The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A230405

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A230405

a(n) = A000217(A230404(n+1)); the first differences of A219650.
(history; published version)

#16 by Michel Marcus at Fri Jan 05 02:54:40 EST 2024

STATUS

reviewed

approved

#15 by Joerg Arndt at Fri Jan 05 00:35:09 EST 2024

STATUS

proposed

reviewed

#14 by Amiram Eldar at Fri Jan 05 00:26:42 EST 2024

STATUS

editing

proposed

#13 by Amiram Eldar at Fri Jan 05 00:22:13 EST 2024

CROSSREFS

Cf. A000217, A007623, A230404.

#12 by Amiram Eldar at Fri Jan 05 00:21:44 EST 2024

CROSSREFS

First differences of A219650. Can be used to compute A219650 and A230412. Cf. also A230413.

Cf. A000217, A230404.

Cf. also A230413.

#11 by Amiram Eldar at Fri Jan 05 00:20:41 EST 2024

FORMULA

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=0..m} a(k) = 2. - Amiram Eldar, Jan 05 2024

STATUS

approved

editing

#10 by N. J. A. Sloane at Fri Nov 08 13:45:11 EST 2013

STATUS

editing

approved

#9 by N. J. A. Sloane at Fri Nov 08 13:45:08 EST 2013

COMMENTS

Construction: Count the number of trailing zeros in the factorial base representation (A007623) of 2n+2 (2, 4, 6, 8, ...) and then take the corresponding triangular number from A000217.

STATUS

proposed

editing

#8 by Antti Karttunen at Sun Nov 03 16:57:05 EST 2013

STATUS

editing

proposed

#7 by Antti Karttunen at Sun Nov 03 14:33:35 EST 2013

COMMENTS

Construction: Count the number of trailing zeros of in the factorial base representation (A007623) of 2n+2 (2, 4, 6, 8, ...) and then take the corresponding triangular number from A000217.