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%I #52 Mar 09 2024 14:54:07
%S 1,3,10,36,133,495,1846,6888,25705,95931,358018,1336140,4986541,
%T 18610023,69453550,259204176,967363153,3610248435,13473630586,
%U 50284273908,187663465045,700369586271,2613814880038,9754889933880,36405744855481,135868089488043,507066613096690
%N a(n) = a(n-3) - 5*a(n-2) + 5*a(n-1), a(0) = 1, a(1) = 3, a(2) = 10.
%C A floretion-generated sequence resulting from a particular transform of the periodic sequence (-1,1).
%C Floretion Algebra Multiplication Program, FAMP Code: .5em[J* ]forseq[ .25( 'i + 'j + 'k + i' + j' + k' + 'ii' + 'jj' + 'kk' + 'ij' + 'ik' + 'ji' + 'jk' + 'ki' + 'kj' + e ) ], em[J]forseq = A001834, vesforseq = (1,-1,1,-1). ForType 1A. Identity used: em[J]forseq + em[J* ]forseq = vesforseq.
%C Also indices of the centered triangular numbers which are triangular numbers - _Richard Choulet_, Oct 09 2007
%C Place a(n) red and b(n) blue balls in an urn; draw 2 balls without replacement. Probability(2 red balls) = 3*Probability(2 blue balls); b(n)=A101265(n). - _Paul Weisenhorn_, Aug 02 2010
%H Harvey P. Dale, <a href="/A102871/b102871.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-5,1).
%F 2*a(n) - A001834(n) = (-1)^(n+1); a(n) = 4*a(n-1) - a(n) - 1;
%F G.f.: (2*x-1)/((x-1)*(x^2-4*x+1)).
%F Superseeker results: a(n+2) - 2a(n+1) + a(n) = A001834(n+1) (from this and the first relation involving A001834 it follows that 4a(n+1) - a(n+2) - a(n) = (-1)^n as well as the recurrence relation given for A001834 ); a(n+1) - a(n) = A001075(n+1); a(n+2) - a(n) = A082841(n+1).
%F a(j+3) - 3*a(j+2) - 3*a(j+1) + a(j) = -2 for all j.
%F a(n+1) = 2*a(n) - 1/2 + (1/2)*(12*a(n)^2 - 12*a(n) + 9)^(1/2). - _Richard Choulet_, Oct 09 2007
%F a(n) = (sqrt(12*b(n)*(b(n)-1) + 1) + 1)/2; b(n) = A101265(n). - _Paul Weisenhorn_, Aug 02 2010
%F a(n) = A001571(n) + 1. - _Johannes Boot_, Jun 17 2011
%F E.g.f.: (exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) + cosh(x) + sinh(x))/2. - _Stefano Spezia_, Sep 19 2023
%e For n=5, a(5)=495; b(5)=286; binomial(495,2) = 122265 = 3*binomial(286,2). - _Paul Weisenhorn_, Aug 02 2010
%p a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=4*a[n-1]-a[n-2]-1 od: seq(a[n], n=1..23); # _Zerinvary Lajos_, Mar 08 2008
%t LinearRecurrence[{5,-5,1},{1,3,10},30] (* _Harvey P. Dale_, Oct 04 2011 *)
%Y Cf. A001075 (first differences), A001834, A082841, A101265.
%K nonn,easy
%O 0,2
%A _Creighton Dement_, Mar 01 2005
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