

A034870


Evennumbered 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
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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 base4, ndigit numbers with nm more 1digits than 2digits. For example s(2,1) = {10,01,13,31} (note that numbers like 1 are leftpadded 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 ndigit, base4 numbers with nm 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), 42464262.
Russell Jay Hendel, Proof that a(n,m) gives the number of ndigit, base4 numbers with nm more 1digits than 2digits.
Wolfdieter Lang, First 9 rows.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2tangle shadows, arXiv:2002.06672 [math.CO], 2020.
Index entries for triangles and arrays related to Pascal's triangle


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)/(1x)^(2*k+1), k>=0; for column m=2*k1 sequence (x^k)*Po(k, x)/(1x)^(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..k1); 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*k3)) / (1  (1+x)^2*x^(4*k1)).
G.f.: A(x) = 1/(1  x*(1+x)^2/(1 + x*(1x^2)*(1+x)^2/(1  x^5*(1+x)^2/(1 + x^3*(1x^4)*(1+x)^2/(1  x^9*(1+x)^2/(1 + x^5*(1x^6)*(1+x)^2/(1  x^13*(1+x)^2/(1 + x^7*(1x^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 leftinverse for A034839.
A034839 * P = A080928. (End)
T(n, k) = GegenbauerC(m, n, 1)) where m = k if k<n else 2*nk.  Peter Luschny, May 08 2016
G.f.: 1/(1x*(y+1)^2).  Vladimir Kruchinin, Nov 22 2020


EXAMPLE

Triangle begins:
1;
1, 2, 1;
1, 4, 6, 4, 1;
1, 6, 15, 20, 15, 6, 1;
...


MAPLE

T := (n, k) > simplify(GegenbauerC(`if`(k<n, k, 2*nk), n, 1));
for n from 0 to 6 do seq(T(n, k), k=0..2*n) od; # Peter Luschny, May 08 2016


MATHEMATICA

Flatten[Table[Binomial[n, k], {n, 0, 20, 2}, {k, 0, n}]] (* Harvey P. Dale, Dec 15 2014 *)


PROG

(Haskell)
a034870 n k = a034870_tabf !! n !! k
a034870_row n = a034870_tabf !! n
a034870_tabf = map a007318_row [0, 2 ..]
 Reinhard Zumkeller, Apr 19 2012, Apr 02 2011
(MAGMA) /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 15 by 2]]; // Vincenzo Librandi, Jul 16 2015
(Maxima)
taylor(1/(1x*(y+1)^2), x, 0, 10, y, 0, 10); /* Vladimir Kruchinin, Nov 22 2020 */


CROSSREFS

Cf. A007318, A034871.
Cf. A000302 (row sums, powers of 4), alternating row sums are 0, except for n=0 which gives 1.
Cf. A003095, A034839, A080928, A086645.
Sequence in context: A190284 A327639 A273891 * A324224 A264622 A275017
Adjacent sequences: A034867 A034868 A034869 * A034871 A034872 A034873


KEYWORD

nonn,tabf,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



