A032130 Shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.
2, 2, 5, 15, 52, 193, 765, 3143, 13323, 57670, 254040, 1134249, 5122124, 23349966, 107310784, 496633774, 2312539465, 10826481544, 50929829953, 240616214596, 1141195080020, 5431477088428, 25933525825389, 124185539096075, 596268057962349, 2869992942831031, 13845453533124431, 66934180769445444, 324218809545624984
Offset: 1
Keywords
Examples
From _Petros Hadjicostas_, Jan 14 2018: (Start) According to Bower's theory in the link above, we have boxes of different sizes and colors. The size of a box is determined by the number of balls it can hold. Two boxes of the same size and color are considered identical (indistinct and unlabeled). We place the boxes on a line that can be read in either direction; i.e., we have a reversible line. Here, a(n) = number of colors a box holding n balls can be, while b(n) = number of ways of placing boxes in a line that can be read in either direction when the total number of balls is n. Since a(1) = 2, a(2) = 2, a(3) = 5, a(4) = 15, etc., a box with 1 ball can be of 2 colors only, a box with 2 balls can be of 2 colors only, a box with 3 balls can be of 5 colors only, a box with 4 balls can be of 15 colors, and so on. When we have n=3 balls, we have b(3) = a(4) = 15. Here, we consider three cases. In the first case, we have one box holding 3 balls and we have 5 possibilities. In the second case, we have a box with 2 balls and a box with 1 ball, and we have 2 x 2 = 4 possibilities here because the line is reversible (i.e., 21 is considered the same as 12). In the third case, we have three identical boxes each holding 1 ball and we have 6 possibilities (if the colors are a and b, we have the possibilities aaa, aab, aba, bba, bab, and bbb). Thus, b(3) = 5 + 4 + 6 = 15 = a(4). When we have n=4 balls, we have b(4) = a(5) = 52. Here we consider 5 cases: a single box with 4 balls (a(4) = 15 possibilities); a box with 3 balls and a box with 1 ball (a(3) x a(1) = 5 x 2 = 10 possibilities); two identical boxes each with 2 balls (3 possibilities, aa, ab, and bb); a box with 2 balls and two identical boxes each with 1 ball (14 possibilities, see below); and 4 identical boxes each with 1 ball (10 possibilities, aaaa, aaab, aaba, aabb, abba, baab, abab, abbb, babb, bbbb). Hence, b(4) = 15 + 10 + 3 + 14 + 10 = 52 = a(5). We explain the fourth case above in more detail. Here, we have a box with 2 balls and two identical boxes each with 1 ball. Let a and b be the two colors for the 1-ball boxes and A and B be the colors for the 2-ball boxes. Then we have the following 14 cases: Aaa, Aab, Abb, Baa, Bab, Bbb, aAa, aAb, bAb, aBa, aBb, bBb, abA, and abB. Note that Aab = baA <> abA = Aba and abB = Bba <> Bab = baB. (End)
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..200
- C. G. Bower, Transforms (2)
Programs
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Mathematica
m = 30; a[1] = 2; A[_] = 0; Do[A[x_] = x(a[1]+(1/2)(A[x]/(1-A[x])+(A[x]+A[x^2])/(1-A[x^2]))) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] // Rest (* Jean-François Alcover, Sep 17 2019 *) -
PARI
BIK(p)={(1/(1-p) + (1+p)/subst(1-p, x, x^2))/2} seq(n)={my(p=O(1));for(i=1, n, p=1+BIK(x*p)); Vec(p)} \\ Andrew Howroyd, Aug 30 2018
Formula
From Petros Hadjicostas, Jan 14 2018: (Start)
The sequence (a(n): n>=1) shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.
G.f.: If A(x) = Sum_{n>=1} a(n)*x^n, then -a(1) + A(x)/x = BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1-A(x^2))). Here a(1) = 2.
In general, if we let a(1) = c, we get:
a(2) = c,
a(3) = (1/2)*(c + 3)*c,
a(4) = (1/2)*(c + 3)*(c + 1)*c,
a(5) = (1/2)*(c^3 + 5*c^2 + 10*c + 4)*c,
a(6) = (1/4)*(2*c^4 + 15*c^3 + 38*c^2 + 37*c + 8)*c,
a(7) = (1/8)*(4*c^4 + 38*c^3 + 103*c^2 + 109*c + 22)*(c + 1)*c,
a(8) = (1/8)*(4*c^6 + 56*c^5 + 251*c^4 + 511*c^3 + 499*c^2 + 201*c + 22)*c,
and so on. No pattern is apparent in these formulae.
(End)
Extensions
Name edited by Petros Hadjicostas, Jan 14 2018
a(24)-a(29) from Petros Hadjicostas, Jan 14 2018
Comments