The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A010875

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A010875

a(n) = n mod 6.
(history; published version)

#82 by Charles R Greathouse IV at Thu Sep 08 08:44:37 EDT 2022

PROG

(MAGMAMagma) [n mod 6: n in [0..100]]; // Wesley Ivan Hurt, Jul 06 2014

Discussion

Thu Sep 08

08:44

OEIS Server: https://oeis.org/edit/global/2944

#81 by Alois P. Heinz at Wed Aug 12 19:45:40 EDT 2020

FORMULA

a(n)=1/6 (2r^(5n + 2)+ 2r^(5n)- 4r^(5n - 2)- 2r^(4n + 2)+ 2r^(4n + 1)+ 2 r^(4n)- 2 r^(4n - 1)- 3r^(3n)- 6r^(2n + 2)- 10r^(2n)+ 4r^(2n - 2)+ 6r^(n + 2)- 2r^(n + 1)- 6r^n+ 2r^(n - 1)+ 15) , where r^5+r^4+r^3+r^2+r+1=0 . - Ammar Khatab, Aug 13 2020

KEYWORD

nonn,easy,changed

STATUS

editing

approved

#80 by Alois P. Heinz at Wed Aug 12 19:45:21 EDT 2020

STATUS

proposed

editing

#79 by Ammar Khatab at Wed Aug 12 19:05:09 EDT 2020

STATUS

editing

proposed

Discussion

Wed Aug 12

19:38

Alois P. Heinz: too complicated ... n mod 6 is short and simple ...

19:45

Alois P. Heinz: ... and formula cannot be verified ...

#78 by Ammar Khatab at Wed Aug 12 19:01:28 EDT 2020

FORMULA

Discussion

Wed Aug 12

19:04

Ammar Khatab: this sequence are the result of "n mod 6"
so we able to write is by using the roots of 1 (without using the real root)
and I used the solution of ( r^5+r^4+r^3+r^2+r+1=0)

#77 by Ammar Khatab at Wed Aug 12 18:57:56 EDT 2020

FORMULA

a(n)=1/6 (2r^(5n + 2)+ 2r^(5n)- 4r^(5n - 2)- 2r^(4n + 2)+ 2r^(4n + 1)+ 2 r^(4n)- 2 r^(4n - 1)- 3r^(3n)- 6r^(2n + 2)- 10r^(2n)+ 4r^(2n - 2)+ 6r^(n + 2)- 2r^(n + 1)- 6r^n+ 2r^(n - 1)+ 15) . - _Ammar Khata_, Aug 13 2020

STATUS

approved

editing

#76 by Bruno Berselli at Thu Dec 12 04:04:41 EST 2019

STATUS

proposed

approved

#75 by Petros Hadjicostas at Thu Dec 12 01:19:36 EST 2019

STATUS

editing

proposed

#74 by Petros Hadjicostas at Thu Dec 12 01:13:38 EST 2019

FORMULA

Complex representation: a(n) = (1/6) * (1 - r^n) * Sum_{k = 1..6} k * Product_{1 <= m < 6, m <> k} (1-r^(n-m)), where r = exp((Pi/3)*i) = (1 + sqrt(3)*i)/2 and i = sqrt(-1).

#73 by Petros Hadjicostas at Thu Dec 12 01:12:47 EST 2019

FORMULA

Complex representation: a(n) = (1/6) * (1 - r^n) *sum Sum_{1<=k< = 1..6, } k * productProduct_{1 <= m < 6, m <> k, } (1-r^(n-m))}}, , where r = exp(Pi/3*i) = (1 + sqrt(3)*i)/2 and i = sqrt(-1).

Trigonometric representation: a(n) = (16/3)^2 * (sin(n*Pi/6))^2 *sum Sum_{1<=k< = 1..6, } k *product Product_{1 <= m < 6, m<>k, } (sin((n-m)*Pi/6))^2}}.