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A265100
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a(n) = 9*A005836(n) + 5, n >= 1.
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5
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5, 14, 32, 41, 86, 95, 113, 122, 248, 257, 275, 284, 329, 338, 356, 365, 734, 743, 761, 770, 815, 824, 842, 851, 977, 986, 1004, 1013, 1058, 1067, 1085, 1094, 2192, 2201, 2219, 2228, 2273, 2282, 2300, 2309, 2435, 2444, 2462, 2471, 2516, 2525
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OFFSET
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1,1
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COMMENTS
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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.
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).
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."
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.
Conjecture 1: The sequence contains all possible block numbers.
Conjecture 2: If m is a block number, then 3*m - 1 is a block number.
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.
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.
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)].
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.
Proof: For (i), the statement obviously follows from the definitions, and (ii) follows from the proof of Conjecture 5.
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LINKS
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FORMULA
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MATHEMATICA
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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}]
5 + 9 Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* Vincenzo Librandi, Dec 03 2015 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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