A281899 a(n) = n + 6*floor(n/3).
0, 1, 2, 9, 10, 11, 18, 19, 20, 27, 28, 29, 36, 37, 38, 45, 46, 47, 54, 55, 56, 63, 64, 65, 72, 73, 74, 81, 82, 83, 90, 91, 92, 99, 100, 101, 108, 109, 110, 117, 118, 119, 126, 127, 128, 135, 136, 137, 144, 145, 146, 153, 154, 155, 162, 163, 164, 171, 172, 173, 180, 181, 182, 189
Offset: 0
Links
- Bruno Berselli, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Crossrefs
Cf. A002264.
The first differences are in A105395.
Cf. similar sequences with formula n+i*floor(n/3): A004773 (i=1), A047217 (i=2), A047240 (i=3), A047354 (i=4), A047469 (i=5), this sequence (i=6).
Cf. numbers that are congruent to {0, 1, 2} mod j: the sequences are listed in the previous row for j = 4..9, respectively.
Programs
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Magma
[n+6*(n div 3): n in [0..70]];
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Maple
A281899:=n->n+6*floor(n/3): seq(A281899(n), n=0..100); # Wesley Ivan Hurt, Feb 09 2017
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Mathematica
Table[n + 6 Floor[n/3], {n, 0, 70}] LinearRecurrence[{1,0,1,-1},{0,1,2,9},90] (* Harvey P. Dale, Feb 25 2018 *) -
Maxima
makelist(n+6*floor(n/3), n, 0, 70);
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PARI
a(n)=n\3*6 + n \\ Charles R Greathouse IV, Feb 07 2017
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Python
[n+6*int(n/3) for n in range(70)]
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Sage
[n+6*floor(n/3) for n in range(70)]
Formula
G.f.: x*(1 + x + 7*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = 3*n - 2*(n mod 3). In general, n + 3*h*floor(n/3) = (h+1)*n - h*(n mod 3).
a(n) + a(n+s) = a(2*n+s-1) + 1, where s is nonnegative and not divisible by 3. Example: for s=14, a(n) + a(n+14) = a(2*n+13) + 1; for n=3, a(3) + a(17) = a(19) + 1 = 9 + 47 = 55 + 1 = 56.
a(6*k+r) = 18*k + a(r), where 0 <= r <= 5.
Comments