|
| |
|
|
A097139
|
|
Convolution of 5^n and floor(n/2).
|
|
2
|
|
|
|
0, 0, 1, 6, 32, 162, 813, 4068, 20344, 101724, 508625, 2543130, 12715656, 63578286, 317891437, 1589457192, 7947285968, 39736429848, 198682149249, 993410746254, 4967053731280, 24835268656410, 124176343282061, 620881716410316
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,4
|
|
|
COMMENTS
|
a(n+1) gives partial sums of A033115 and second partial sums of A015531.
Partial sums of (1/4)*floor(5^n/6) = (1/3)*floor(5^n/8). - Mircea Merca, Dec 27 2010
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = 5^(n+1)/96 -n/8 -3/32 +(-1)^n/24. - R. J. Mathar, Jan 08 2011
G.f.: x^2/((1-x)*(1-5*x)*(1-x^2)).
a(n) = 6*a(n-1) - 4*a(n-2) - 6*a(n-3) + 5*a(n-4).
a(n) = Sum_{k=0..n} floor((n-k)/2)*4^k = Sum_{k=0..n} floor(k/2)*4^(n-k).
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).
a(n) = a(n-2) + (5^(n-1)-1)/4, n>1. (End)
a(n) = (floor(5^(n+1)/24) - floor((n+1)/2))/4. - Seiichi Manyama, Dec 22 2023
|
|
|
MAPLE
|
|
|
|
MATHEMATICA
|
f[n_] := Floor[5^n/6]/4; Accumulate@ Array[f, 24, 0]
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]
CoefficientList[ Series[x^2/((1 - x) (1 - 5 x) (1 - x^2)), {x, 0, 23}], x] (* Robert G. Wilson v, Jan 02 2011 *)
LinearRecurrence[{6, -4, -6, 5}, {0, 0, 1, 6}, 30] (* Harvey P. Dale, Mar 16 2019 *)
|
|
|
PROG
|
(Magma) [5^(n+1)/96 -n/8 -3/32 +(-1)^n/24: n in [0..30]]; // Vincenzo Librandi, Jun 25 2011
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
