The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A130520 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A130520

a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)
(history; published version)

#46 by N. J. A. Sloane at Fri Dec 15 19:18:29 EST 2023

STATUS

proposed

approved

#45 by Michael Somos at Wed Nov 15 23:21:07 EST 2023

STATUS

editing

proposed

#44 by Michael Somos at Wed Nov 15 23:20:55 EST 2023

COMMENTS

Given a sequence b(n) defined by variables ab(0) to ab(5) and recursion b(n) = -(b(n-6) * a(n-2) * (b(n-4) * b(n-2)^3 - b(n-3)^3 * b(n-1)) - b(n-5) * b(n-3) * b(n-1) * (b(n-5) * b(n-2)^2 - b(n-4)^2 * b(n-1)))/(b(n-4) * (b(n-5) * b(n-3)^3 - b(n-4)^3 * b(n-2))). The denominator of b(n+1) has a factor of (b(1) * b(3)^3 - b(2)^3 * b(4))^a(n+1). For example, if gb(0) = 2, gb(1) = gb(2) = gb(3) = 1, gb(4) = 1+x, gb(5) = 4, then the denominator of b(n+1) is x^a(n+1). - Michael Somos, Nov 15 2023

Discussion

Wed Nov 15

23:21

Michael Somos: Added more info.

#43 by Michael Somos at Wed Nov 15 23:19:30 EST 2023

COMMENTS

Given a sequence b(n) defined by variables a(0) to a(5) and recursion b(n) = -(b(n-6) * a(n-2) * (b(n-4) * b(n-2)^3 - b(n-3)^3 * b(n-1)) - b(n-5) * b(n-3) * b(n-1) * (b(n-5) * b(n-2)^2 - b(n-4)^2 * b(n-1)))/(b(n-4) * (b(n-5) * b(n-3)^3 - b(n-4)^3 * b(n-2))). The denominator of b(n+1) has a factor of (b(1) * b(3)^3 - b(2)^3 * b(4))^a(n+1). For example, if g(0) = 2, g(1) = g(2) = g(3) = 1, g(4) = 1+x, g(5) = 4, then the denominator of b(n+1) is x^a(n+1). - Michael Somos, Nov 15 2023

STATUS

approved

editing

#42 by Joerg Arndt at Sat Sep 17 03:45:15 EDT 2022

STATUS

reviewed

approved

#41 by Michel Marcus at Sat Sep 17 03:39:38 EDT 2022

STATUS

proposed

reviewed

#40 by Amiram Eldar at Sat Sep 17 02:39:06 EDT 2022

STATUS

editing

proposed

#39 by Amiram Eldar at Sat Sep 17 02:28:34 EDT 2022

FORMULA

From Amiram Eldar, Sep 17 2022: (Start)

Sum_{n>=5} 1/a(n) = 518/45 - 2*sqrt(2*(sqrt(5)+5))*Pi/3.

Sum_{n>=5} (-1)^(n+1)/a(n) = 8*sqrt(5)*arccoth(3/sqrt(5))/3 + 92*log(2)/15 - 418/45. (End)

STATUS

approved

editing

#38 by Charles R Greathouse IV at Thu Sep 08 08:45:30 EDT 2022

PROG

(MAGMAMagma) [Round(n*(n-3)/10): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

Discussion

Thu Sep 08

08:45

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

#37 by Joerg Arndt at Sun Sep 01 01:38:49 EDT 2019

STATUS

reviewed

approved