The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A166984

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A166984

a(n) = 20*a(n-1) - 64*a(n-2) for n > 1; a(0) = 1, a(1) = 20.
(history; published version)

#58 by Hugo Pfoertner at Thu Oct 03 06:09:54 EDT 2024

STATUS

reviewed

approved

#57 by Joerg Arndt at Thu Oct 03 01:56:37 EDT 2024

STATUS

proposed

reviewed

#56 by G. C. Greubel at Wed Oct 02 15:02:43 EDT 2024

STATUS

editing

proposed

#55 by G. C. Greubel at Wed Oct 02 14:57:51 EDT 2024

COMMENTS

Limit_{n -> oo} a(n)/a(n-1) = 16.

a(n) = A115490(n+1)/3.

FORMULA

Limit_{n -> oo} a(n)/a(n-1) = 16.

a(n) = A115490(n+1)/3.

From _Sum_{n>=0} a(n) x^(2*n+4)/(2*n+4)! = ( sinh(x) )^4/4!. - _Robert A. Russell_, Apr 03 2013: (Start)

E.g.f.: sinh(x)^4/4!.

a(n) = Sum{n>=0, a(n) x^(2n+4)/(2n+4)!}. (End)

E.g.f.: (4/3)*exp(10*x)*sinh(6*x + log(2)). - G. C. Greubel, Oct 02 2024

PROG

(Magma) [ n le 2 select 19*n-18 else 20*Self(n-1)-64*Self(n-2): n in [1..17] ];

(SageMath)

A166984=BinaryRecurrenceSequence(20, -64, 1, 20)

[A166984(n) for n in range(31)] # G. C. Greubel, Oct 02 2024

CROSSREFS

Cf. A000302, A002450, A002452, A002453, A006105, A079598, A083584.

Cf. A006105, A166965, A115490, A166914, A166917, A166927, A002452, A002453, A166965, A307695.

STATUS

approved

editing

Discussion

Wed Oct 02

15:02

G. C. Greubel: Corrected RAR's formula. The form a(n) = Sum_{n} a(n) * x^(2*n+4)/(2*n+4)! required investigation.

#54 by Alois P. Heinz at Mon Aug 08 18:58:20 EDT 2022

STATUS

editing

approved

#53 by Alois P. Heinz at Mon Aug 08 18:57:42 EDT 2022

DATA

1, 20, 336, 5440, 87296, 1397760, 22368256, 357908480, 5726601216, 91625881600, 1466015154176, 23456246661120, 375299963355136, 6004799480791040, 96076791961092096, 1537228672451215360, 24595658763514413056, 393530540233410478080, 6296488643803287126016

FORMULA

a(0) = 1, a(n) = 16*a(n-1) + 4^n with a(0) = 1. - Nadia Lafreniere, Aug 08 2022

STATUS

proposed

editing

#52 by Nadia Lafreniere at Mon Aug 08 18:40:24 EDT 2022

STATUS

editing

proposed

#51 by Nadia Lafreniere at Mon Aug 08 18:39:40 EDT 2022

FORMULA

a(0) = 1, a(n) = 16*a(n-1) + 4^n. - Nadia Lafreniere, Aug 08 2022

Discussion

Mon Aug 08

18:40

Nadia Lafreniere: Sorry about that. I meant a(n) = 16*a(n-1)+4^n. I made the edit.

#50 by Alois P. Heinz at Mon Aug 08 18:06:40 EDT 2022

STATUS

proposed

editing

#49 by Nadia Lafreniere at Mon Aug 08 15:25:05 EDT 2022

STATUS

editing

proposed

Discussion

Mon Aug 08

18:06

Alois P. Heinz: a(n) = 16*a(n) + 4^n ? really ?