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A034870
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Even-numbered rows of Pascal's triangle.
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24
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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
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OFFSET
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0,3
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COMMENTS
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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 0's 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
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LINKS
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FORMULA
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T(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.
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)
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 a generalized Riordan array (1, (1 + x)^2) as defined in the Bala link.
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.
T(n, k) = GegenbauerC(m, -n, -1)) where m = k if k<n else 2*n-k. - Peter Luschny, May 08 2016
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EXAMPLE
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Triangle begins:
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;
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MAPLE
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T := (n, k) -> simplify(GegenbauerC(`if`(k<n, k, 2*n-k), -n, -1));
for n from 0 to 6 do seq(T(n, k), k=0..2*n) od; # Peter Luschny, May 08 2016
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MATHEMATICA
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Flatten[Table[Binomial[n, k], {n, 0, 20, 2}, {k, 0, n}]] (* Harvey P. Dale, Dec 15 2014 *)
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PROG
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(Haskell)
a034870 n k = a034870_tabf !! n !! k
a034870_row n = a034870_tabf !! n
a034870_tabf = map a007318_row [0, 2 ..]
(Magma) /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 15 by 2]]; // Vincenzo Librandi, Jul 16 2015
(Maxima)
(Sage) flatten([[binomial(2*n, k) for k in (0..2*n)] for n in (0..12)]) # G. C. Greubel, Mar 18 2022
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CROSSREFS
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Cf. A000302 (row sums, powers of 4), alternating row sums are 0, except for n=0 which gives 1.
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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