The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A070782

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A070782

a(n) = Sum_{k=0..n} binomial(5*n,5*k).
(history; published version)

#26 by Harvey P. Dale at Sun Jun 18 14:56:36 EDT 2023

STATUS

editing

approved

#25 by Harvey P. Dale at Sun Jun 18 14:56:34 EDT 2023

MATHEMATICA

LinearRecurrence[{21, 353, -32}, {1, 2, 254}, 20] (* Harvey P. Dale, Jun 18 2023 *)

STATUS

approved

editing

#24 by Sean A. Irvine at Mon May 27 17:15:48 EDT 2019

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reviewed

approved

#23 by Michel Marcus at Mon May 27 16:57:51 EDT 2019

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proposed

reviewed

#22 by Jon E. Schoenfield at Mon May 27 16:22:55 EDT 2019

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editing

proposed

#21 by Jon E. Schoenfield at Mon May 27 16:22:53 EDT 2019

FORMULA

a(n) = (1/5)*32^n + (2/5)*(-11/2 + (5/2)*sqrt(5))^n + (2/5)*(-11/2 - (5/2)*sqrt(5))^n.

Let b(n) = a(n) - 2^(5n)/5 ; then b(n) + 11*b(n-1) - b(n-2) = 0 . - Benoit Cloitre, May 27 2004

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proposed

editing

#20 by Colin Barker at Mon May 27 08:55:34 EDT 2019

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editing

proposed

#19 by Colin Barker at Mon May 27 08:53:58 EDT 2019

PROG

(PARI) Vec((1 - 19*x - 141*x^2) / ((1 - 32*x)*(1 + 11*x - x^2)) + O(x^4020)) \\ Colin Barker, May 27 2019

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proposed

editing

#18 by Colin Barker at Mon May 27 08:48:02 EDT 2019

STATUS

editing

proposed

#17 by Colin Barker at Mon May 27 08:47:29 EDT 2019

LINKS

<a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (21,353,-32).

FORMULA

From Colin Barker, May 27 2019: (Start)

G.f.: (1 - 19*x - 141*x^2) / ((1 - 32*x)*(1 + 11*x - x^2)).

a(n) = 21*a(n-1) + 353*a(n-2) - 32*a(n-3) for n>2.

(End)

PROG

(PARI) Vec((1 - 19*x - 141*x^2) / ((1 - 32*x)*(1 + 11*x - x^2)) + O(x^40)) \\ Colin Barker, May 27 2019

STATUS

approved

editing