VOOZH about

URL: https://oeis.org/A239894

⇱ A239894 - OEIS

login
A239894
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.
3
1, 1, 1, 4, 2, 1, 25, 9, 3, 1, 217, 58, 15, 4, 1, 2470, 500, 100, 22, 5, 1, 35647, 5574, 861, 152, 30, 6, 1, 637129, 78595, 9435, 1313, 215, 39, 7, 1, 13843948, 1376162, 130159, 14192, 1870, 290, 49, 8, 1, 360022957, 29417919, 2232792, 191850, 20001, 2547, 378, 60, 9, 1
OFFSET
1,4
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
Kreweras, G.; Dumont, D. , Sur les anagrammes alternés. (French) [On alternating anagrams] Discrete Math. 211 (2000), no. 1-3, 103--110. MR1735352 (2000h:05013).
FORMULA
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.
T(n,k) = Sum_{i=1..n-k+1} A218826(i)*T(n-i, k-1) for k > 1. - Andrew Howroyd, Feb 24 2020
EXAMPLE
Triangle begins:
1
1 1
4 2 1
25 9 3 1
217 58 15 4 1
...
PROG
(PARI) \\ here G(n) is A000366(n).
G(n)={(-1/2)^(n-2)*sum(k=0, n, binomial(n, k)*(1-2^(n+k+1))*bernfrac(n+k+1))}
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}
{my(T=A(10)); for(n=1, #T, print(T[n, 1..n]))} \\ Andrew Howroyd, Feb 24 2020
CROSSREFS
Row sums are A000366.
First column is A218826.
Sequence in context: A111536 A111559 A224798 * A152406 A105623 A364870
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Apr 04 2014
EXTENSIONS
Terms a(16) and beyond from Andrew Howroyd, Feb 19 2020
STATUS
approved