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A260492 Pascal's triangle aerated with columns of zeros.

%I A260492 #31 Jan 16 2023 09:03:44
%S A260492 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,
%T A260492 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,
%U A260492 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
%N A260492 Pascal's triangle aerated with columns of zeros.
%C A260492 To obtain this array we take Pascal's triangle A007318 and insert a column of zeros at columns 1, 3, 5, .... Thus the n-th row of this array gives the coefficients in the expansion of (1 + x^2)^n.
%C A260492 The transpose of this array is a generalized Riordan array (1, 1 + x^2) as defined by Wilson in 2005 (see the Bala link for details). Note that these are not the same as the generalized Riordan arrays introduced by Wang and Wang in 2008.
%C A260492 Call this array C and let p(x) = 1 + x^2. Then C^2 gives the coefficients in the expansion of the polynomials ( p(p(x)) )^n, C^3 gives the coefficients in the expansion of the polynomials ( p(p(p(x))) )^n and so on.
%C A260492 See A204293 for Pascal's triangle aerated by both row and columns.
%C A260492 The triangle is read by rows of lengths equal to the odd numbers 2*n + 1 = A005408(n), n >= 0; although it is then (from the 3rd line of comments & formula on) considered as an infinite square matrix C, the upper right being filled with zeros. - _M. F. Hasler_, Aug 19 2015
%H A260492 P. Bala, <a href="/A260492/a260492.pdf">Notes on generalized Riordan arrays</a>
%H A260492 Evgeniy Burlachenko, <a href="https://arxiv.org/abs/1709.02229">Riordan arrays and generalized Euler polynomials</a>, arXiv:1709.02229 [math.NT], 2017.
%H A260492 Evgeniy Burlachenko, <a href="https://arxiv.org/abs/1903.10435">Riordan arrays, Chebyshev polynomials, Fibonacci bases</a>, arXiv:1903.10435 [math.NT], 2019.
%H A260492 Alan D. Sokal, <a href="https://arxiv.org/abs/1804.08919">How to generalize (and not to generalize) the Chu-Vandermonde identity</a>, arXiv:1804.08919 [math.CO], 2018.
%H A260492 W. Wang and T. Wang, <a href="https://doi.org/10.1016/j.disc.2007.12.037">Generalized Riordan array</a>, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.
%F A260492 T(n,k) = (1 - k mod 2) binomial(n,floor(k/2)).
%F A260492 O.g.f.: 1/(1 - (1 + x^2)*t) = 1 + (1 + x^2)*t + (1 + 2*x^2 + x^4)*t^2 + ....
%F A260492 Let C denote this array.
%F A260492 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.
%F A260492 First column of C^2 = 2^n; first column of C^3 = 5^n, first column of C^4 = 26^n and so on.
%F A260492 Let P denote Pascal's triangle A007318. Then C * transpose(C) = P * transpose(P) = the square symmetric Pascal matrix.
%F A260492 For n >= 0, (P^n)*C is the array P^(n+1) aerated by columns.
%F A260492 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.
%F A260492 Let R equal Pascal's triangle aerated with rows of zeros. Then C*R = P^2.
%F A260492 R*P*C is P^3 aerated by both rows and columns
%F A260492 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.
%F A260492 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
%e A260492 Table begins
%e A260492 n\k 0  1  2  3  4  5  6  7  8  9 10 11 12
%e A260492 - - - - - - - - - - - - - - - - - - - - -
%e A260492 0   1
%e A260492 1   1  0  1
%e A260492 2   1  0  2  0  1
%e A260492 3   1  0  3  0  3  0  1
%e A260492 4   1  0  4  0  6  0  4  0  1
%e A260492 5   1  0  5  0 10  0 10  0  5  0  1
%e A260492 6   1  0  6  0 15  0 20  0 15  0  6  0  1
%e A260492 ...
%p A260492 #A260492
%p A260492 #define the aerated Pascal matrix (with indexing starting at 1)
%p A260492 C := Matrix(30, 30, (i,j) -> (1 - mod(j-1, 2))*binomial(i-1, floor((j-1)/2))):
%p A260492 for n from 1 to 12 do seq(C(n, k), k = 1 .. 2*n-1) end do;
%p A260492 # alternative program using row polynomials
%p A260492 rowpoly:= proc(n) option remember; expand((1 + x^2)^n) end:
%p A260492 T := (n, k)-> coeff(rowpoly(n), x, k):
%p A260492 seq(seq(T(n, k), k = 0..2*n), n = 0..11);
%t A260492 T[n_, k_] := If[OddQ[k], 0, Binomial[n, k/2]];
%t A260492 Table[T[n, k], {n, 0, 10}, {k, 0, 2n}] // Flatten (* _Jean-François Alcover_, Mar 23 2018 *)
%o A260492 (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
%Y A260492 Cf. A003095, A027826, A007318, A034870, A072191, A161869, A204293.
%K A260492 nonn,tabf,easy
%O A260492 0,7
%A A260492 _Peter Bala_, Aug 15 2015