A218826 Number of indecomposable (by concatenation) alternating n-anagrams.
1, 1, 4, 25, 217, 2470, 35647, 637129, 13843948, 360022957, 11054457253, 396003680518, 16377463914091, 774714094061221, 41572230979229284, 2512149910125036865, 169831839578092130017, 12769241823369505582150, 1062122471116082751430087, 97264621940872013476357969
Offset: 1
Keywords
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..200
- G. Kreweras and J. Barraud, Anagrammes alternés, European Journal of Combinatorics,Volume 18, Issue 8, November 1997, Pages 887-891.
- G. Kreweras and D. Dumont, Sur les anagrammes alternés, Discrete Mathematics, Volume 211, Issues 1-3, 28 January 2000, Pages 103-110.
Programs
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PARI
a000366(n)= {return((-1/2)^(n-2)*sum(k=0, n, binomial(n, k)*(1-2^(n+k+1))*bernfrac(n+k+1)));} bi(n,k) = {if (matb[n,k] == 0, if (n==k, v=1, if (k==1, v = b(n),v = sum(i=1, n-k+1, b(i)*bi(n-i,k-1)););); matb[n,k] = v;); return (matb[n,k]);} b(n) = {if (n==1, return(a000366(n+1)), return(a000366(n+1) - sum(i=2, n, bi(n,i))));} allb(m) = {matb = matrix(m,m); for (i=1, m, print1(b(i), ", "););}
Formula
G.f.: (1-Q(0))/x, where Q(k) = 1 - ((k+1)*(k+2)/2)*x/(1 - ((k+1)*(k+2)/2)*x/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: 1/Q(0), where Q(k) = 1 - ((k+1)*(k+2)/2)*x/(1 - ((k+2)*(k+3)/2)*x/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2013
Comments