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A243661
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Triangle read by rows: the x = 1+q Narayana triangle at m=3.
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3
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1, 4, 3, 22, 33, 12, 140, 315, 231, 55, 969, 2907, 3213, 1547, 273, 7084, 26565, 39270, 28560, 10200, 1428, 53820, 242190, 448500, 437000, 235980, 66861, 7752, 420732, 2208843, 4916457, 6009003, 4351347, 1864863, 437437, 43263, 3362260, 20173560, 52451256, 77134200, 70122000, 40320150, 14307150, 2861430, 246675
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OFFSET
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1,2
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COMMENTS
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See Novelli-Thibon (2014) for precise definition.
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LINKS
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FORMULA
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T(n,k) = binomial(4*n+1-k,n-k) * binomial(3*n,k-1) / n.
More generally: T_m(n,k) = binomial((m+1)*n+1-k,n-k) * binomial(m*n,k-1) / n where m = 3.
Sum_{k=1..n} (-1)^k * T(n,k) = -1. (End)
Sum_{k = 1..n} (-1)^(k+1)*T(n,k)*binomial(x + 4*n - k + 1, 4*n - k + 1) = (x + 1) * ( Product_{k = 2..n} (x + k)^2 ) * ( Product_{k = 1..2*n+1} (x + n + k) ) / (n!*(3*n + 1)!) for n >= 1. Cf. A126216 and A243660. - Peter Bala, Oct 08 2022
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EXAMPLE
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Triangle begins:
1;
4, 3;
22, 33, 12;
140, 315, 231, 55;
969, 2907, 3213, 1547, 273;
7084, 26565, 39270, 28560, 10200, 1428;
...
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MATHEMATICA
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polrecip[P_, x_] := P /. x -> 1/x // Together // Numerator;
P[n_, m_] := Sum[Binomial[m n + 1, k] Binomial[(m + 1) n - k, n - k] (1 - x)^k x^(n - k), {k, 0, n}]/(m n + 1);
T[m_] := Reap[For[i=1, i <= 20, i++, z = polrecip[P[i, m], x] /. x -> 1+q; Sow[CoefficientList[z, q]]]][[2, 1]];
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PROG
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(PARI)
N(n, m)=sum(k=0, n, binomial(m*n+1, k)*binomial((m+1)*n-k, n-k)*(1-x)^k*x^(n-k))/(m*n+1);
T(m)=for(i=1, 20, z=subst(polrecip(N(i, m)), x, 1+q); print(Vecrev(z)));
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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