%I #24 Feb 25 2020 12:13:43

%S 1,1,1,4,2,1,25,9,3,1,217,58,15,4,1,2470,500,100,22,5,1,35647,5574,

%T 861,152,30,6,1,637129,78595,9435,1313,215,39,7,1,13843948,1376162,

%U 130159,14192,1870,290,49,8,1,360022957,29417919,2232792,191850,20001,2547,378,60,9,1

%N Triangle read by rows: T(n,k) (n>=1, 1 <= k <= n) = number of alternating anagrams on n letters (of length 2n) which are decomposable into at most k slices.

%H Andrew Howroyd, <a href="/A239894/b239894.txt">Table of n, a(n) for n = 1..1275</a> (first 50 rows)

%H Kreweras, G.; Dumont, D. , <a href="http://dx.doi.org/10.1016/S0012-365X(99)00238-1">Sur les anagrammes alternés</a>. (French) [On alternating anagrams] Discrete Math. 211 (2000), no. 1-3, 103--110. MR1735352 (2000h:05013).

%F T(n,k) = b(1)*T(n-1,k-1)+b(2)*T(n-2,k-1)+...+b(n-k+1)*T(k-1,k-1), where b(i) = A218826(i) for k > 1.

%F T(n,k) = Sum_{i=1..n-k+1} A218826(i)*T(n-i, k-1) for k > 1. - _Andrew Howroyd_, Feb 24 2020

%e Triangle begins:

%e 1

%e 1 1

%e 4 2 1

%e 25 9 3 1

%e 217 58 15 4 1

%e ...

%o (PARI) \\ here G(n) is A000366(n).

%o G(n)={(-1/2)^(n-2)*sum(k=0, n, binomial(n, k)*(1-2^(n+k+1))*bernfrac(n+k+1))}

%o A(n)={my(M=matrix(n,n)); for(n=1, n, for(k=2, n, M[n,k] = sum(i=1, n-k+1, M[i,1]*M[n-i, k-1])); M[n,1]=G(n+1)-sum(i=2, n, M[n,i])); M}

%o {my(T=A(10)); for(n=1, #T, print(T[n, 1..n]))} \\ _Andrew Howroyd_, Feb 24 2020

%Y Row sums are A000366.

%Y First column is A218826.

%K nonn,tabl

%O 1,4

%A _N. J. A. Sloane_, Apr 04 2014

%E Terms a(16) and beyond from _Andrew Howroyd_, Feb 19 2020