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A215005 a(n) = a(n-2) + a(n-1) + floor(n/2) + 1 for n > 1 and a(0)=0, a(1)=1. 2

%I #26 Oct 05 2017 02:54:31

%S 0,1,3,6,12,21,37,62,104,171,281,458,746,1211,1965,3184,5158,8351,

%T 13519,21880,35410,57301,92723,150036,242772,392821,635607,1028442,

%U 1664064,2692521,4356601,7049138,11405756,18454911,29860685,48315614,78176318,126491951,204668289

%N a(n) = a(n-2) + a(n-1) + floor(n/2) + 1 for n > 1 and a(0)=0, a(1)=1.

%C If the seed is {1,1}: 1, 1, 4, 7, 14, 24, 42, 70, 117, 192, 315, 513, 835, 1355, 2198, 3561, 5768, 9338, 15116, 24464, 39591, 64066, 103669, 167747, ...

%C If the seed is {1,2}: A129696.

%C Same seed, but -1 in the formula instead of +1: b(n)=a(n-2)+1 for n>=2, i.e. 0, 1, 1, 2, 4, 7, 13, 22, 38, 63, 105, 172, 282, 459, 747, 1212, 1966, 3185, 5159, 8352, 13520, 21881, 35411, 57302, 92724, 150037, 242773, 392822, ...

%H Colin Barker, <a href="/A215005/b215005.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-3,0,1).

%F a(n) = 2*F(n+2)-n/2-9/4+(-1)^n/4, where F is Fibonacci number. - _Vaclav Kotesovec_, Aug 11 2012

%F From _Colin Barker_, Sep 16 2015: (Start)

%F a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5) for n>4.

%F G.f.: x*(x^2-x-1) / ((x-1)^2*(x+1)*(x^2+x-1)).

%F (End)

%t LinearRecurrence[{2, 1, -3, 0, 1}, {0, 1, 3, 6, 12}, 39] (* _Jean-François Alcover_, Oct 05 2017 *)

%o (Python)

%o prpr = 0

%o prev = 1

%o for n in range(2,100):

%o print prpr,

%o curr = prpr+prev + 1 + n//2

%o prpr = prev

%o prev = curr

%o (PARI) concat(0, Vec(x*(x^2-x-1) / ((x-1)^2*(x+1)*(x^2+x-1)) + O(x^100))) \\ _Colin Barker_, Sep 16 2015

%Y Cf. A129696 (same formula, seed {1,2}).

%Y Cf. A000071 (a(n+1) = a(n-1) + a(n) + 1).

%K nonn,easy

%O 0,3

%A _Alex Ratushnyak_, Jul 31 2012

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