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Revision History for A208950 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
a(4*n) = n*(16*n^2-1)/3, a(2*n+1) = n*(n+1)*(2*n+1)/6, a(4*n+2) = (4*n+1)*(4*n+2)*(4*n+3)/6.
(history; published version)
#62 by Michael De Vlieger at Fri Aug 12 09:22:34 EDT 2022
STATUS

proposed

approved

#61 by Michel Marcus at Fri Aug 12 06:09:30 EDT 2022
STATUS

editing

proposed

#60 by Michel Marcus at Fri Aug 12 06:09:25 EDT 2022
COMMENTS

a(n+2) is divisible by A060819(floor(n/3)); a(n) is divisible by A176672(floor(n/3)).

a(n) is divisible by A176672(floor(n/3)).

STATUS

proposed

editing

#59 by Amiram Eldar at Fri Aug 12 04:27:33 EDT 2022
STATUS

editing

proposed

#58 by Amiram Eldar at Fri Aug 12 04:12:48 EDT 2022
FORMULA

G.f. : (x^2 + x^3 + 5*x^4 + 5*x^5 + 31*x^6 + 10*x^7 + 22*x^8 + 10*x^9 + 31*x^10 + 5*x^11 + 5*x^12 + x^13 + x^14) / ((1-x)^4*(1+x)^4*(1 + 4*x^2 + 6*x^4 + 4*x^6 + x^8)). - R. J. Mathar, Mar 10 2012

#57 by Amiram Eldar at Fri Aug 12 04:12:16 EDT 2022
COMMENTS

a(n)/a(n+4) = n*(n^2-1)/((n+3)*(n+4)*(n+5)).

a(n)/a(n+12) = (n-1)*n*(n+1)/((n+11)*(n+12)*(n+13)).

FORMULA

a(n)/a(n+4) = n*(n^2-1)/((n+3)*(n+4)*(n+5)).

a(n)/a(n+12) = (n-1)*n*(n+1)/((n+11)*(n+12)*(n+13)).

#56 by Amiram Eldar at Fri Aug 12 04:09:36 EDT 2022
FORMULA

Sum_{n>=2} 1/a(n) = 12 - 27*log(2)/2. - Amiram Eldar, Aug 12 2022

PROG

(MAGMAMagma) [Binomial(n+1, 3)*GCD(n+2, 4)/4: n in [0..50]]; // G. C. Greubel, Sep 20 2018

STATUS

approved

editing

#55 by Michel Marcus at Sat Oct 19 04:16:01 EDT 2019
STATUS

reviewed

approved

#54 by Joerg Arndt at Sat Oct 19 03:23:50 EDT 2019
STATUS

proposed

reviewed

#53 by Michel Marcus at Sat Oct 19 00:26:01 EDT 2019
STATUS

editing

proposed