A353327 -id:A353327 - cp's OEIS frontend

A353314 If n is of the form 3k, then a(n) = n, and if n is of the form 3k+r, with r = 1 or 2, then a(n) = 5*k + 3 + r.

0, 4, 5, 3, 9, 10, 6, 14, 15, 9, 19, 20, 12, 24, 25, 15, 29, 30, 18, 34, 35, 21, 39, 40, 24, 44, 45, 27, 49, 50, 30, 54, 55, 33, 59, 60, 36, 64, 65, 39, 69, 70, 42, 74, 75, 45, 79, 80, 48, 84, 85, 51, 89, 90, 54, 94, 95, 57, 99, 100, 60, 104, 105, 63, 109, 110, 66, 114, 115, 69, 119, 120, 72, 124, 125, 75, 129, 130

Offset: 0

Antti Karttunen, Apr 14 2022

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Nicholas Drozd, A Busy Beaver Champion Derived from Scratch
Nicholas Drozd, Feedback to Doron Zeilberger's opinion #155, Jan. 4, 2022.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A353313 (variant), A349876 (the first multiple of 3 reached when iterating this sequence), A349877 (number of iterations to reach the first multiple of 3), A353327 (A102899).

Array[If[#2 == 0, #1, 5 #1 + 3 + #2 & @@ QuotientRemainder[#1, 3]] & @@ {#, Mod[#, 3]} &, 78, 0] (* Michael De Vlieger, Apr 14 2022 *)

A353314(n) = { my(r=(n%3)); if(!r,n,((5*((n-r)/3)) + r + 3)); };

a(n) = n + A353327(n) = n + A102899(3+n).
From Chai Wah Wu, Jul 27 2022: (Start)
a(n) = 2*a(n-3) - a(n-6) for n > 5.
G.f.: x*(x^3 + 3*x^2 + 5*x + 4)/(x^6 - 2*x^3 + 1). (End)

A102899 a(n) = ceiling(n/3)^2 - floor(n/3)^2.

Original entry on oeis.org

0, 1, 1, 0, 3, 3, 0, 5, 5, 0, 7, 7, 0, 9, 9, 0, 11, 11, 0, 13, 13, 0, 15, 15, 0, 17, 17, 0, 19, 19, 0, 21, 21, 0, 23, 23, 0, 25, 25, 0, 27, 27, 0, 29, 29, 0, 31, 31, 0, 33, 33, 0, 35, 35, 0, 37, 37, 0, 39, 39, 0, 41, 41, 0, 43, 43, 0, 45, 45, 0, 47, 47, 0, 49, 49, 0, 51, 51, 0, 53, 53, 0

Offset: 0

Paul Barry, Jan 17 2005

If n is a multiple of 3, then a(n) = 0, and if n is of the form 3k+r, with r = 1 or 2, then a(n) = 2*k + 1. - Antti Karttunen, Apr 14 2022

Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A003417, A004396, A011655, A120691, A353314, A353327.

I:=[0,1,1,0,3,3]; [n le 6 select I[n] else 2*Self(n-3) - Self(n-6): n in [1..91]]; // G. C. Greubel, Dec 09 2022

LinearRecurrence[{0,0,2,0,0,-1}, {0,1,1,0,3,3}, 90] (* G. C. Greubel, Dec 09 2022 *)

A102899(n)=(n\3*2+1)*(0M. F. Hasler, Dec 13 2007

def A102899(n): return (1+2*(n//3))*((n%3)>0)
[A102899(n) for n in range(91)] # G. C. Greubel, Dec 09 2022

G.f.: x*(1+x+x^3+x^4)/(1-2*x^3+x^6).
a(n) = A011655(n)*A004396(n).
a(n) = (2/3)*floor((2*n+1)/3)*(1-cos(2*Pi*n/3)).
From M. F. Hasler, Dec 13 2007: (Start)
a(n) = |A120691(n+1)| for n>0.
a(n) = ([n/3]*2 + 1)*dist(n,3Z). (End)
a(n) = 2*sin(n*Pi/3)*(4*n*sin(n*Pi/3)-sqrt(3)*cos(n*Pi))/9. - Wesley Ivan Hurt, Sep 24 2017
a(n) = 2*a(n-3) - a(n-6), for n > 5. - Chai Wah Wu, Jul 27 2022

cp's OEIS Frontend

A353314 If n is of the form 3k, then a(n) = n, and if n is of the form 3k+r, with r = 1 or 2, then a(n) = 5*k + 3 + r.

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