn
%O A265100 1,1
%A A265100 _L. Edson Jeffery_, Dec 01 2015