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A032132
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Number of series-reduced dyslexic planted planar trees with n leaves.
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4
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1, 1, 2, 6, 17, 57, 191, 684, 2482, 9275, 35127, 135156, 525545, 2064329, 8173895, 32600082, 130823306, 527888023, 2140454687, 8716907165, 35638352814, 146221542191, 601870210193, 2484682879348, 10285116277096, 42679973961811, 177514171393035, 739881841810694, 3089914920914855, 12927860306782626
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OFFSET
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1,3
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COMMENTS
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Apparently, beginning with a(3), number of non-equivalent canonical forms of separation coordinates on the hyperspheres. Cf. Schöbel and Veselov for this and other interpretations. - Tom Copeland, Nov 21 2017
Let A(x) = Sum_{n>=1} a(n)*x^n. For a derivation of the formula BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1-A(x^2))), see the comments for sequence A001224 and the weblink below containing Bower's theory of transforms.
We clarify the comment by T. Copeland above. Consider the material in Section 3 of Devadoss and Read (2001). According to their terminology, let b(m,n) be "the number of A-clusters having m cells and n outside edges not counting the root edge." Let B(x,y) = Sum_{m>=0, n>=0} b(m,n)*x^m*y^n. (See p. 78 in their paper, where they use the notations a_{m,n} and A(x,y) rather than b(m,n) and B(x,y), respectively, that we use here.)
On p. 79 (Eq. (3.1)) of their paper, they prove that B(x,y) = y + (x/2)*(B(x,y)^2/(1-B(x,y)) + (1 + B(x,y))*B(x^2, y^2)/(1-B(x^2,y^2))). Unfortunately, the factor x in the previous formula is left out (i.e., it is a typo), and the same typo is reproduced in Schöbel and Veselov (2014, 2015).
Using Table 2 (p. 92) from Devadoss and Read (2001) (and the material on p. 79), we get that B(x,y) = y+ x*y^2 + (x^2 + x)*y^3 + (2*x^3 + 3*x^2 + x)*y^4 + (3*x^4 + 8*x^3 + 5*x^2 + x)*y^5 + ...
We claim that a(n) = Sum_{m>=0} b(m,n) and A(y) = Sum_{n>=1} a(n)*y^n = B(x=1, y). To prove these claims, note that, for x=1, the above series becomes B(x=1,y) = y + y^2 + 2*y^3 + 6*y^4 + 17*y^5 + ..., while the functional equation above becomes B(1, y) = y + (1/2)*(B(1,y)^2/(1-B(1,y)) + (1 + B(1,y))*B(1,y^2)/(1-B(1,y^2))), which is equivalent to 2*B(1,y) = y + (1/2)*(B(1,y)/(1-B(1,y)) + (B(1,y) + B(1,y^2))/(1-B(1,y^2))). The latter formula is the one given in the formula section below (derived from Bower's theory) with x replaced with y and A(x) replaced with B(1,y). This proves that B(x=1, y) = A(y), from which we can easily get that a(n) = Sum_{m>=0} b(m,n).
Note that b(m=0, n) = 0 for n <> 1, but b(m=0, n=1) = 1; b(m,n) = 0 when m >= n >= 1; and b(m=1, n) = 1 for n>=2. Also, b(m,m+1) = A001190(m+1) for m>=1, which are the Wedderburn-Etherington numbers, and apparently b(m=2, n) = A024206(n-1) for n>=2 (conjecture).
In Section 6 of their paper, Schöbel and Veselov (2014, 2015) prove that b(m,n) is the "number of non-equivalent faces of [the Stasheff polytope] K_n of codimension m-1." Apparently then, for n>=2 and k>=0, b(n-k,n+1) is the "number of canonical forms for separation coordinates of [hypersphere] S^n" with k "independent continuous parameters". For k=0 and n>=2, b(n,n+1) = A001190(n+1) = "number of canonical forms for separation coordinates" of hypersphere S^n with 0 continuous parameters.
It turns out that for k, the number of continuous parameters of S^n, we have 0 <= k <= n-1 (see pp. 1269-1270 in Shobel and Veselov (2015)). Hence, for n>=2, Sum_{k=0..n-1} b(n-k, n+1) = Sum_{m=1..n} b(m, n+1) = Sum_{m=0..n} b(m, n+1) = a(n+1) (see above). As a result, for n>=2, a(n+1) is the "total number of [non-equivalent] canonical forms for separation coordinates on [hypersphere] S^n", which is the comment made by T. Copeland above.
(End)
For an explanation on the meaning of clusters of types A, B, and C see Section 3 (pp. 78-81) in Devadoos and Read (2001). See also the comments for sequence A232206. - Petros Hadjicostas, Mar 02 2018
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LINKS
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FORMULA
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Doubles (index 2+) under "BIK" (reversible, indistinct, unlabeled) transform.
G.f.: If A(x) = Sum_{n>=1} a(n)*x^n, then 2*A(x) = x + BIK(A(x)) = x + (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1-A(x^2))). - Petros Hadjicostas, Jan 17 2018
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MATHEMATICA
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BIK[p_] := (1/(1-p) + (1+p)/(1-p /. x -> x^2))/2;
seq[n_] := Module[{p=x}, For[i=2, i <= n, i++, p += x^i Coefficient[BIK[p] + x O[x]^i // Normal, x, i]]; CoefficientList[p/x, x]];
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PROG
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(PARI)
BIK(p)={(1/(1-p) + (1+p)/subst(1-p, x, x^2))/2}
seq(n)={my(p=x); for(i=2, n, p+=x^i*polcoeff(BIK(p) + O(x*x^i), i)); Vecrev(p/x)} \\ Andrew Howroyd, Aug 30 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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