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A007758
a(n) = 2^n*n^2.
50
0, 2, 16, 72, 256, 800, 2304, 6272, 16384, 41472, 102400, 247808, 589824, 1384448, 3211264, 7372800, 16777216, 37879808, 84934656, 189267968, 419430400, 924844032, 2030043136, 4437573632, 9663676416, 20971520000, 45365592064, 97844723712, 210453397504
OFFSET
0,2
COMMENTS
"The traveling salesman problem can be solved in time O(n^2 2^n) (where n is the size of the network to visit)." [Wikipedia] - Jonathan Vos Post, Apr 10 2006
Satisfies Benford's law [Theodore P. Hill, Personal communication, Feb 06, 2017]. - N. J. A. Sloane, Feb 08 2017
REFERENCES
Arno Berger and Theodore P. Hill. An Introduction to Benford's Law. Princeton University Press, 2015.
Konrad Knopp, Theory and Application of Infinite Series, Dover, p. 269.
LINKS
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Wikipedia, Complexity.
FORMULA
From Henry Bottomley, Jun 13 2001: (Start)
a(n) = 2*A014477(n-1).
G.f.: 2*x(1+2*x)/(1-2*x)^3.
Binomial transform of A002939.
Inverse binomial transform of A062189. (End)
Sum_{n>=1} 1/a(n) = Pi^2/12 - (1/2)*(log(2))^2. - Benoit Cloitre, Apr 05 2002
a(n) = Sum_{k=1..n} k*2^k. - Zerinvary Lajos, Oct 09 2006
E.g.f.: exp(2*x)*(2*x + 4*x^2). - Geoffrey Critzer, Aug 28 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = -Li_2(-1/2) (A355234). - Amiram Eldar, Jun 28 2022
MAPLE
seq(seq(k^n*n^k, k=2..2), n=0..25); and seq(2^n*n^2, n=0..25); # Zerinvary Lajos, Jul 01 2007
MATHEMATICA
Table[n^2 * 2^n, {n, 0, 31}] (* Alonso del Arte, Oct 22 2014 *)
LinearRecurrence[{6, -12, 8}, {0, 2, 16}, 30] (* Harvey P. Dale, Jan 27 2017 *)
PROG
(Magma) [2^n*n^2: n in [0..30]]; // Vincenzo Librandi, Oct 27 2011
(PARI) a(n)=n^2<<n \\ Charles R Greathouse IV, Oct 28 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David J. Snook (ua532(AT)freenet.victoria.bc.ca)
STATUS
approved