A142978 Table of figurate numbers for the n-dimensional cross polytopes.
1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 19, 16, 5, 1, 10, 33, 44, 25, 6, 1, 12, 51, 96, 85, 36, 7, 1, 14, 73, 180, 225, 146, 49, 8, 1, 16, 99, 304, 501, 456, 231, 64, 9, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10
Offset: 1
Examples
The square array A(n, k) begins: n\k| 1 2 3 4 5 6 ---+------------------------------- 1 | 1 2 3 4 5 6 A000027 2 | 1 4 9 16 25 36 A000290 3 | 1 6 19 44 85 146 A005900 4 | 1 8 33 96 225 456 A014820 5 | 1 10 51 180 501 1182 A069038 6 | 1 12 73 304 985 2668 A069039 7 | 1 14 99 476 1765 5418 A099193
Links
- Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened
- Peter Bala, A142978 and a family of continued fraction expansions for log(2)
- Valentin Bonzom and Etera R. Livine, Self-duality of the 6j-symbol and Fisher zeros for the Tetrahedron, arXiv:1905.00348 [math-ph], 2019.
- Steven Edwards and William Griffiths, Generalizations of Delannoy and cross polytope numbers, Fib. Q., Vol. 55, No. 4 (2017), pp. 356-366.
- Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
- Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
- Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
- Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
- Eric W. Weisstein, Meixner polynomial of the first kind
- Eric W. Weisstein, Mittag-Leffler polynomial
- Yutong Yang, From Simplest Recursion to the Recursion of Generalizations of Cross Polytope Numbers, (2017), Honors College Capstone and Theses, 13.
Crossrefs
Programs
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Haskell
a142978 n k = a142978_tabl !! (n-1) !! (k-1) a142978_row n = a142978_tabl !! (n-1) a142978_tabl = map reverse a104698_tabl -- Reinhard Zumkeller, Jul 17 2015
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Maple
with(combinat): T:=(n,k) -> add(binomial(n-1,i)*binomial(k+i,n),i = 0..n-1); for n from 1 to 10 do seq(T(n,k),k = 1..10) end do; # Program restored by Peter Bala, Oct 02 2008 A := (n, k) -> k*hypergeom([1 - n, 1 - k], [2], 2): seq(print(seq(simplify(A(n, k)), k = 1..9)), n=1..7); # Peter Luschny, Mar 23 2023
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Mathematica
t[n_, k_] := Sum[ Binomial[n-1, i]*Binomial[k+i, n], {i, 0, n-1}]; Table[t[n-k, k], {n, 1, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)
Formula
T(n,k) = Sum_{i = 0..n-1} C(n-1,i)*C(k+i,n).
Reciprocity law: n*T(n,k) = k*T(k,n).
Recurrence relation: T(n,1) = 1, T(1,k) = k, T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k), n,k > 1.
O.g.f. row n: x*(1 + x)^(n-1)/(1 - x)^(n+1).
O.g.f. for array: Sum_{n >= 1, k >= 1} T(n, k)*x^k*y^n = x*y/((1 - x)*(1 - x - y - x*y)).
The n-th row entries are the values [p_n(k)], k >= 1, of the polynomial function p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(x,k). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_n(x) is the unique polynomial solution of the difference equation x*( f(x+1) - f(x-1) ) = 2*n*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_n(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (-1)^(n+1) * (2*n)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n)). For example, n = 3 gives log(2) = 4/6 + (1/6)*(1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ...). See A142983 for further details.
From Peter Bala, Oct 02 2008: (Start)
The odd-indexed columns of this array form the array A142992 of crystal ball sequences for lattices of type C_n.
Conjectural congruences for main diagonal entries: Put A(n) = T(n,n). Calculation suggests the following congruences: for prime p > 3 and m, r >= 1, A(m*p^r) == A(m*p^(r-1)) (mod p^(3*r));
Sum_{k = 0..p-1} A(k)^2 == 0 (mod p) if p is a prime of the form 8*n+1 or 8*n+7;
Sum_{k = 0..p-1} A(k)^2 == -1 (mod p) if p is a prime of the form 8*n+3 or 8*n+5.
(End)
From Peter Bala, Sep 27 2021: (Start)
T(n,k) = (1/2)*Sum_{i = 0..k} binomial(k,i)*binomial(n+k-1-i,k-1).
T(n,k) = (1/2)*[x^n] ((1+x)/(1-x))^k = (1/2)*(k/n)*[x^k] ((1+x)/(1-x))^n.
n*T(n,k) = 2*k*T(n-1,k) + (n - 2)*T(n-2,k). (End)
A(n,k) = k*hypergeom([1 - n, 1 - k], [2], 2). - Peter Luschny, Mar 23 2023
T(n,k) = 2*(Sum_{j=1..k-1} T(n-1,j)) + T(n-1,k) for n > 1. - Robert FERREOL, Jun 25 2024
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