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A295380
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Number of canonical forms for separation coordinates on hyperspheres S_n, ordered by increasing number of independent continuous parameters.
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1
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1, 1, 1, 2, 3, 1, 3, 8, 5, 1, 6, 20, 22, 8, 1, 11, 49, 73, 46, 11, 1, 23, 119, 233, 206, 87, 15, 1, 46, 288, 689, 807, 485, 147, 19, 1, 98, 696, 1988, 2891, 2320, 1021, 236, 24, 1, 207, 1681, 5561, 9737, 9800, 5795, 1960, 356, 29, 1, 451, 4062, 15322, 31350, 38216, 28586, 13088, 3525, 520, 35, 1, 983, 9821, 41558, 97552, 139901, 127465, 74280, 27224, 5989, 730, 41, 1
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OFFSET
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1,4
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COMMENTS
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Table 1 of the Schöbel and Veselov paper with initial 1 added. Reverse of Table 2 of the Devadoss and Read paper.
Apparently A032132 contains the row sums.
In this triangle, which is read by rows, for 0 <= k <= n-1 and n>=1, let T(n,k) be the number of inequivalent canonical forms for separation coordinates of the hypersphere S^n with k independent continuous parameters. It is the mirror image of sequence A232206, that is, T(n, k) = A232206(n+1, n-k) for 0 <= k <= n-1 and n>=1. (Triangular array A232206(N, K) is defined for N >= 2 and 1 <= K <= N-1.)
If B(x,y) = Sum_{n,k>=0} T(n,k)*x^n*y^k (with T(0,0) = 1, T(0,k) = 0 for k>=1, and T(n,k) = 0 for 1 <= n <= k), then B(x,y) = 1 + (x/2)*(B(x,y)^2/(1-x*y*B(x,y)) + (1 + x*y*B(x,y))*B(x^2,y^2)/(1-x^2*y^2*B(x^2,y^2))). This can be derived from the bivariate g.f. of A232206. See the comments for that sequence.
Let S(n) := Sum_{k>=0} T(n,k). The g.f. of S(n) is B(x, y=1). If we let y=1 in the above functional equation, we get x*B(x,1) = x + (1/2)*((x*B(x,1))^2/(1-x*B(x,1)) + (1 + x*B(x,1))*x^2*B(x^2,1)/(1-x^2*B(x^2,1))). After some algebra, we get 2*x*B(x,1) = x + (1/2)(x*B(x,1)/(1-x*B(x,1)) + (x*B(x,1) + x^2*B(x^2,1))/(1-x^2*B(x,1))), i.e., 2*x*B(x,1) = x + BIK(x*B(x,1)), where we have the "BIK" (reversible, indistinct, unlabeled) transform of C. G. Bower. This proves that S(n) = A032132(n+1) for n>=0, which is Copeland's claim above.
Note that for the second column we have T(n,k=2) = A048739(n-2) for 2 <= n < = 10, but T(11,2) = 4062 <> 4059 = A048739(9). In any case, they have different g.f.s (see the formula section below).
(End)
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LINKS
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FORMULA
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G.f.: If B(x,y) = Sum_{n,k>=0} T(n,k)*x^n*y^k (with T(0,0) = 1, T(0,k) = 0 for k>=1, and T(n,k) = 0 for 1 <= n <= k), then B(x,y) = 1 + (x/2)*(B(x,y)^2/(1-x*y*B(x,y)) + (1 + x*y*B(x,y))*B(x^2,y^2)/(1-x^2*y^2*B(x^2,y^2))).
If c(N,K) = A232206(N,K) and C(x,y) = Sum_{N,K>=0} c(N,K)*x^N*y^K (with c(1,0) = 1 and c(N,K) = 0 for 0 <= N <= K), then C(x,y) = x*B(x*y, 1/y) and B(x,y) = C(x*y, 1/y)/(x*y).
Setting y=0 in the above functional equation, we get x*B(x,0) = x + (1/2)*((x*B(x,0))^2 + x^2*B(x^2,0)), which is the functional equation for the g.f. of the first column. This proves that T(n,k=0) = A001190(n+1) for n>=0 (assuming T(0,0) = 1).
The g.f. of the second column is B_1(x,0) = Sum_{n>=0} T(n,2)*x^n = lim_{y->0} (B(x,y)-B(x,0))/y, where B(x,0) = 1 + x + x^2 + ... is the g.f. of the first column. We get B_1(x,0) = x*B(x,0)*(B(x,0) - 1)/(1 - x*B(x,0)).
(End)
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EXAMPLE
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Triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9
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(S^1) 1,
(S^2) 1, 1,
(S^3) 2, 3, 1,
(S^4) 3, 8, 5, 1,
(S^5) 6, 20, 22, 8, 1,
(S^6) 11, 49, 73, 46, 11, 1,
(S^7) 23, 119, 233, 206, 87, 15, 1,
(S^8) 46, 288, 689, 807, 485, 147, 19, 1,
(S^9) 98, 696, 1988, 2891, 2320, 1021, 236, 24, 1,
(S^10) 207, 1681, 5561, 9737, 9800, 5795, 1960, 356, 29, 1,
...
(End)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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