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A125275
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Eigensequence of triangle A039599: a(n) = Sum_{k=0..n-1} A039599(n-1,k)*a(k) for n>0 with a(0)=1.
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3
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1, 1, 2, 7, 31, 162, 968, 6481, 47893, 386098, 3364562, 31460324, 313743665, 3320211313, 37124987124, 436985496790, 5397178181290, 69748452377058, 940762812167126, 13213888481979449, 192891251215160017
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Starting with offset 1 = row sums of triangle A147294. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 05 2008]
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LINKS
| Guo-Niu Han, Enumeration of Standard Puzzles
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FORMULA
| a(n) = Sum_{k=0..n-1} a(k) * C(2*n-1,n-k-1)*(2*k+1)/(2*n-1) for n>0 with a(0)=1.
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EXAMPLE
| a(3) = 2*(1) + 3*(1) + 1*(2) = 7;
a(4) = 5*(1) + 9*(1) + 5*(2) + 1*(7) = 31;
a(5) = 14*(1) + 28*(1) + 20*(2) + 7*(7) + 1*(31) = 162.
Triangle A039599(n,k) = C(2*n+1,n-k)*(2*k+1)/(2*n+1) begins:
1;
1, 1;
2, 3, 1;
5, 9, 5, 1;
14, 28, 20, 7, 1;
42, 90, 75, 35, 9, 1; ...
where g.f. of column k = G000108(x)^(2*k+1)
and G000108(x) = (1 - sqrt(1-4*x))/(2x) is the Catalan function.
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PROG
| (PARI) a(n)=if(n==0, 1, sum(k=0, n-1, a(k)*binomial(2*n-1, n-k-1)*(2*k+1)/(2*n-1)))
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CROSSREFS
| Cf. A039599, A000108; A125276 (variant).
A147294 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 05 2008]
Sequence in context: A030882 A030966 A009132 * A007446 A002872 A105216
Adjacent sequences: A125272 A125273 A125274 * A125276 A125277 A125278
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KEYWORD
| nonn
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Nov 26 2006
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