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Odd-numbered columns of Losanitsch triangle A034851 formatted as triangle with an additional first column.
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%I #26 Sep 28 2024 21:56:29

%S 1,0,1,0,1,1,0,2,2,1,0,2,6,3,1,0,3,10,12,4,1,0,3,19,28,20,5,1,0,4,28,

%T 66,60,30,6,1,0,4,44,126,170,110,42,7,1,0,5,60,236,396,365,182,56,8,1,

%U 0,5,85,396,868,1001,693,280,72,9,1

%N Odd-numbered columns of Losanitsch triangle A034851 formatted as triangle with an additional first column.

%C Because the sequence of column m=2*k, k >= 1, of A034851 is the partial sum sequence of the one of column m=2*k-1 the present triangle is essentially Losanitsch's triangle A034851.

%C Row sums give A051450 with A051450(0) := 1. Column sequences (without leading zeros) are for m=0..6: A000007, A008619, A005993, A005995, A018211, A018213, A062136.

%H Michael De Vlieger, <a href="/A062135/b062135.txt">Table of n, a(n) for n = 0..11475</a> (rows n = 0..150, flattened)

%F T(n, m) = A034851(n-1+m, n-m), n >= m >= 0; A034851(n-1, n) := 0, n >= 1, A034851(-1, 0) := 1.

%F T(n, m) = 0 if n<m; T(0, 0)=1, T(n, 0)=0 if n >= 1; T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1) if n+m even and T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1)-binomial((n+m-3)/2, m-1) if n+m odd, n >= m >= 1.

%F G.f. for column m: x^m*Pe(m, x^2)/(((1-x)^(2*m))*(1+x)^m), m >= 0, with Pe(m, x^2)= sum(A034839(m, k)*x^(2*k), k=0..floor(n/2)), the row polynomial of array A034839 (even-indexed entries of the rows of Pascal's triangle).

%e Triangle begins:

%e {1};

%e {0,1};

%e {0,1,1};

%e {0,2,2,1};

%e ...

%e Pe(4,x^2)=1+6*x^2+x^4.

%t t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2; Flatten[Table[t[n - 1 + m, n - m], {n, 0, 12}, {m, 0, n}]] (* _Michael De Vlieger_, Sep 28 2024, after _Jean-François Alcover_ at A034851 *)

%Y Cf. A034851, A034839.

%K nonn,easy,tabl

%O 0,8

%A _Wolfdieter Lang_, Jun 19 2001

%E More terms from _Michael De Vlieger_, Sep 28 2024