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A125273 Eigensequence of triangle A085478: a(n) = Sum_{k=0..n-1} A085478(n-1,k)*a(k) for n > 0 with a(0) = 1.

%I A125273 #31 May 31 2022 11:21:25
%S A125273 1,1,2,6,23,106,567,3434,23137,171174,1376525,11934581,110817423,
%T A125273 1095896195,11487974708,127137087319,1480232557526,18075052037054,
%U A125273 230855220112093,3076513227516437,42686898298650967,615457369662333260
%N A125273 Eigensequence of triangle A085478: a(n) = Sum_{k=0..n-1} A085478(n-1,k)*a(k) for n > 0 with a(0) = 1.
%H A125273 Seiichi Manyama, <a href="/A125273/b125273.txt">Table of n, a(n) for n = 0..517</a>
%H A125273 Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>. [Cached copy]
%H A125273 Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H A125273 Jeffrey B. Remmel, <a href="https://doi.org/10.37236/3210">Consecutive Up-down Patterns in Up-down Permutations</a>, Electron. J. Combin., 21 (2014), #P3.2.
%F A125273 a(n) = Sum_{k=0..n-1} binomial(n+k-1, n-k-1)*a(k) for n > 0 with a(0) = 1.
%F A125273 G.f. satisfies: A(x) = 1 + x*A(x/(1-x)^2) / (1-x). - _Paul D. Hanna_, Aug 15 2007
%e A125273 a(3) = 1*(1) + 3*(1) + 1*(2) = 6;
%e A125273 a(4) = 1*(1) + 6*(1) + 5*(2) + 1*(6) = 23;
%e A125273 a(5) = 1*(1) + 10*(1) + 15*(2) + 7*(6) + 1*(23) = 106.
%e A125273 Triangle A085478(n,k) = binomial(n+k, n-k) (with rows n >= 0 and columns k = 0..n) begins:
%e A125273   1;
%e A125273   1,  1;
%e A125273   1,  3,  1;
%e A125273   1,  6,  5,  1;
%e A125273   1, 10, 15,  7,  1;
%e A125273   1, 15, 35, 28,  9,  1;
%e A125273   ...
%e A125273 where g.f. of column k = 1/(1-x)^(2*k+1).
%t A125273 A125273=ConstantArray[0,20]; A125273[[1]]=1; Do[A125273[[n]]=1+Sum[A125273[[k]]*Binomial[n+k-1, n-k-1],{k,1,n-1}];,{n,2,20}]; Flatten[{1,A125273}] (* _Vaclav Kotesovec_, Dec 10 2013 *)
%o A125273 (PARI) a(n)=if(n==0,1,sum(k=0,n-1, a(k)*binomial(n+k-1, n-k-1)))
%Y A125273 Cf. A085478, A125274 (variant), A351813.
%K A125273 nonn
%O A125273 0,3
%A A125273 _Paul D. Hanna_, Nov 26 2006