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Revision History for A153060

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

A153060

a(0)=4, a(n)=a(n-1)^2 - n for n > 0.
(history; published version)

#17 by Andrey Zabolotskiy at Mon Mar 06 20:32:01 EST 2023

STATUS

editing

approved

#16 by Andrey Zabolotskiy at Mon Mar 06 20:31:57 EST 2023

PROG

Cf. A153059.

CROSSREFS

Cf. A153059.

STATUS

approved

editing

#15 by Joerg Arndt at Fri Jul 15 02:30:21 EDT 2022

STATUS

editing

approved

#14 by Joerg Arndt at Fri Jul 15 02:30:19 EDT 2022

NAME

a(0)=4, a(n)=a(n-1)^2 - n for n > 0.

MATHEMATICA

a=2; lst={}; Do[a=a^2-n; AppendTo[lst, a], {n, 0, 10}]; lst

STATUS

reviewed

editing

#13 by Michel Marcus at Fri Jul 15 02:01:05 EDT 2022

STATUS

proposed

reviewed

#12 by Georg Fischer at Fri Jul 15 00:39:34 EDT 2022

STATUS

editing

proposed

Discussion

Fri Jul 15

00:40

Georg Fischer: a(-1) is not in the sequence.

#11 by Georg Fischer at Fri Jul 15 00:39:13 EDT 2022

NAME

a(-10)=2, 4, a(n)=a(n-1)^2-n for n > 0.

PROG

Cf. A153059.

EXTENSIONS

Definition amended by Georg Fischer, Jul 15 2022

STATUS

approved

editing

#10 by Vaclav Kotesovec at Thu Dec 18 08:23:51 EST 2014

STATUS

editing

approved

#9 by Vaclav Kotesovec at Thu Dec 18 08:23:47 EST 2014

FORMULA

a(n) ~ c^(2^n), where c = 3.8643187364012066048723545708047003493006832597186603613771125549936188672... . - Vaclav Kotesovec, Dec 18 2014

#8 by Vaclav Kotesovec at Thu Dec 18 08:22:38 EST 2014

MATHEMATICA

RecurrenceTable[{a[0]==4, a[n]==a[n-1]^2-n}, a, {n, 0, 10}] (* Vaclav Kotesovec, Dec 18 2014 *)

STATUS

approved

editing