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 A034870 Even-numbered rows of Pascal's triangle. 24
 1, 1, 2, 1, 1, 4, 6, 4, 1, 1, 6, 15, 20, 15, 6, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS The sequence of row lengths of this array is [1,3,5,7,9,11,13,...]= A005408(n), n>=0. Equals X^n * [1,0,0,0,...] where X = an infinite tridiagonal matrix with (1,1,1,...) in the main and subsubdiagonal and (2,2,2,...) in the main diagonal. X also = a triangular matrix with (1,2,1,0,0,0,...) in each column. - Gary W. Adamson, May 26 2008 a(n,m) has the following interesting combinatoric interpretation. Let s(n,m) equal the set of all base-4, n-digit numbers with n-m more 1-digits than 2-digits. For example s(2,1) = {10,01,13,31} (note that numbers like 1 are left-padded with 0s to ensure that they have 2 digits). Notice that #s(2,1) = a(2,1) with # indicating cardinality. This is true in general. a(n,m)=#s(n,m). In words, a(n,m) gives the number of n-digit, base-4 numbers with n-m more 1 digits than 2 digits. A proof is provided in the Links section. - Russell Jay Hendel, Jun 23 2015 LINKS Reinhard Zumkeller, Rows n=0..150 of triangle, flattened E. H. M. Brietzke, An identity of Andrews and a new method for the Riordan array proof of combinatorial identities, Discrete Math., 308 (2008), 4246-4262. Wolfdieter Lang, First 9 rows. Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018. FORMULA a(n, m) = binomial(2*n, m), 0<= m <= 2*n, 0<=n, else 0. G.f. for column m=2*k sequence: (x^k)*Pe(k, x)/(1-x)^(2*k+1), k>=0; for column m=2*k-1 sequence (x^k)*Po(k, x)/(1-x)^(2*k), k>=1, with the row polynomials Pe(k, x) := sum(A091042(k, m)*x^m, m=0..k) and Po(k, x) := 2*sum(A091044(k, m)*x^m, m=0..k-1); see also triangle A091043. From Paul D. Hanna, Apr 18 2012: (Start) Let A(x) be the g.f. of the flattened sequence, then: G.f.: A(x) = Sum_{n>=0} x^(n^2) * (1+x)^(2*n). G.f.: A(x) = Sum_{n>=0} x^n*(1+x)^(2*n) * Product_{k=1..n} (1 - (1+x)^2*x^(4*k-3)) / (1 - (1+x)^2*x^(4*k-1)). G.f.: A(x) = 1/(1 - x*(1+x)^2/(1 + x*(1-x^2)*(1+x)^2/(1 - x^5*(1+x)^2/(1 + x^3*(1-x^4)*(1+x)^2/(1 - x^9*(1+x)^2/(1 + x^5*(1-x^6)*(1+x)^2/(1 - x^13*(1+x)^2/(1 + x^7*(1-x^8)*(1+x)^2/(1 - ...))))))))), a continued fraction. (End) From Peter Bala, Jul 14 2015: (Start) Denote this array by P. Then P * transpose(P) is the square array ( binomial(2*n + 2*k, 2*k) )n,k>=0, which, read by antidiagonals, is A086645. Transpose(P) is the generalized Riordan array (1, (1 + x)^2). Let p(x) = (1 + x)^2. P^2 gives the coefficients in the expansion of the polynomials ( p(p(x)) )^n, P^3 gives the coefficients in the expansion of the polynomials ( p(p(p(x))) )^n and so on. Row sums are 2^(2*n); row sums of P^2 are 5^(2*n), row sums of P^3 are 26^(2*n). In general, the row sums of P^k, k = 0,1,2,..., are equal to A003095(k)^(2*n). The signed version of this array ( (-1)^k*binomial(2*n,k) )n,k>=0 is a left-inverse for A034839. A034839 * P = A080928. (End) T(n, k) = GegenbauerC(m, -n, -1)) where m = k if k simplify(GegenbauerC(`if`(k

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Last modified December 10 04:04 EST 2019. Contains 329885 sequences. (Running on oeis4.)