The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A124391 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A124391

Numbers m that divide A123269(m) = Sum_{i=1..m, j=1..m, k=1..m} i^j^k.
(history; published version)

#19 by N. J. A. Sloane at Sat Feb 27 21:37:10 EST 2021

STATUS

proposed

approved

#18 by Jon E. Schoenfield at Thu Feb 25 19:57:33 EST 2021

STATUS

editing

proposed

#17 by Jon E. Schoenfield at Thu Feb 25 19:57:25 EST 2021

COMMENTS

A123269(m) = Sum_{i=21..m, j=1..m, k=1..m} i^j^k = {1, 28, 7625731729896, ...}.

STATUS

reviewed

editing

Discussion

Thu Feb 25

19:57

Jon E. Schoenfield: Argh!  Sorry!  :-(

#16 by Joerg Arndt at Thu Feb 25 01:57:11 EST 2021

STATUS

proposed

reviewed

Discussion

Thu Feb 25

02:30

Michel Marcus: Sum_{i=2..m, :   should be Sum_{i=1..m,   ?

#15 by Joerg Arndt at Thu Feb 25 01:56:44 EST 2021

STATUS

editing

proposed

#14 by Joerg Arndt at Thu Feb 25 01:56:35 EST 2021

COMMENTS

Primes terms are listed in A039787, primes p such that p-1 is squarefree.

#13 by Joerg Arndt at Thu Feb 25 01:56:16 EST 2021

COMMENTS

Primes terms are {2, 3, 7, 11, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, ...} = listed in A039787, primes p such that p-1 is squarefree.

STATUS

proposed

editing

#12 by Michel Marcus at Thu Feb 25 01:43:35 EST 2021

STATUS

editing

proposed

#11 by Michel Marcus at Thu Feb 25 01:42:55 EST 2021

NAME

Numbers m that divide A123269(nm) = Sum_{i=1..m, j=1..m, k=1..m} i^j^k.

Discussion

Thu Feb 25

01:43

Michel Marcus: and 2nd comment could/should simply be : Primes terms are A039787,

#10 by Michel Marcus at Thu Feb 25 01:41:58 EST 2021

COMMENTS

Primes in {a(n)} terms are {2, 3, 7, 11, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, ...} = {A039787(n)}, , primes p such that p-1 is squarefree.

STATUS

proposed

editing

Discussion

Thu Feb 25

01:42

Michel Marcus: for me the 1st comment should go