The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A081614 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A081614

Subsequence of A005428 with state = 1.
(history; published version)

#61 by Alois P. Heinz at Mon May 31 05:30:03 EDT 2021

STATUS

proposed

approved

#60 by Alois P. Heinz at Sat May 29 12:22:58 EDT 2021

STATUS

editing

proposed

#59 by Alois P. Heinz at Sat May 29 12:22:51 EDT 2021

FORMULA

a(n) = [(n+1)-th even number of A061419]/2. -_ _John-Vincent Saddic_, May 29 2021

STATUS

proposed

editing

#58 by John-Vincent Saddic at Sat May 29 12:18:43 EDT 2021

STATUS

editing

proposed

#57 by John-Vincent Saddic at Sat May 29 12:18:26 EDT 2021

COMMENTS

One way to calculate this is a(n) = [(n+1)th even number of A061419]/2. - John-Vincent Saddic, May 28 2021

FORMULA

a(n) = [(n+1)-th even number of A061419]/2. -John-Vincent Saddic, May 29 2021

STATUS

proposed

editing

#56 by John-Vincent Saddic at Sat May 29 01:29:16 EDT 2021

STATUS

editing

proposed

Discussion

Sat May 29

11:18

Michel Marcus: why not simply make a formula like : a(n) = [(n+1)-th even number of A061419]/2.

#55 by John-Vincent Saddic at Sat May 29 01:27:51 EDT 2021

COMMENTS

A really easy One way to calculate this is a(n) = [(n+1)th even number of A061419]/2. - John-Vincent Saddic, May 28 2021

STATUS

proposed

editing

#54 by Kevin Ryde at Sat May 29 01:23:50 EDT 2021

STATUS

editing

proposed

#53 by Kevin Ryde at Sat May 29 01:18:46 EDT 2021

COMMENTS

A really easy way to calculate is a(n) = [(n+1)th even number of A061419]/2 _. - _John-Vincent Saddic_, May 28 2021

CROSSREFS

Cf. A005428, A073941, A081615, A061419.

STATUS

proposed

editing

Discussion

Sat May 29

01:23

Kevin Ryde: Can usually leave out "easy".  The A061419 calculation is just about lurking in the PARI code ... maybe :).

#52 by John-Vincent Saddic at Fri May 28 23:25:44 EDT 2021

STATUS

editing

proposed