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%I #36 Jan 04 2024 08:57:55
%S 1,0,1,0,1,1,0,1,4,1,0,1,10,10,1,0,1,20,50,20,1,0,1,35,175,175,35,1,0,
%T 1,56,490,980,490,56,1,0,1,84,1176,4116,4116,1176,84,1,0,1,120,2520,
%U 14112,24696,14112,2520,120,1,0,1,165,4950,41580,116424,116424,41580,4950,165,1
%N Triangle read by rows. The coefficients of the Baxter polynomials p(0, x) = 1 and p(n, x) = x*hypergeom([-1 - n, -n, 1 - n], [2, 3], -x) for n >= 1.
%C This triangle is a member of a family of Pascal-like triangles. Let T(n, k, m) = sf(m)*F(n - 1) / (F(k - 1)*F(n - k)) if k > 0 and otherwise k^n, where F(n) = Product_{j=0..m} (n + j)! and sf(m) are the superfactorials A000178. The case m = 2 gives this triangle, some other cases are given in the crossreferences. See also A342889 for a related representation of generalized binomial coefficients.
%F T(n, k) = [x^k] p(n, x).
%F T(n, k) = 2*F(n-1)/(F(k-1)*F(n-k)) for k > 0 where F(n) = n!*(n+1)!*(n+2)!.
%F p(n, 1) = A001181(n), i.e. the Baxter numbers are the values of the Baxter polynomials at x = 1.
%F (-1)^(n + 1)*p(2*n + 1, -1) = A217800(n) .
%e Triangle T(n, k) starts:
%e [0] 1
%e [1] 0, 1
%e [2] 0, 1, 1
%e [3] 0, 1, 4, 1
%e [4] 0, 1, 10, 10, 1
%e [5] 0, 1, 20, 50, 20, 1
%e [6] 0, 1, 35, 175, 175, 35, 1
%e [7] 0, 1, 56, 490, 980, 490, 56, 1
%e [8] 0, 1, 84, 1176, 4116, 4116, 1176, 84, 1
%e [9] 0, 1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1
%e .
%e Let p = (p1, p2,..., pn) denote a permutation of {1, 2,..., n}. The pair (p(i), p(i+1)) is a 'rise' if p(i) < p(i+1). Additionally a conventional rise is counted at the beginning of p.
%e T(n, k) is the number of Baxter permutations of {1,2,...,n} with k rises. For example for n = 4, [T(n, k) for k = 0..n] = [0, 1, 10, 10, 1]. The permutations, with preceding number of rises, are:
%e .
%e 1 [4, 3, 2, 1], 3 [2, 3, 4, 1], 2 [3, 4, 2, 1], 3 [2, 3, 1, 4],
%e 2 [3, 2, 4, 1], 3 [2, 1, 3, 4], 2 [3, 2, 1, 4], 3 [1, 3, 4, 2],
%e 2 [2, 4, 3, 1], 3 [1, 3, 2, 4], 2 [4, 2, 3, 1], 3 [3, 4, 1, 2],
%e 2 [2, 1, 4, 3], 3 [3, 1, 2, 4], 2 [4, 2, 1, 3], 3 [1, 2, 4, 3],
%e 2 [1, 4, 3, 2], 3 [1, 4, 2, 3], 2 [4, 1, 3, 2], 3 [4, 1, 2, 3],
%e 2 [4, 3, 1, 2], 4 [1, 2, 3, 4].
%p p := (n, x) -> ifelse(n = 0, 1, x*hypergeom([-1-n, -n, 1-n], [2, 3], -x)):
%p seq(seq(coeff(simplify(p(n, x)), x, k), k = 0..n), n = 0..10);
%p # Alternative:
%p T := proc(n, k) local F; F := n -> n!*(n+1)!*(n+2)!;
%p ifelse(k = 0, k^n, 2*F(n-1)/(F(k-1)*F(n-k))) end:
%p for n from 0 to 9 do seq(T(n, k), k = 0..n) od;
%o (PARI) C=binomial;
%o T(n, k) = if(n==0 && k==0, 1, ( C(n+1,k-1) * C(n+1,k) * C(n+1,k+1) ) / ( C(n+1,1) * C(n+1,2) ) );
%o for(n=0,10,for(k=0,n,print1(T(n,k),", "));print());
%o \\ _Joerg Arndt_, Jan 04 2024
%o (SageMath)
%o def A359363(n):
%o if n == 0: return SR(1)
%o h = x*hypergeometric([-1 - n, -n, 1 - n], [2, 3], -x)
%o return h.series(x, n + 1).polynomial(SR)
%o for n in range(10): print(A359363(n).list())
%o def PolyA359363(n, t): return Integer(A359363(n)(x=t).n())
%o # _Peter Luschny_, Jan 04 2024
%o (Python)
%o from functools import cache
%o from math import factorial
%o @cache
%o def A359363Row(n: int) -> list[int]:
%o @cache
%o def F(n: int): return factorial(n) ** 3 * ((n+1) * (n+1) * (n+2))
%o if n == 0: return [1]
%o return [0] + [(2*F(n-1))//(F(k-1) * F(n-k)) for k in range(1, n+1)]
%o for n in range(0, 10): print(A359363Row(n))
%o # _Peter Luschny_, Jan 04 2024
%Y Special cases of the general formula: A097805 (m = 0), (0,1)-Pascal triangle; A090181 (m = 1), triangle of Narayana; this triangle (m = 2); A056940 (m = 3), with 1,0,0...; A056941 (m = 4), with 1,0,0...; A142465 (m = 5), with 1,0,0....
%Y Variant: A056939. Diagonals: A000292, A006542, A047819.
%Y Cf. A000178, A001181, A046996, A217800, A342889.
%K nonn,tabl
%O 0,9
%A _Peter Luschny_, Dec 28 2022
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