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A059502
a(n) = (3*n*F(2n-1) + (3-n)*F(2n))/5 where F() = Fibonacci numbers A000045.
11
0, 1, 3, 9, 27, 80, 234, 677, 1941, 5523, 15615, 43906, 122868, 342409, 950727, 2631165, 7260579, 19982612, 54865566, 150316973, 411015705, 1121818311, 3056773383, 8316416134, 22593883752, 61301547025, 166118284299, 449639574897, 1215751720491, 3283883157848
OFFSET
0,3
COMMENTS
Substituting x(1-x)/(1-2x) into x/(1-x)^2 yields g.f. of sequence.
Variation of A059216 (and of Boustrophedon transform) applied to 1,2,3,4,...: fill an array by diagonals, each time in the same direction, say the 'up' direction. The first column is 1,2,3,4,... For the next element of a diagonal, add to the previous element the elements of the row the new element is in. The first row gives a(n).
FORMULA
a(n) = 2*a(n-1) + Sum{m<=n-2}a(m) + A001519(n-2).
G.f.: x(1-x)(1-2x)/(1-3x+x^2)^2. - Emeric Deutsch, Oct 07 2002
a(n) = A147703(n,1). - Philippe Deléham, Nov 29 2008
a(n) = A001871(n-1)-3*A001871(n-2)+2*A001871(n-3). - R. J. Mathar, Apr 09 2019
EXAMPLE
The array (see A059503) begins
1 3 9 27 80 ...
2 5 14 40 ...
3 7 19 ...
4 9 5 ...
MATHEMATICA
Table[(3n Fibonacci[2n-1]+(3-n)Fibonacci[2n])/5, {n, 0, 30}] (* or *) CoefficientList[Series[x(1-x)(1-2x)/(1-3x+x^2)^2, {x, 0, 30}], x] (* Harvey P. Dale, Apr 23 2011 *)
PROG
(PARI) a(n)=(3*n*fibonacci(2*n-1)+(3-n)*fibonacci(2*n))/5
(PARI) { for (n = 0, 200, write("b059502.txt", n, " ", (3*n*fibonacci(2*n - 1) + (3 - n)*fibonacci(2*n))/5); ) } \\ Harry J. Smith, Jun 27 2009
(Magma) [(3*n*Fibonacci(2*n-1)+(3-n)*Fibonacci(2*n))/5: n in [0..100]]; // Vincenzo Librandi, Apr 23 2011
KEYWORD
easy,nonn,nice
AUTHOR
Floor van Lamoen, Jan 19 2001
STATUS
approved