A107756 Numbers k such that Sum_{j=1..k} Catalan(j) == 1 (mod 3).
1, 4, 5, 6, 7, 10, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 28, 31, 32, 33, 34, 37, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 82, 85, 86, 87, 88, 91, 94, 95
Offset: 1
Links
- Y. More, Problem 11165, Amer. Math. Monthly, 112 (2005), 568.
Programs
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Mathematica
Select[ Range[ 100], Mod[ Sum[(2 n)!/n!/(n + 1)!, {n, #}], 3] == 1 &] (* Robert G. Wilson v, Jun 14 2005 *) Position[Accumulate[CatalanNumber[Range[100]]],?(Mod[#,3]==1&)]//Flatten (* _Harvey P. Dale, Aug 26 2022 *)
Formula
Equivalently, numbers k such that base-3 expansion of k+1 contains a 2.