|
|
A062189
|
|
a(n) = 2 * 3^(n-2)*n*(1+2*n).
|
|
2
|
|
|
0, 2, 20, 126, 648, 2970, 12636, 51030, 198288, 747954, 2755620, 9959598, 35429400, 124357194, 431530092, 1482720390, 5050815264, 17075199330, 57338232372, 191385721566, 635369601960, 2099044209402, 6903833113980
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Define a triangle with left (first) column T(n,0)=n^2 for n=0,1,2,3.. and the remaining terms T(r,c) = T(r-1,c-1) + 2*T(r,c-1). Then T(n,n) = a(n) on the diagonal. T(n,1) = A056105(n). - J. M. Bergot, Jan 26 2013
|
|
LINKS
|
|
|
FORMULA
|
|
|
MATHEMATICA
|
Table[2*3^(n-2)*n*(1+2*n), {n, 0, 30}] (* G. C. Greubel, Jun 06 2019 *)
LinearRecurrence[{9, -27, 27}, {0, 2, 20}, 30] (* Harvey P. Dale, Jun 08 2022 *)
|
|
PROG
|
(PARI) { for (n=0, 200, write("b062189.txt", n, " ", n*(4*n + 2)*3^(n - 2)) ) } \\ Harry J. Smith, Aug 02 2009
(Magma) [2*3^(n-2)*n*(1+2*n): n in [0..30]]; // G. C. Greubel, Jun 06 2019
(Sage) [2*3^(n-2)*n*(1+2*n) for n in (0..30)] # G. C. Greubel, Jun 06 2019
(GAP) List([0..30], n-> 2*3^(n-2)*n*(1+2*n)) # G. C. Greubel, Jun 06 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|