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%I #27 Jan 18 2021 17:03:59
%S 0,0,0,1,3,9,27,63,147,343,735,1575,3375,6975,14415,29791,60543,
%T 123039,250047,504063,1016127,2048383,4112895,8258175,16581375,
%U 33227775,66585855,133432831,267126783,534776319,1070599167,2142244863,4286583807,8577357823,17158905855,34326194175,68669157375
%N a(n) = Product_{i=0..2} (2^floor((n+i)/3)-1).
%C From _Tom Karzes_, Jul 05 2016: (Start)
%C This is a three-dimensional analog of the holes-in-sheet-of-paper sequence A274230.
%C In d dimensions, assuming the axes for folding are selected in a round-robin fashion, the number of times a given dimension is folded is:
%C floor((n+i)/d)
%C where i runs from 0 (for the last dimension to be folded) through d-1 (for the first dimension to be folded).
%C The corresponding number of internal dividing lines/planes/etc. is (2^floor((n+i)/d)-1). The number of internal d-way intersections, which corresponds to the number of holes, is:
%C Product_{i=0..d-1}(2^floor((n+i)/d)-1)
%C where d is the number of dimensions and n is the total number of folds.
%C Note that the first several nonzero entries in these sequences are the powers of 3. Specifically, in d dimensions, the first d entries are 0, followed by the first (d+1) powers of 3.
%C It's not hard to see why this is so. The first nonzero entry occurs at d folds, and the value is 1. This is when you've folded once along each dimension.
%C After that, the next d folds each divide 2 old partitions into 4 new ones, i.e., they change the internal folds from 1 to 3. So for the next d entries you just multiply the previous entry by 3 (or more generally, by 3/1).
%C After that you multiply by 7/3 for the next d entries, then 15/7, then 31/15, etc. Each time you're just replacing one of the old factors with a new one, where each factor is one less than a power of two.
%C Here's an alternative formulation that avoids the iterated product.
%C For a given number of folds, there are only two factors, each raised to some exponent (with the sum of the exponents being the dimension d):
%C v1 = 2^(n/d)-1
%C v2 = 2^(n/d+1)-1
%C p1 = d-mod(n,d)
%C p2 = mod(n,d)
%C holes = (v1^p1)*(v2^p2)
%C This flattens to:
%C ((2^(n/d)-1)^(d-mod(n,d))) * ((2^(n/d+1)-1)^(mod(n,d)))
%C (End)
%D Tom Karzes, Posting to Math Fun Mailing List, Jul 05 2016.
%F Empirical g.f.: x^3*(1+2*x^2) / ((1-x)*(1-2*x)*(1-2*x^3)*(1-4*x^3)). - _Colin Barker_, Jul 06 2016
%p f:=(n,d) -> mul(2^floor((n+i)/d)-1, i = 0 .. d-1);
%p [seq(f(n,3),n=0..40)];
%o (PARI) a(n) = prod(i=0, 2, 2^floor((n+i)/3)-1) \\ _Colin Barker_, Jul 06 2016
%Y Cf. A274230, A274627.
%K nonn
%O 0,5
%A _N. J. A. Sloane_, Jul 05 2016
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