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G.f. A(x) satisfies: [x^n] A(x)^n / (x*A(x))' = 0 for n > 1.
(history; published version)
#35 by Vaclav Kotesovec at Tue Oct 20 03:31:38 EDT 2020
STATUS

editing

approved

#34 by Vaclav Kotesovec at Tue Oct 20 03:31:32 EDT 2020
FORMULA

a(n) ~ c * 2^n * (n-1)!, where c = 0.1261880758068409567445... - Vaclav Kotesovec, Oct 20 2020

STATUS

approved

editing

#33 by Alois P. Heinz at Mon Oct 08 19:32:57 EDT 2018
STATUS

editing

approved

#32 by Alois P. Heinz at Mon Oct 08 17:56:50 EDT 2018
COMMENTS

More generally, [x^n] G(x,k)^(k*(n+1)-1) / (x*G(x,k)^k)' = 0 is satisfied by an integer sequence series G(x,k) when k is a fixed positive integer.

STATUS

approved

editing

Discussion
Mon Oct 08
17:56
Alois P. Heinz: ...
#31 by Jon E. Schoenfield at Sun Oct 07 20:43:16 EDT 2018
STATUS

proposed

approved

#30 by Jon E. Schoenfield at Sun Oct 07 19:53:02 EDT 2018
STATUS

editing

proposed

#29 by Jon E. Schoenfield at Sun Oct 07 19:52:59 EDT 2018
NAME

G.f. A(x) satisfies: [x^n] A(x)^n / (x*A(x))' = 0 for n > 1.

COMMENTS

Odd terms seem to occur only at positions 0, 1, and 2*A118113(k) for k >= 0.

More generally, [x^n] G(x,k)^(k*(n+1)-1) / (x*G(x,k)^k)' = 0 is satisfied by an integer series sequence G(x,k) when k is a fixed positive integer.

FORMULA

G.f. A(x) satisfies: [x^n] A(x)^n / (A(x) + x*A'(x)) = 0 for n > 1.

EXAMPLE

such that [x^n] A(x)^n / (x*A(x))' = 0 for n > 1.

in which the main diagonal consists of all zeros after the initial terms, illustrating that [x^n] A(x)^n / (x*A(x))' = 0 for n > 1.

STATUS

approved

editing

#28 by Paul D. Hanna at Thu Apr 05 11:53:25 EDT 2018
STATUS

editing

approved

#27 by Paul D. Hanna at Thu Apr 05 11:53:23 EDT 2018
COMMENTS

Compare to identity: [x^n] (x*F(x))' / F(x)^(n+1) = 0 holds when F(0) = 1.

STATUS

approved

editing

#26 by Paul D. Hanna at Sat Mar 31 17:01:47 EDT 2018
STATUS

editing

approved