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1, 2, 6, 24, 115, 618, 3591, 22088, 141903, 943590, 6452490, 45159480, 322305165, 2339100078, 17223121350, 128428689888, 968383277791, 7374380672718, 56655414930642, 438741242896680, 3422125459579869, 26866961380274598, 212191772351034249, 1685036376746788392
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OFFSET
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0,2
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COMMENTS
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a(n) is the number of permutations of length n+1 avoiding the partially ordered pattern (POP) {1>2>3>4} of length 5. That is, the number of length n+1 permutations having no subsequences of length 5 in which the element in position 1 is larger than the element in position 2, which in turn is larger than the element in position 3, and that element is larger than the element in position 4. - Sergey Kitaev, Dec 13 2020
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LINKS
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FORMULA
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a(n) = (n+1)*A005802(n), where A005802(n) = number of permutations in S_n with longest increasing subsequence of length <= 3.
a(n) = Sum_{k=0..n} C(2k,k)*C(n,k)*C(n+1,k)/(k+1)^2.
Recurrence: (n+2)^2*a(n) = (n+1)*(7*n+2)*a(n-1) + 3*(n-2)*(7*n-4)*a(n-2) - 27*(n-2)*(n-1)*a(n-3) . - Vaclav Kotesovec, Oct 20 2012
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EXAMPLE
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a(2) = 1*(1) + 1*(3) + 2*(1) = 6;
a(3) = 1*(1) + 1*(6) + 2*(6) + 5*(1) = 24;
a(4) = 1*(1) + 1*(10)+ 2*(20)+ 5*(10)+ 14*(1) = 115.
The Narayana triangle A001263(n+1,k+1) = C(n,k)*C(n+1,k)/(k+1) begins:
1;
1, 1;
1, 3, 1;
1, 6, 6, 1;
1, 10, 20, 10, 1;
1, 15, 50, 50, 15, 1; ...
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MAPLE
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a := n -> hypergeom([1/2, -n - 1, -n], [2, 2], 4):
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MATHEMATICA
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Table[Sum[Binomial[2*k, k]*Binomial[n, k]*Binomial[n+1, k]/(k+1)^2, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 20 2012 *)
Table[HypergeometricPFQ[{1/2, -1 - n, -n}, {2, 2}, 4], {n, 0, 20}] (* Vaclav Kotesovec, May 14 2016 *)
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PROG
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(PARI) {a(n)=sum(k=0, n, binomial(2*k, k)*binomial(n, k)*binomial(n+1, k)/(k+1)^2)}
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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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