A176204 - OEIS

%I #7 Sep 08 2022 08:45:52

%S 1,1,1,1,13,1,1,41,41,1,1,101,261,101,1,1,225,1205,1205,225,1,1,477,

%T 4761,9661,4761,477,1,1,985,17169,62473,62473,17169,985,1,1,2005,

%U 58429,352933,624757,352933,58429,2005,1,1,4049,191357,1820765,5241413,5241413,1820765,191357,4049,1

%N Triangle T(n, k) = 4 * A008292(n+1, k) - 3, read by rows.

%C  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).

%C Row sums are: {1, 2, 15, 84, 465, 2862, 20139, 161256, 1451493, 14515170, 159667167, ...}.

%C Former title: A recursive symmetrical triangular sequence based on Eulerian numbers: q=2: T(n, m, q) = 2*T(n, m, q-1) - 1.

%H G. C. Greubel, <a href="/A176204/b176204.txt">Rows n = 0..100 of the triangle, flattened</a>

%F T(n, m, q) = 2*T(n, m, q-1) - 1, with T(n, m, 0) = A008292(n+1, m).

%F From _G. C. Greubel_, Mar 12 2020: (Start)

%F T(n, k, q) = 2^q * A008292(n+1, k) - (2^q - 1).

%F Sum_{k=0..n} T(n, k, q) = (n+1)*( 2^q * n! - 2^q + 1) (row sums). (End)

%e Triangle begins as:

%e   1;

%e   1,    1;

%e   1,   13,      1;

%e   1,   41,     41,       1;

%e   1,  101,    261,     101,       1;

%e   1,  225,   1205,    1205,     225,       1;

%e   1,  477,   4761,    9661,    4761,     477,       1;

%e   1,  985,  17169,   62473,   62473,   17169,     985,      1;

%e   1, 2005,  58429,  352933,  624757,  352933,   58429,   2005,    1;

%e   1, 4049, 191357, 1820765, 5241413, 5241413, 1820765, 191357, 4049, 1;

%p A008292:= (n,k) -> add((-1)^j*binomial(n+1,j)*(k-j+1)^n, j=0..k+1);

%p A176204:= (n,k,q) -> 2^q*( A008292(n+1,k) -1) + 1;

%p seq(seq( A176204(n,k,2), k=0..n), n=0..12); # _G. C. Greubel_, Mar 12 2020

%t Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1, j]*(k-j+1)^n, {j,0,k+1}];

%t T[n_, m_, q_]:= 2^q*Eulerian[n+1, m] - 2^q +1;

%t Table[T[n, m, 2], {n,0,12}, {m,0,n}]//Flatten (* modified by _G. C. Greubel_, Mar 12 2020 *)

%o (PARI) Eulerian(n,k) = sum(j=0,k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n);

%o T(n,k,q) = 2^q*Eulerian(n+1,k) - (2^q - 1); \\ _G. C. Greubel_, Mar 12 2020

%o (Magma) Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >;

%o [[4*Eulerian(n+1,k) -3: k in [0..n]]: n in [0..12]]; // _G. C. Greubel_, Mar 12 2020

%o (Sage)

%o def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1))

%o def T(n,k,q): return 2^q*Eulerian(n+1,k) - 2^q + 1

%o [[T(n,k,2) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Mar 12 2020

%Y Cf. A007318, A109128, A131061, A168625, A176200.

%K nonn,tabl

%O 0,5

%A _Roger L. Bagula_, Apr 11 2010

%E Edited by _G. C. Greubel_, Mar 12 2020