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Revision History for A347261 (Underlined text is an addition; strikethrough text is a deletion.)

Showing entries 1-10 | older changes
A347261 Numbers k for which sigma(k)/k = 32/11.
(history; published version)
#13 by Alois P. Heinz at Thu Aug 26 15:30:36 EDT 2021
STATUS

proposed

approved

#12 by Michel Marcus at Thu Aug 26 00:49:24 EDT 2021
STATUS

editing

proposed

#11 by Michel Marcus at Thu Aug 26 00:49:20 EDT 2021
DATA

924, 2970, 16368, 18018, 268224, 1107161088, 283465678848, 4535476813824, 76092819268618420224, 87729047721804447573604856326476791808

EXTENSIONS

a(9)-a(10) from Michel Marcus, Aug 26 2021

STATUS

proposed

editing

#10 by Jon E. Schoenfield at Wed Aug 25 23:55:33 EDT 2021
STATUS

editing

proposed

#9 by Jon E. Schoenfield at Wed Aug 25 23:55:30 EDT 2021
COMMENTS

Conjecture: a(n) = 33*A341623(n) for n >= 1. Motivation: If no term of A341623 is divisible by 11 (which appears to be the case), then sigma[(33*A341623(n)]/[))/(33*A341623(n)] = )) = sigma(11)*sigma[(3*A341623(n)]/[))/(33*A341623(n)] = )) = 12*[*(8*A341623(n)]/[))/(33*A341623(n)] = )) = 32/11. Does this sequence, though, contain any additional terms that are not generated by A341623?

STATUS

proposed

editing

#8 by Timothy L. Tiffin at Wed Aug 25 23:14:15 EDT 2021
STATUS

editing

proposed

#7 by Timothy L. Tiffin at Wed Aug 25 23:14:12 EDT 2021
COMMENTS

Conjecture: a(n) = 33*A341623(n) for n >= 1. Motivation: If no term of A341623 is divisible by 11 (which appears to be the case), then sigma[33*A341623(n)]/[33*A341623(n)] = sigma(11)*sigma[3*A341623(n)]/[33*A341623(n)] = 12*[8*A341623(n)]/[33*A341623(n)] = 32/11. Does this sequence, though, contain any additional terms that are not indexedgenerated toby A341623?

STATUS

proposed

editing

#6 by Timothy L. Tiffin at Wed Aug 25 23:13:12 EDT 2021
STATUS

editing

proposed

#5 by Timothy L. Tiffin at Wed Aug 25 23:09:58 EDT 2021
COMMENTS

Terms ending in "24" or "x8" (where x is an even digit) have this form. Example: a(1) = 33*A000396(2), a(3) = 33*A000396(3), and a(5n) = 33*A000396(4), ...n-1) for 5, 6, 7, 8.

Conjecture: a(n) = 33*A341623(n) for n >= 1. Motivation: If no term of A341623 is divisible by 11 (which appears to be the case), then sigma[33*A341623(n)]/[33*A341623(n)] = sigma(11)*sigma[3*A341623(n)]/[33*A341623(n)] = 12*[8*A341623(n)]/[33*A341623(n)] = 32/11. Does this sequence, though, contain any additional terms that are not indexed to A341623?

FORMULA

a(n) = 33*A341623(n) for n >= 1.

CROSSREFS

Cf. A000203, A000396, A341623.

#4 by Timothy L. Tiffin at Wed Aug 25 18:13:16 EDT 2021
DATA

924, 2970, 16368, 18018, 268224, 1107161088, 283465678848, 4535476813824

FORMULA

a(n) = 33*A341623(n) for n >= 1.

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Last modified August 30 21:29 EDT 2024. Contains 375550 sequences. (Running on oeis4.)