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%I #37 Apr 11 2024 16:30:36
%S 0,1,1,7,26,57,99,159,244,353,485,647,846,1081,1351,1663,2024,2433,
%T 2889,3399,3970,4601,5291,6047,6876,7777,8749,9799,10934,12153,13455,
%U 14847,16336,17921,19601,21383,23274,25273,27379,29599,31940,34401,36981,39687
%N a(n) + a(n+2) = n^3.
%C If the offset were 0, the formula would be: a(0)=0, a(1)=1, for n>=2: a(n) = (n-1)^3 - a(n-2).
%H Vincenzo Librandi, <a href="/A206481/b206481.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-7,8,-7,4,-1).
%F a(n) = (1/2)*((n-3)*n^2-4*cos((Pi*n)/2)+2). - _Harvey P. Dale_, Sep 14 2012
%F G.f.: x^2*(1 - 3*x + 10*x^2 - 3*x^3 + x^4)/((1-x)^4*(1+x^2)). - _Paul D. Hanna_, Sep 14 2012
%t LinearRecurrence[{4, -7, 8, -7, 4, -1}, {0, 1, 1, 7, 26, 57}, 60]
%t RecurrenceTable[{a[1]==0,a[2]==1,a[n]==(n-2)^3-a[n-2]},a,{n,50}] (* _Harvey P. Dale_, Sep 14 2012 *)
%o (Python)
%o prpr = 0
%o prev = 1
%o for n in range(1,77):
%o print(prpr, end=',')
%o curr = n*n*n - prpr # a(n+1)
%o prpr = prev
%o prev = curr
%Y Cf. A144129 (bisection).
%K nonn,easy,changed
%O 1,4
%A _Vladimir Joseph Stephan Orlovsky_, Feb 08 2012
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