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Revision History for A367903 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Number of sets of nonempty subsets of {1..n} contradicting a strict version of the axiom of choice.
(history; published version)
#20 by Michael De Vlieger at Tue Jul 30 14:41:26 EDT 2024
STATUS

reviewed

approved

#19 by Michel Marcus at Tue Jul 30 11:07:45 EDT 2024
STATUS

proposed

reviewed

#18 by Andrew Howroyd at Tue Jul 30 10:22:49 EDT 2024
STATUS

editing

proposed

#17 by Andrew Howroyd at Tue Jul 30 10:22:45 EDT 2024
FORMULA

a(n) + A367904(n) + A367772(n) = 2^(2^n-1) = A058891(n+1) = 2^(2^n-1).

#16 by Andrew Howroyd at Tue Jul 30 10:18:06 EDT 2024
FORMULA

a(n) + A367904(n) + A367772(n) = 2^(2^n-1) = A058891(n+1).

STATUS

proposed

editing

Discussion
Tue Jul 30
10:22
Andrew Howroyd: We can include both. I don't think it matters if it is 'a(n) = ' or otherwise - particularly in a case like this when the formula isn't necessarily used to calculate terms from a practical standpoint - it's just expressing a relation between sequences.
#15 by Michel Marcus at Fri Jul 26 11:53:32 EDT 2024
STATUS

editing

proposed

#14 by Michel Marcus at Fri Jul 26 11:53:22 EDT 2024
FORMULA

A367903a(n) + A367904(n) + A367772(n) = A058891(n).

STATUS

proposed

editing

#13 by Christian Sievers at Fri Jul 26 11:49:59 EDT 2024
STATUS

editing

proposed

#12 by Christian Sievers at Fri Jul 26 11:44:37 EDT 2024
DATA

0, 0, 1, 67, 30997, 2147296425, 9223372036784737528, 170141183460469231731687303625772608225, 57896044618658097711785492504343953926634992332820282019728791606173188627779

EXTENSIONS

a(5)-a(8) from Christian Sievers, Jul 26 2024

STATUS

approved

editing

Discussion
Fri Jul 26
11:49
Christian Sievers: Formula seems wrong, RHS should be A058891(n+1), and maybe then better just write 2^(2^n-1). And shouldn't it be solved for this sequence?
#11 by Michael De Vlieger at Thu Dec 28 09:22:15 EST 2023
STATUS

proposed

approved