The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A138041

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A138041

a(1) = 1, a(2) = 10; for n>2, a(n+1) = 4*a(n) + 6*a(n-1). Also a(n) = upper left term in the 2 X 2 matrix [1,3; 3,3].
(history; published version)

#16 by Joerg Arndt at Tue Jan 02 08:59:29 EST 2024

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#15 by Paolo P. Lava at Tue Jan 02 08:30:06 EST 2024

FORMULA

a(n)=(1/2)*(2+sqrt(10))^n-(13/20)*(2-sqrt(10))^n*sqrt(10)+(13/20)*(2+sqrt(10))^n*sqrt(10)+(1/2)*(2-sqrt(10))^n. - Paolo P. Lava, Jun 03 2008

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#14 by Jon E. Schoenfield at Wed Sep 16 13:17:20 EDT 2015

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#13 by Jon E. Schoenfield at Wed Sep 16 13:17:18 EDT 2015

FORMULA

a(n)=(1/2)*(2+sqrt(10))^n-(13/20)*(2-sqrt(10))^n*sqrt(10)+(13/20)*(2+sqrt(10))^n*sqrt(10)+(1/2)*(2-sqrt(10))^n . - Paolo P. Lava, Jun 03 2008

EXAMPLE

a(4) = 244 = upper left term in [1,3; 3,3]^4.

MATHEMATICA

a = {1, 10}; Do[AppendTo[a, 4*a[[ -1]] + 6*a[[ -2]]], {25}]; a - _(* _Stefan Steinerberger_ *)

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#12 by Ray Chandler at Fri Jul 31 21:26:57 EDT 2015

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#11 by Ray Chandler at Fri Jul 31 21:26:52 EDT 2015

LINKS

<a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4, 6).

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#10 by Harvey P. Dale at Sun Mar 09 09:19:18 EDT 2014

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#9 by Harvey P. Dale at Sun Mar 09 09:19:12 EDT 2014

LINKS

Harvey P. Dale, <a href="/A138041/b138041.txt">Table of n, a(n) for n = 1..1000</a>

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#8 by Harvey P. Dale at Sun Mar 09 09:18:03 EDT 2014

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#7 by Harvey P. Dale at Sun Mar 09 09:17:57 EDT 2014

MATHEMATICA

LinearRecurrence[{4, 6}, {1, 10}, 30] (* Harvey P. Dale, Mar 09 2014 *)

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