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Numbers n for which there is a permutation p of (1,2,3,...,n) such that k+p(k) is a Catalan number for 1<=k<=n.
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%I #36 Nov 20 2016 09:04:12

%S 0,1,3,4,9,10,12,13,28,29,31,32,37,38,40,41,90,91,93,94,99,100,102,

%T 103,118,119,121,122,127,128,130,131,297,298,300,301,306,307,309,310,

%U 325,326,328,329,334,335,337,338,387,388,390,391,396,397,399,400,415,416

%N Numbers n for which there is a permutation p of (1,2,3,...,n) such that k+p(k) is a Catalan number for 1<=k<=n.

%C A001453 is a subsequence. - _Altug Alkan_, Sep 29 2016

%C n>=1 is in the sequence if and only if there is a Catalan number c such that c/2 <= n < c and c-n-1 is in the sequence. - _Robert Israel_, Nov 20 2016

%H Robert Israel, <a href="/A276986/b276986.txt">Table of n, a(n) for n = 1..10000</a>

%F a(i) + a(2^n+1-i) = A000108(n+1)-1 for 1<=i<=2^n. - _Robert Israel_, Nov 20 2016

%e 3 is in the sequence because the permutation (1,3,2) added termwise to (1,2,3) yields (2,5,5) and both 2 and 5 are Catalan numbers.

%p S:= {0}:

%p for i from 1 to 8 do

%p c:= binomial(2*i,i)/(i+1);

%p S:= S union map(t -> c - t - 1, S);

%p od:

%p sort(convert(S,list)); # _Robert Israel_, Nov 20 2016

%t CatalanTo[n0_] :=

%t Module[{n = n0}, k = 1; L = {};

%t While[CatalanNumber[k] <= 2*n, L = {L, CatalanNumber[k]}; k++];

%t L = Flatten[L]]

%t perms[n0_] := Module[{n = n0, S, func, T, T2},

%t func[k_] := Cases[CatalanTo[n], x_ /; 1 <= x - k <= n] - k;

%t T = Tuples[Table[func[k], {k, 1, n}]];

%t T2 = Cases[T, x_ /; Length[Union[x]] == Length[x]];

%t Length[T2]]

%t Select[Range[41], perms[#] > 0 &]

%Y Cf. A000108, A073364.

%K nonn

%O 1,3

%A _Gary E. Davis_, Sep 24 2016

%E More terms from _Alois P. Heinz_, Sep 28 2016

%E a(23)-a(58) from _Robert Israel_, Nov 18 2016