The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History forA228036

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A228036

(10^n)-th primitive antiharmonic number.
(history; published version)

#14byMichael De Vliegerat Sat Oct 22 16:19:14 EDT 2022

STATUS

proposed

approved

#13byJon E. Schoenfieldat Sat Oct 22 15:31:53 EDT 2022

STATUS

editing

proposed

#12byJon E. Schoenfieldat Sat Oct 22 15:31:50 EDT 2022

NAME

(10^n)-th primitive antiharmonic number.

COMMENTS

We conjecture that lim_{n->inftyoo} a(n)/100^n = lim_{n->inftyoo}A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.10, 5.84, 5.99, 6.05 for n = 1..5, as well as by the values ofA228023(n)/n^2.

STATUS

approved

editing

#11byJoerg Arndtat Wed Sep 04 05:50:11 EDT 2013

STATUS

proposed

approved

#10byCharles R Greathouse IVat Tue Sep 03 17:12:45 EDT 2013

STATUS

editing

proposed

Discussion

Tue Sep 03

17:13

Charles R Greathouse IV:> 700 CPU-hours to find a(5).

#9byCharles R Greathouse IVat Tue Sep 03 17:12:41 EDT 2013

COMMENTS

We conjecture that lim_{n->infty} a(n)/100^n = lim_{n->infty}A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.1005,10,5.837732,84,5.9940738,__99,6.05for n = 1..5, as well as by the values ofA228023(n)/n^2.

#8byCharles R Greathouse IVat Tue Sep 03 17:11:51 EDT 2013

DATA

1, 605, 51005, 5837732, 599407380,60462121402

COMMENTS

We conjecture that lim_{n->infty} a(n)/100^n = lim_{n->infty}A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.1005, 5.837732, 5.994073809940738,__for n = 1,2,3,4,..5,as well as by the values ofA228023(n)/n^2.

EXTENSIONS

a(5) fromCharles R Greathouse IV,Sep 03 2013

STATUS

approved

editing

#7byBruno Berselliat Mon Aug 05 18:09:21 EDT 2013

STATUS

proposed

approved

#6byCharles R Greathouse IVat Mon Aug 05 01:24:57 EDT 2013

STATUS

editing

proposed

#5byCharles R Greathouse IVat Mon Aug 05 01:24:36 EDT 2013

COMMENTS

We conjecture that lim_{n->infty} a(n)/(10100^n)^2= lim_{n->infty}A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.1005, 5.837732, 5.99407380 for n = 1, 2, 3, 4, as well as by the values ofA228023(n)/n^2.

STATUS

proposed

editing