A097139 - OEIS

%I #39 May 09 2024 09:08:06

%S 0,0,1,6,32,162,813,4068,20344,101724,508625,2543130,12715656,

%T 63578286,317891437,1589457192,7947285968,39736429848,198682149249,

%U 993410746254,4967053731280,24835268656410,124176343282061,620881716410316

%N Convolution of 5^n and floor(n/2).

%C a(n+1) gives partial sums of A033115 and second partial sums of A015531.

%C Partial sums of (1/4)*floor(5^n/6) = (1/3)*floor(5^n/8). - _Mircea Merca_, Dec 27 2010

%H Vincenzo Librandi, <a href="/A097139/b097139.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6,-4,-6,5).

%F a(n) = 5^(n+1)/96 -n/8 -3/32 +(-1)^n/24. - _R. J. Mathar_, Jan 08 2011

%F G.f.: x^2/((1-x)*(1-5*x)*(1-x^2)).

%F a(n) = 6*a(n-1) - 4*a(n-2) - 6*a(n-3) + 5*a(n-4).

%F a(n) = Sum_{k=0..n} floor((n-k)/2)*4^k = Sum_{k=0..n} floor(k/2)*4^(n-k).

%F From _Mircea Merca_, Dec 27 2010: (Start)

%F 4*a(n) = round((5*5^n-12*n-9)/24) = floor((5*5^n-12*n-5)/24) = ceiling((5*5^n-12*n-13)/24) = round((5*5^n-12*n-5)/24).

%F a(n) = a(n-2) + (5^(n-1)-1)/4, n>1. (End)

%F a(n) = (floor(5^(n+1)/24) - floor((n+1)/2))/4. - _Seiichi Manyama_, Dec 22 2023

%p A097139 := proc(n) 5^(n+1)/96 -n/8 -3/32 +(-1)^n/24 ; end proc: # _R. J. Mathar_, Jan 08 2011

%t f[n_] := Floor[5^n/6]/4; Accumulate@ Array[f, 24, 0]

%t a[n_] := a[n] = 6 a[n - 1] - 4 a[n - 2] - 6 a[n - 3] + 5 a[n - 4]; a[0] = a[1] = 0; a[2] = 1; a[3] = 6; Array[a, 24, 0]

%t CoefficientList[ Series[x^2/((1 - x) (1 - 5 x) (1 - x^2)), {x, 0, 23}], x] (* _Robert G. Wilson v_, Jan 02 2011 *)

%t LinearRecurrence[{6,-4,-6,5},{0,0,1,6},30] (* _Harvey P. Dale_, Mar 16 2019 *)

%o (Magma) [5^(n+1)/96 -n/8 -3/32 +(-1)^n/24: n in [0..30]]; // _Vincenzo Librandi_, Jun 25 2011

%Y Column k=5 of A368296.

%K easy,nonn

%O 0,4

%A _Paul Barry_, Jul 29 2004