OFFSET
0,5
COMMENTS
From Tom Karzes, Jul 05 2016: (Start)
This is a three-dimensional analog of the holes-in-sheet-of-paper sequence A274230.
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:
floor((n+i)/d)
where i runs from 0 (for the last dimension to be folded) through d-1 (for the first dimension to be folded).
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:
Product_{i=0..d-1}(2^floor((n+i)/d)-1)
where d is the number of dimensions and n is the total number of folds.
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.
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.
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).
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.
Here's an alternative formulation that avoids the iterated product.
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):
v1 = 2^(n/d)-1
v2 = 2^(n/d+1)-1
p1 = d-mod(n,d)
p2 = mod(n,d)
holes = (v1^p1)*(v2^p2)
This flattens to:
((2^(n/d)-1)^(d-mod(n,d))) * ((2^(n/d+1)-1)^(mod(n,d)))
(End)
REFERENCES
Tom Karzes, Posting to Math Fun Mailing List, Jul 05 2016.
FORMULA
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
MAPLE
f:=(n, d) -> mul(2^floor((n+i)/d)-1, i = 0 .. d-1);
[seq(f(n, 3), n=0..40)];
PROG
(PARI) a(n) = prod(i=0, 2, 2^floor((n+i)/3)-1) \\ Colin Barker, Jul 06 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Jul 05 2016
STATUS
approved