A034870 - OEIS

%I #97 Aug 07 2024 09:18:44

%S 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,

%T 120,210,252,210,120,45,10,1,1,12,66,220,495,792,924,792,495,220,66,

%U 12,1,1,14,91,364,1001,2002,3003,3432,3003,2002,1001,364,91,14,1

%N Even-numbered rows of Pascal's triangle.

%C The sequence of row lengths of this array is [1,3,5,7,9,11,13,...]= A005408(n), n>=0.

%C 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

%C 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

%H Reinhard Zumkeller, <a href="/A034870/b034870.txt">Rows n=0..150 of triangle, flattened</a>

%H Peter Bala, <a href="/A260492/a260492.pdf">Notes on generalized Riordan arrays</a>

%H E. H. M. Brietzke, <a href="http://dx.doi.org/10.1016/j.disc.2007.08.050">An identity of Andrews and a new method for the Riordan array proof of combinatorial identities</a>, Discrete Math., 308 (2008), 4246-4262.

%H Russell Jay Hendel, <a href="/A034870/a034870_1.txt">Proof that a(n,m) gives the number of n-digit, base-4 numbers with n-m more 1-digits than 2-digits.</a>

%H Wolfdieter Lang, <a href="/A034870/a034870.txt">First 9 rows</a>.

%H Franck Ramaharo, <a href="https://arxiv.org/abs/1802.07701">Statistics on some classes of knot shadows</a>, arXiv:1802.07701 [math.CO], 2018.

%H Franck Ramaharo, <a href="https://arxiv.org/abs/2002.06672">A bracket polynomial for 2-tangle shadows</a>, arXiv:2002.06672 [math.CO], 2020.

%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>

%F T(n, m) = binomial(2*n, m), 0<= m <= 2*n, 0<=n, else 0.

%F 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.

%F From _Paul D. Hanna_, Apr 18 2012: (Start)

%F Let A(x) be the g.f. of the flattened sequence, then:

%F G.f.: A(x) = Sum_{n>=0} x^(n^2) * (1+x)^(2*n).

%F 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)).

%F 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.

%F (End)

%F From _Peter Bala_, Jul 14 2015: (Start)

%F 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.

%F Transpose(P) is a generalized Riordan array (1, (1 + x)^2) as defined in the Bala link.

%F 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.

%F 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).

%F The signed version of this array ( (-1)^k*binomial(2*n,k) )n,k>=0 is a left-inverse for A034839.

%F A034839 * P = A080928. (End)

%F T(n, k) = GegenbauerC(m, -n, -1) where m = k if k<n else 2*n-k. - _Peter Luschny_, May 08 2016

%F G.f.: 1/(1-x*(y+1)^2). - _Vladimir Kruchinin_, Nov 22 2020

%e Triangle begins:

%e   1;

%e   1,  2,  1;

%e   1,  4,  6,   4,   1;

%e   1,  6, 15,  20,  15,   6,   1;

%e   1,  8, 28,  56,  70,  56,  28,   8,   1;

%e   1, 10, 45, 120, 210, 252, 210, 120,  45,  10,  1;

%e   1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1;

%p T := (n,k) -> simplify(GegenbauerC(`if`(k<n,k,2*n-k),-n, -1));

%p for n from 0 to 6 do seq(T(n,k),k=0..2*n) od; # _Peter Luschny_, May 08 2016

%t Flatten[Table[Binomial[n,k],{n,0,20,2},{k,0,n}]] (* _Harvey P. Dale_, Dec 15 2014 *)

%o (Haskell)

%o a034870 n k = a034870_tabf !! n !! k

%o a034870_row n = a034870_tabf !! n

%o a034870_tabf = map a007318_row [0, 2 ..]

%o -- _Reinhard Zumkeller_, Apr 19 2012, Apr 02 2011

%o (Magma) /* As triangle: */ [[Binomial(n,k): k in [0..n]]: n in [0.. 15 by 2]]; // _Vincenzo Librandi_, Jul 16 2015

%o (Maxima)

%o taylor(1/(1-x*(y+1)^2),x,0,10,y,0,10); /* _Vladimir Kruchinin_, Nov 22 2020 */

%o (Sage) flatten([[binomial(2*n, k) for k in (0..2*n)] for n in (0..12)]) # _G. C. Greubel_, Mar 18 2022

%Y Cf. A007318, A034871.

%Y Cf. A000302 (row sums, powers of 4), alternating row sums are 0, except for n=0 which gives 1.

%Y Cf. A003095, A034839, A080928, A086645.

%K nonn,tabf,easy

%O 0,3

%A _N. J. A. Sloane_