A260492 Pascal's triangle aerated with columns of zeros.
1, 1, 0, 1, 1, 0, 2, 0, 1, 1, 0, 3, 0, 3, 0, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 0, 6, 0, 15, 0, 20, 0, 15, 0, 6, 0, 1, 1, 0, 7, 0, 21, 0, 35, 0, 35, 0, 21, 0, 7, 0, 1, 1, 0, 8, 0, 28, 0, 56, 0, 70, 0, 56, 0, 28, 0, 8, 0, 1, 1, 0, 9, 0, 36, 0, 84, 0, 126, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 1, 0, 10, 0, 45, 0, 120, 0, 210, 0, 252, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1
Offset: 0
Examples
Table begins n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 - - - - - - - - - - - - - - - - - - - - - 0 1 1 1 0 1 2 1 0 2 0 1 3 1 0 3 0 3 0 1 4 1 0 4 0 6 0 4 0 1 5 1 0 5 0 10 0 10 0 5 0 1 6 1 0 6 0 15 0 20 0 15 0 6 0 1 ...
Links
- P. Bala, Notes on generalized Riordan arrays
- Evgeniy Burlachenko, Riordan arrays and generalized Euler polynomials, arXiv:1709.02229 [math.NT], 2017.
- Evgeniy Burlachenko, Riordan arrays, Chebyshev polynomials, Fibonacci bases, arXiv:1903.10435 [math.NT], 2019.
- Alan D. Sokal, How to generalize (and not to generalize) the Chu-Vandermonde identity, arXiv:1804.08919 [math.CO], 2018.
- W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.
Programs
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Maple
#A260492 #define the aerated Pascal matrix (with indexing starting at 1) C := Matrix(30, 30, (i,j) -> (1 - mod(j-1, 2))*binomial(i-1, floor((j-1)/2))): for n from 1 to 12 do seq(C(n, k), k = 1 .. 2*n-1) end do; # alternative program using row polynomials rowpoly:= proc(n) option remember; expand((1 + x^2)^n) end: T := (n, k)-> coeff(rowpoly(n), x, k): seq(seq(T(n, k), k = 0..2*n), n = 0..11);
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Mathematica
T[n_, k_] := If[OddQ[k], 0, Binomial[n, k/2]]; Table[T[n, k], {n, 0, 10}, {k, 0, 2n}] // Flatten (* Jean-François Alcover, Mar 23 2018 *) -
PARI
C=matrix(20,20,m,n,if(n%2,binomial(m-1,n\2))); a=concat(vector(5,i,vector(i*2-1,j,C[i,j]))) \\ (C is read by rows, not antidiagonals.) - M. F. Hasler, Aug 19 2015
Formula
T(n,k) = (1 - k mod 2) binomial(n,floor(k/2)).
O.g.f.: 1/(1 - (1 + x^2)*t) = 1 + (1 + x^2)*t + (1 + 2*x^2 + x^4)*t^2 + ....
Let C denote this array.
Row sums of C = 2^n; Row sums of C^2 = 5^n; Row sums of C^3 = 26^n; Row of sums C^4 = 677^n. In general the row sums of C^m = A003095(m)^n.
First column of C^2 = 2^n; first column of C^3 = 5^n, first column of C^4 = 26^n and so on.
Let P denote Pascal's triangle A007318. Then C * transpose(C) = P * transpose(P) = the square symmetric Pascal matrix.
For n >= 0, (P^n)*C is the array P^(n+1) aerated by columns.
First column of P*C = 2^n; first column of (P*C)^2 = 6^n; first column of (P*C)^3 = 38^n, and so on, where [2, 6, 38, ...] is A072191.
Let R equal Pascal's triangle aerated with rows of zeros. Then C*R = P^2.
R*P*C is P^3 aerated by both rows and columns
Conjecturally, the limit of R^n * C^n as n -> oo has as its first column an aerated version of A027826, with zeros elsewhere in the array.
Aeration by rows/columns amounts to multiplication to the left/right by the identity matrix aerated by rows/column: R = J*P, C = P*J'. This makes obvious that P^n*C = P^(n+1)*J', R*P^n = J*P^(n+1), R*P^n*C = J*P^(n+2)*J' (aerated by both rows and columns). - M. F. Hasler, Aug 19 2015
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