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A176204
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Triangle T(n, k) = 4 * A008292(n+1, k) - 3, read by rows.
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1
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1, 1, 1, 1, 13, 1, 1, 41, 41, 1, 1, 101, 261, 101, 1, 1, 225, 1205, 1205, 225, 1, 1, 477, 4761, 9661, 4761, 477, 1, 1, 985, 17169, 62473, 62473, 17169, 985, 1, 1, 2005, 58429, 352933, 624757, 352933, 58429, 2005, 1, 1, 4049, 191357, 1820765, 5241413, 5241413, 1820765, 191357, 4049, 1
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OFFSET
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0,5
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COMMENTS
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This sequence belongs to the class defined by T(n, m, q) = 2*T(n, m, q-1) - 1. The first few q values gives the sequences: A008292(n+1, k) (q=0), A176200 (q=1), this sequence (q=2).
Row sums are: {1, 2, 15, 84, 465, 2862, 20139, 161256, 1451493, 14515170, 159667167, ...}.
Former title: A recursive symmetrical triangular sequence based on Eulerian numbers: q=2: T(n, m, q) = 2*T(n, m, q-1) - 1.
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LINKS
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FORMULA
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T(n, m, q) = 2*T(n, m, q-1) - 1, with T(n, m, 0) = A008292(n+1, m).
T(n, k, q) = 2^q * A008292(n+1, k) - (2^q - 1).
Sum_{k=0..n} T(n, k, q) = (n+1)*( 2^q * n! - 2^q + 1) (row sums). (End)
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EXAMPLE
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Triangle begins as:
1;
1, 1;
1, 13, 1;
1, 41, 41, 1;
1, 101, 261, 101, 1;
1, 225, 1205, 1205, 225, 1;
1, 477, 4761, 9661, 4761, 477, 1;
1, 985, 17169, 62473, 62473, 17169, 985, 1;
1, 2005, 58429, 352933, 624757, 352933, 58429, 2005, 1;
1, 4049, 191357, 1820765, 5241413, 5241413, 1820765, 191357, 4049, 1;
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MAPLE
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A008292:= (n, k) -> add((-1)^j*binomial(n+1, j)*(k-j+1)^n, j=0..k+1);
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MATHEMATICA
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Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1, j]*(k-j+1)^n, {j, 0, k+1}];
T[n_, m_, q_]:= 2^q*Eulerian[n+1, m] - 2^q +1;
Table[T[n, m, 2], {n, 0, 12}, {m, 0, n}]//Flatten (* modified by G. C. Greubel, Mar 12 2020 *)
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PROG
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(PARI) Eulerian(n, k) = sum(j=0, k+1, (-1)^j*binomial(n+1, j)*(k-j+1)^n);
T(n, k, q) = 2^q*Eulerian(n+1, k) - (2^q - 1); \\ G. C. Greubel, Mar 12 2020
(Magma) Eulerian:= func< n, k | (&+[(-1)^j*Binomial(n+1, j)*(k-j+1)^n: j in [0..k+1]]) >;
[[4*Eulerian(n+1, k) -3: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Mar 12 2020
(Sage)
def Eulerian(n, k): return sum((-1)^j*binomial(n+1, j)*(k-j+1)^n for j in (0..k+1))
def T(n, k, q): return 2^q*Eulerian(n+1, k) - 2^q + 1
[[T(n, k, 2) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 12 2020
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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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