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A099628
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Numbers m where m-th Catalan number A000108(m) = binomial(2m,m)/(m+1) is divisible by 2 but not by 4, i.e., where A048881(m) = 1.
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3
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2, 4, 5, 8, 9, 11, 16, 17, 19, 23, 32, 33, 35, 39, 47, 64, 65, 67, 71, 79, 95, 128, 129, 131, 135, 143, 159, 191, 256, 257, 259, 263, 271, 287, 319, 383, 512, 513, 515, 519, 527, 543, 575, 639, 767, 1024, 1025, 1027, 1031, 1039, 1055, 1087, 1151, 1279, 1535, 2048
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OFFSET
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1,1
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COMMENTS
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Also, there is exactly one digit position in which both a(n)+1 and a(n)-1, written in binary, have a 1; i.e., the bitwise AND of a(n)-1 and a(n)+1 is 2^k, with k > 0. - Wouter Meeussen, Nov 24 2007
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LINKS
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FORMULA
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As triangle, T(n,k) = 2^(n+1) + 2^k - 1 = A099627(n+1, k).
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EXAMPLE
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As triangle, rows start
2;
4, 5;
8, 9, 11;
16, 17, 19, 23;
32, 33, 35, 39, 47;
...
5 is in the sequence since 10!/(5!6!) = 42 is divisible by 2 but not 4;
6 is not in the sequence since 12!/(6!7!) = 132 is divisible by 4;
7 is not in the sequence since 14!/(7!8!) = 429 is not divisible by 2.
Table showing the binary expansion of a(n) for n = 1..15, replacing 0 with "." to accentuate the pattern of bits:
n a(n) a(n)_2
----------------
1 2 1.
2 4 1..
3 5 1.1
4 8 1...
5 9 1..1
6 11 1.11
7 16 1....
8 17 1...1
9 19 1..11
10 23 1.111
11 32 1.....
12 33 1....1
13 35 1...11
14 39 1..111
15 47 1.1111 (End)
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MATHEMATICA
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Select[Range[2048], IntegerQ[Log[2, BitAnd[ #+1, #-1]]]&] (* Wouter Meeussen, Nov 24 2007 *)
Table[2^(n + 1) + 2^k - 1, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 28 2022 *)
Select[Range[2100], Boole[Divisible[CatalanNumber[#], {2, 4}]]=={1, 0}&] (* Harvey P. Dale, Jan 31 2024 *)
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PROG
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(Magma) /* As triangle */ [[2^(n+1) + 2^k - 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 27 2017
(Python)
from itertools import count, islice
def A099628_gen(): # generator of terms
m = 1
for n in count(1):
m *= 2
r, k = m-1, 1
for _ in range(n):
yield r+k
k *= 2
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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