A274626 - OEIS

%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