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a(n) is the square of the n-th partial sum minus the n-th partial sum of the squares, divided by a(n-1), for all n>=1, starting with a(0)=1, a(1)=2.
(history; published version)
#5 by Harvey P. Dale at Fri Mar 24 14:21:36 EDT 2017
STATUS

editing

approved

#4 by Harvey P. Dale at Fri Mar 24 14:21:32 EDT 2017
LINKS

<a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-1).

MATHEMATICA

LinearRecurrence[{1, 3, -1}, {1, 2, 2, 8}, 40] (* Harvey P. Dale, Mar 24 2017 *)

STATUS

approved

editing

#3 by Russ Cox at Fri Mar 30 18:36:39 EDT 2012
AUTHOR

_Paul D. Hanna (pauldhanna(AT)juno.com), _, Sep 16 2003

Discussion
Fri Mar 30
18:36
OEIS Server: https://oeis.org/edit/global/213
#2 by N. J. A. Sloane at Sat Nov 10 03:00:00 EST 2007
KEYWORD

nonn,new

nonn

AUTHOR

Paul D . Hanna (pauldhanna(AT)juno.com), Sep 16 2003

#1 by N. J. A. Sloane at Thu Feb 19 03:00:00 EST 2004
NAME

a(n) is the square of the n-th partial sum minus the n-th partial sum of the squares, divided by a(n-1), for all n>=1, starting with a(0)=1, a(1)=2.

DATA

1, 2, 2, 8, 12, 34, 62, 152, 304, 698, 1458, 3248, 6924, 15210, 32734, 71440, 154432, 336018, 727874, 1581496, 3429100, 7445714, 16151518, 35059560, 76068400, 165095562, 358241202, 777459488, 1687087532, 3661224794, 7945027902

OFFSET

0,2

FORMULA

a(n) = a(n-1) + 3a(n-2) - a(n-3) for n>3; G.f.: (1+x-3x^2+x^3)/(1-x-3x^2+x^3).

EXAMPLE

a(4)=12 since ((1+2+2+8)^2 - (1^2+2^2+2^2+8^2))/8 = (13^2-73)/8 = 12.

KEYWORD

nonn

AUTHOR

Paul D Hanna (pauldhanna(AT)juno.com), Sep 16 2003

STATUS

approved