

A274626


a(n) = Product_{i=0..2} (2^floor((n+i)/3)1).


2



0, 0, 0, 1, 3, 9, 27, 63, 147, 343, 735, 1575, 3375, 6975, 14415, 29791, 60543, 123039, 250047, 504063, 1016127, 2048383, 4112895, 8258175, 16581375, 33227775, 66585855, 133432831, 267126783, 534776319, 1070599167, 2142244863, 4286583807, 8577357823, 17158905855, 34326194175, 68669157375
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OFFSET

0,5


COMMENTS

Tom Karzes, Jul 05 2016: (Start)
This is a threedimensional analog of the holesinsheetofpaper sequence A274230.
In d dimensions, assuming the axes for folding are selected in a roundrobin 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 d1 (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 dway intersections, which corresponds to the number of holes, is:
Product_{i=0..d1}(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 = dmod(n,d)
p2 = mod(n,d)
holes = (v1^p1)*(v2^p2)
This flattens to:
((2^(n/d)1)^(dmod(n,d))) * ((2^(n/d+1)1)^(mod(n,d)))
(End)


REFERENCES

Tom Karzes, Posting to Math Fun Mailing List, Jul 05 2016.


LINKS

Table of n, a(n) for n=0..36.


FORMULA

Empirical g.f.: x^3*(1+2*x^2) / ((1x)*(12*x)*(12*x^3)*(14*x^3)).  Colin Barker, Jul 06 2016


MAPLE

f:=(n, d) > mul(2^floor((n+i)/d)1, i = 0 .. d1);
[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

Cf. A274230, A274627.
Sequence in context: A227097 A201202 A260938 * A161712 A280466 A137368
Adjacent sequences: A274623 A274624 A274625 * A274627 A274628 A274629


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Jul 05 2016


STATUS

approved



