A369405 Context-free language 1^n.0^(2n).
100, 110000, 111000000, 111100000000, 111110000000000, 111111000000000000, 111111100000000000000, 111111110000000000000000, 111111111000000000000000000, 111111111100000000000000000000, 111111111110000000000000000000000, 111111111111000000000000000000000000
Offset: 1
Links
- Paolo Xausa, Table of n, a(n) for n = 1..300
- Index entries for linear recurrences with constant coefficients, signature (1100,-100000).
Programs
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Maple
a:= n-> convert(4^n*(2^n-1), binary): seq(a(n), n=1..15); # Alois P. Heinz, Feb 04 2024
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Mathematica
Array[(10^#-1)*10^(2*#)/9 &, 20] (* or *) LinearRecurrence[{1100, -100000}, {100, 110000}, 20] (* Paolo Xausa, Feb 27 2024 *) -
Python
def A369405(n): return (10**n-1)//9*10**(n<<1) # Chai Wah Wu, Feb 11 2024
Formula
From Robert Israel, Jan 22 2024: (Start)
a(n) = (10^n-1)*10^(2*n)/9.
G.f.: 100*x/(100000*x^2 - 1100*x + 1). (End)
From Alois P. Heinz, Feb 04 2024: (Start)
a(n) = 10 * A138119(n).
a(n) = 100 * A147816(n). (End)
Comments