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A218826
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Number of indecomposable (by concatenation) alternating n-anagrams.
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3
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1, 1, 4, 25, 217, 2470, 35647, 637129, 13843948, 360022957, 11054457253, 396003680518, 16377463914091, 774714094061221, 41572230979229284, 2512149910125036865, 169831839578092130017, 12769241823369505582150, 1062122471116082751430087, 97264621940872013476357969
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OFFSET
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1,3
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COMMENTS
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Alternating anagrams enumeration is related to A000366 by a(n) = A000366(n+1).
For all n, a(n) are periodically congruent to 1, 1 and 4 modulo 6.
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LINKS
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G. Kreweras and J. Barraud, Anagrammes alternés, European Journal of Combinatorics,Volume 18, Issue 8, November 1997, Pages 887-891.
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FORMULA
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G.f.: (1-Q(0))/x, where Q(k)= 1 - 1/2*(k+1)*(k+2)*x/(1 - 1/2*(k+1)*(k+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
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PROG
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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), ", "); ); }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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