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Revision History for A262997 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
a(n+3) = a(n) + 24*n + 40, a(0)=0, a(1)=5, a(2)=19.
(history; published version)
#25 by N. J. A. Sloane at Mon Nov 16 11:38:07 EST 2015
STATUS

proposed

approved

#24 by Michel Marcus at Fri Nov 06 11:37:41 EST 2015
STATUS

editing

proposed

Discussion
Fri Nov 06
12:44
Paul Curtz: Thanks Michel. Unfortunely I am not allowed to submit more than three sequences or comments.
#23 by Michel Marcus at Fri Nov 06 11:37:04 EST 2015
COMMENTS

(Corrections must be made in A262397: the third row is A248438(n+1), not d(n+1) and the fifth one is d(n+1), not A240438(n+1). In It appears ..., A240438(n+1) and d(n+1) must be swapped.)

STATUS

proposed

editing

Discussion
Fri Nov 06
11:37
Michel Marcus: (Corrections must be made in A262397: the third row is A248438(n+1), not d(n+1) and the fifth one is d(n+1), not A240438(n+1). In  It appears ..., A240438(n+1) and d(n+1) must be swapped.)
#22 by Paul Curtz at Mon Oct 19 09:27:19 EDT 2015
STATUS

editing

proposed

#21 by Paul Curtz at Mon Oct 19 09:26:20 EDT 2015
FORMULA

a(2n+2) = -a(2n+1) + 4*(n+1).

#20 by Paul Curtz at Mon Oct 19 09:17:35 EDT 2015
FORMULA

a(2n+1) = -a(2n) + 6*n + 3.

STATUS

proposed

editing

#19 by Michael De Vlieger at Fri Oct 09 10:27:40 EDT 2015
STATUS

editing

proposed

#18 by Michael De Vlieger at Fri Oct 09 10:27:37 EDT 2015
MATHEMATICA

a[0] = 0; a[1] = 5; a[2] = 19; a[n_] := a[n] = a[n - 3] + 24 (n - 3) + 40; Table[a@ n, {n, 0, 46}] (* Michael De Vlieger, Oct 09 2015 *)

STATUS

proposed

editing

#17 by Paul Curtz at Fri Oct 09 04:59:13 EDT 2015
STATUS

editing

proposed

#16 by Paul Curtz at Fri Oct 09 04:56:02 EDT 2015
COMMENTS

(Corrections must be made in A262397: the third row is A248438(n+1), not d(n+1) and the fifth one is d(n+1), not A240438(n+1). In It appears ..., A240438(n+1) and d(n+1) must be swapped.)

Discussion
Fri Oct 09
04:58
Paul Curtz: Thanks. Done.