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A035607
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Table a(d,m) of number of points of L1 norm m in cubic lattice Z^d, read by antidiagonals (d >= 1, m >= 0).
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28
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1, 1, 2, 1, 4, 2, 1, 6, 8, 2, 1, 8, 18, 12, 2, 1, 10, 32, 38, 16, 2, 1, 12, 50, 88, 66, 20, 2, 1, 14, 72, 170, 192, 102, 24, 2, 1, 16, 98, 292, 450, 360, 146, 28, 2, 1, 18, 128, 462, 912, 1002, 608, 198, 32, 2, 1, 20, 162, 688, 1666, 2364, 1970, 952, 258, 36, 2, 1, 22, 200, 978, 2816
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OFFSET
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0,3
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COMMENTS
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Table also gives coordination sequences of same lattices.
a(d,m) also gives the number of ways to choose m squares from a 2 X (d-1) grid so that no two squares in the selection are (horizontally or vertically) adjacent. - Jacob A. Siehler, May 13 2006
A035607 is jointly generated with the Delannoy triangle A008288 as an array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 05 2012
Also, the polynomial v(n,x) above is x + (x + 1)*f(n-1,x), where f(0,x) = 1. - Clark Kimberling, Oct 24 2014
Rows also give the coefficients of the independence polynomial of the n-ladder graph. - Eric W. Weisstein, Dec 29 2017
Considering both sequences as square arrays (offset by one row), the rows of A035607 are the first differences of the rows of A008288, and the rows of A008288 are the partial sums of the rows of A035607. - Shel Kaphan, Feb 23 2023
Considering only points with nonnegative coordinates, the number of points at L1 distance = m in d dimensions is the same as the number of ways of putting m indistinguishable balls into d distinguishable urns, binomial(m+d-1, d-1). This is one facet of the cross-polytope. Allowing for + and - coordinates, there are binomial(d,i)*2^i facets containing points with up to i nonzero coordinates. Eliminating double counting of points with any coordinates = 0, there are Sum_{i=1..d} (-1)^(d-i)*binomial(m+i-1,i-1)*binomial(d,i)*2^i points at distance m in d dimensions. One may avoid the alternating sum by using binomial(m-1,i-1) to count only the points per facet with exactly i nonzero coordinates, avoiding any double counting, but the result is the same. - Shel Kaphan, Mar 04 2023
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LINKS
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J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
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FORMULA
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f(d,m) = Sum_{j=0..d-1} binomial(floor((d-1+j)/2), d-m-1)*binomial(d-m-1, floor((d-1-j)/2)), d >= 1 and 0 <= m <= d-1.
f(d,m) = f(d-1,m-1) + f(d-1,m) + f(d-2,m-1) (d >= 3 and 1 <= m <= d-1) with f(d,0) = 1 (d >= 1) and f(d,d-1) = 2 (d>=2). (End)
The generating function G(x,y) of this double sequence is the sum of a(n,p)*x^n*y^p, n=1..oo, p=0..oo, which is G(x,y) = x*(1+y)/(1-x-y-x*y).
The horizontal generating function H_n(y), which generates the rows of the table: (1, 2, 2, 2, 2, ...), (1, 4, 8, 12, 16, ...), (1, 6, 18, 38, 66, ...), is the sum of a(n,p)*y^p, p=0..oo, for each fixed n. This is H_n(y) = ((1+y)^n)/((1-y)^n)).
The vertical generating function V_p(x), which generates the columns of the table: (1, 1, 1, 1, 1, ...}, (2, 4, 6, 8, 10, ...), (2, 8, 18, 32, 50, ...), is the sum of a(n,p)*x^n, n=1..oo, for each fixed p. This is V_p(x) = 2*((1+x)^(p-1))/((1-x)^(p+1)) for p >= 1 and V_0(x) = x/(1-x). (End)
G.f.: (1+x)/(1-x-x*y-x^2*y). - Vladeta Jovovic, Apr 02 2002 (But see previous lines!)
Seen as a triangle read by rows: T(n,0) = 1, for n > 1: T(n,n-1) = 2, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-1), 0 < k < n. - Reinhard Zumkeller, Jul 20 2013
Seen as a triangle T(n,k) with 0 <= k < n read by rows: T(n,0)=1 for n > 0 and T(n,k) = Sum_{i=0..k-1} binomial(n-k,i+1)*binomial(k-1,i)*2^(i+1) for k > 0. - Werner Schulte, Feb 22 2018
With p >= 1 and q >= 0, as a square array a(p,q) = T(p+q-1,q) = 2*p*Hypergeometric2F1[1-p, 1-q, 2, 2] for q >= 1. Consequently, a(p,q) = a(q,p)*p/q. - Shel Kaphan, Feb 14 2023
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EXAMPLE
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As a triangle of coefficients in polynomials v(n,x) in Comments, the first 6 rows are
1
1 2
1 4 2
1 6 8 2
1 8 18 12 2
1 10 32 38 16 2
... (End)
For d=3, m=4:
There are binomial(3,1)*2^1 = 6 facets (vertices) of binomial(4+1-1,1-1) = 1 point with <= one nonzero coordinate.
There are binomial(3,2)*2^2 = 12 facets (edges) of binomial(4+2-1,2-1) = 5 points with <= two nonzero coordinates.
There are binomial(3,3)*2^3 = 8 facets (faces) of binomial(4+3-1,3-1) = 15 points with <= three nonzero coordinates.
a(3,4) = 8*15 - 12*5 + 6*1 = 120 - 60 + 6 = 66. (End)
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MAPLE
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A035607 := proc(d, m) local j: add(binomial(floor((d-1+j)/2), d-m-1)*binomial(d-m-1, floor((d-1-j)/2)), j=0..d-1) end: seq(seq(A035607(d, m), m=0..d-1), d=1..11); # d=dimension, m=norm # Johannes W. Meijer, Aug 05 2011
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MATHEMATICA
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u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Reverse /@ CoefficientList[CoefficientList[Series[(1 + x)/(1 - x - x y - x^2 y), {x, 0, 10}], x], y] // Flatten (* Eric W. Weisstein, Dec 29 2017 *)
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PROG
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(Haskell)
a035607 n k = a035607_tabl !! n !! k
a035607_row n = a035607_tabl !! n
a035607_tabl = map fst $ iterate
(\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 2])
(Sage)
@cached_function
def prec(n, k):
if k==n: return 1
if k==0: return 0
return prec(n-1, k-1)+2*sum(prec(n-i, k-1) for i in (2..n-k+1))
return [prec(n, n-k) for k in (0..n-1)]
(PARI) T(n, k) = if (k==0, 1, sum(i=0, k-1, binomial(n-k, i+1)*binomial(k-1, i)*2^(i+1)));
tabl(nn) = for (n=1, nn, for (k=0, n-1, print1(T(n, k), ", ")); print); \\ as a triangle; Michel Marcus, Feb 27 2018
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CROSSREFS
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Cf. A008288, which has g.f. 1/(1-x-x*y-x^2*y).
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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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