The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A001672

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A001672

a(n) = floor(Pi^n).
(history; published version)

#23 by Alois P. Heinz at Mon May 28 14:22:40 EDT 2018

STATUS

reviewed

approved

#22 by Joerg Arndt at Mon May 28 13:24:07 EDT 2018

STATUS

proposed

reviewed

#21 by Jon E. Schoenfield at Sun May 27 19:56:10 EDT 2018

STATUS

editing

proposed

#20 by Jon E. Schoenfield at Sun May 27 19:56:05 EDT 2018

LINKS

T. D. Noe, <a href="/A001672/b001672.txt">Table of n, a(n) for n = 0..300</a>

FORMULA

a(n)^(1/n) converges to Pi because |1 - a(n)/piPi^n| = |piPi^n - a(n)|/piPi^n < 1/piPi^n and so a(n)^(1/n) = (Pi^n*(1+o(1)))^(1/n) = Pi*(1+o(1)). - Hieronymus Fischer, Jan 22 2006

AUTHOR

N. J. A. Sloane.

STATUS

proposed

editing

#19 by Alois P. Heinz at Sun May 27 11:18:15 EDT 2018

STATUS

editing

proposed

#18 by Alois P. Heinz at Sun May 27 11:17:26 EDT 2018

NAME

Floora(n) = floor(Pi^n).

FORMULA

a(n)^(1/n) converges to pi Pi because |1-a(n)/pi^n|=|pi^n-a(n)|/pi^n<1/pi^n and so a(n)^(1/n)=(piPi^n*(1+o(1)))^(1/n)=piPi*(1+o(1)). - Hieronymus Fischer, Jan 22 2006

CROSSREFS

Cf. A001673.

STATUS

proposed

editing

#17 by M. F. Hasler at Sun May 27 11:04:37 EDT 2018

STATUS

editing

proposed

Discussion

Sun May 27

11:11

Michel Marcus: also add  Cf. A001673 ?

#16 by M. F. Hasler at Sun May 27 11:04:21 EDT 2018

PROG

(PARI) A001672(n)=Pi^n\1 \\ An error message will say when so if default(realprecision) must be increased. - M. F. Hasler, May 27 2018

#15 by M. F. Hasler at Sun May 27 11:02:20 EDT 2018

PROG

(PARI) A001672(n)=Pi^n\1 \\ An error will say when default(realprecision) must be increased. - M. F. Hasler, May 27 2018

CROSSREFS

See also A002160: closest integer to Pi^n.

STATUS

approved

editing

#14 by T. D. Noe at Thu Aug 09 16:58:58 EDT 2012

STATUS

editing

approved