

A276986


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.


1



0, 1, 3, 4, 9, 10, 12, 13, 28, 29, 31, 32, 37, 38, 40, 41, 90, 91, 93, 94, 99, 100, 102, 103, 118, 119, 121, 122, 127, 128, 130, 131, 297, 298, 300, 301, 306, 307, 309, 310, 325, 326, 328, 329, 334, 335, 337, 338, 387, 388, 390, 391, 396, 397, 399, 400, 415, 416
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OFFSET

1,3


COMMENTS

n>=1 is in the sequence if and only if there is a Catalan number c such that c/2 <= n < c and cn1 is in the sequence.  Robert Israel, Nov 20 2016


LINKS



FORMULA



EXAMPLE

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.


MAPLE

S:= {0}:
for i from 1 to 8 do
c:= binomial(2*i, i)/(i+1);
S:= S union map(t > c  t  1, S);
od:


MATHEMATICA

CatalanTo[n0_] :=
Module[{n = n0}, k = 1; L = {};
While[CatalanNumber[k] <= 2*n, L = {L, CatalanNumber[k]}; k++];
L = Flatten[L]]
perms[n0_] := Module[{n = n0, S, func, T, T2},
func[k_] := Cases[CatalanTo[n], x_ /; 1 <= x  k <= n]  k;
T = Tuples[Table[func[k], {k, 1, n}]];
T2 = Cases[T, x_ /; Length[Union[x]] == Length[x]];
Length[T2]]
Select[Range[41], perms[#] > 0 &]


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



