OFFSET
1,2
COMMENTS
This sequence gives a lower bound on the number of ways of combining n 2 X 4 LEGO blocks.
The formula as given was found at the LEGO Company in 1974 and the numbers a(2), a(3), a(6) were used in communication until the emergence of A112389. - Søren Eilers, Aug 02 2018
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..602
S. Eilers, The LEGO counting problem, Amer. Math. Monthly, 123 (May 2016), 415-426.
Jørgen Kirk Kristiansen, Taljonglering med klodser - eller talrige klodser, Klodshans 1974 [In Danish].
Fabien Pazuki, Combinatoire des briques LEGO, Images des Mathématiques, CNRS, 2016. [In French]
Index entries for linear recurrences with constant coefficients, signature (48,-92).
FORMULA
From Colin Barker, May 31 2016: (Start)
a(n) = 2^(n-2)*(23+23^n)/23.
a(n) = 48*a(n-1) - 92*a(n-2) for n > 2.
G.f.: x*(1-24*x) / ((1-2*x)*(1-46*x)).
(End)
First formula follows by simplifying the formula in the definition, and the other two follow immediately. - Rick L. Shepherd, Jun 02 2016
Since there are 46 ways to attach one such brick on top of another, 2 of which are self-symmetric, the number of buildings with n 2 X 4 LEGO bricks of maximal height becomes a(n) = (46^(n-1) + 2^(n-1))/2 when adjusted for rotation in the XY-plane. That this is the same as the original formula found at LEGO follows by isolating a finite geometric series. - Søren Eilers, Aug 02 2018
MAPLE
t1:=n->22*add(46^i*2^(n-2-i), i=0..n-2)+2^(n-1);
t2:=[seq(t1(n), n=1..20)];
MATHEMATICA
Table[22*Sum[46^k * 2^(n-k-2), {k, 0, n-2}] + 2^(n-1), {n, 1, 25}] (* G. C. Greubel, May 31 2016 *)
PROG
(Ruby)
def A272690(n)
22 * (0..n - 2).inject(0){|s, i| s + 46 ** i * 2 ** (n - 2 - i)} + 2 ** (n - 1)
end # Seiichi Manyama, May 31 2016
(PARI) A272690(n) = 2^(n - 2)*(1 + 23^(n - 1)) \\ Rick L. Shepherd, Jun 02 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, May 31 2016
STATUS
approved