A272690 a(n) = 22*Sum_{i=0..n-2} 46^i*2^(n-2-i) + 2^(n-1).
1, 24, 1060, 48672, 2238736, 102981504, 4737148480, 217908828672, 10023806116096, 461095081334784, 21210373741388800, 975677192103862272, 44881150836777619456, 2064532938491770404864, 94968515170621438443520, 4368551697848586168041472, 200953378101034963729186816
Offset: 1
Keywords
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 entry for sequences related to LEGO blocks
- Index entries for linear recurrences with constant coefficients, signature (48,-92).
Programs
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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)];
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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 *) -
PARI
A272690(n) = 2^(n - 2)*(1 + 23^(n - 1)) \\ Rick L. Shepherd, Jun 02 2016
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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
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
Comments