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A346415
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Triangle T(n,k), n>=0, 0<=k<=n, read by rows, where column k is (1/k!) times the k-fold exponential convolution of Fibonacci numbers with themselves.
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5
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 3, 11, 6, 1, 0, 5, 35, 35, 10, 1, 0, 8, 115, 180, 85, 15, 1, 0, 13, 371, 910, 630, 175, 21, 1, 0, 21, 1203, 4494, 4445, 1750, 322, 28, 1, 0, 34, 3891, 22049, 30282, 16275, 4158, 546, 36, 1, 0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1
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OFFSET
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0,8
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COMMENTS
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The sequence of column k>0 satisfies a linear recurrence with constant coefficients of order k+1.
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LINKS
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Alois P. Heinz, Rows n = 0..140, flattened
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EXAMPLE
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Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 3, 11, 6, 1;
0, 5, 35, 35, 10, 1;
0, 8, 115, 180, 85, 15, 1;
0, 13, 371, 910, 630, 175, 21, 1;
0, 21, 1203, 4494, 4445, 1750, 322, 28, 1;
0, 34, 3891, 22049, 30282, 16275, 4158, 546, 36, 1;
0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1;
...
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MAPLE
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b:= proc(n) option remember; `if`(n=0, 1, add(expand(x*b(n-j)
*binomial(n-1, j-1)*(<<0|1>, <1|1>>^j)[1, 2]), j=1..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..12);
# second Maple program:
b:= proc(n, k) option remember; `if`(k=0, 0^n, `if`(k=1,
combinat[fibonacci](n), (q-> add(binomial(n, j)*
b(j, q)*b(n-j, k-q), j=0..n))(iquo(k, 2))))
end:
T:= (n, k)-> b(n, k)/k!:
seq(seq(T(n, k), k=0..n), n=0..12);
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MATHEMATICA
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b[n_, k_] := b[n, k] = If[k == 0, 0^n, If[k == 1, Fibonacci[n], With[{q = Quotient[k, 2]}, Sum[Binomial[n, j] b[j, q] b[n-j, k-q], {j, 0, n}]]]];
T[n_, k_] := b[n, k]/k!;
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Nov 06 2021, after 2nd Maple program *)
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CROSSREFS
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Columns k=0-4 give: A000007, A000045, A014335, A014337, A014341.
T(n+j,n) for j=0-2 give: A000012, A000217, A000914.
Row sums give A256180.
Sequence in context: A130717 A137396 A244213 * A352067 A178245 A338753
Adjacent sequences: A346412 A346413 A346414 * A346416 A346417 A346418
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KEYWORD
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nonn,tabl
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AUTHOR
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Alois P. Heinz, Jul 15 2021
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STATUS
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approved
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