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A141697 T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n - p*binomial(n-1, k))/2 where p=12 and q=14.

%I A141697 #43 Aug 12 2022 20:17:05
%S A141697 1,1,1,1,16,1,1,59,59,1,1,158,426,158,1,1,369,2054,2054,369,1,1,804,
%T A141697 8247,16792,8247,804,1,1,1687,29925,109123,109123,29925,1687,1,1,3466,
%U A141697 102088,617302,1092910,617302,102088,3466,1,1,7037,334664,3185840,9171722,9171722,3185840,334664,7037,1
%N A141697 T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n - p*binomial(n-1, k))/2 where p=12 and q=14.
%C A141697 Row n is made of coefficients from 7*(1 - x)^(n+1) * polylog(-n,x)/x - 6*(1 + x)^(n-1). - _Thomas Baruchel_, Jun 03 2018
%H A141697 G. C. Greubel, <a href="/A141697/b141697.txt">Rows n = 1..100 of triangle, flattened</a>
%H A141697 Thomas Baruchel, <a href="https://math.stackexchange.com/q/2806590/">A conjectured formula for the polylogarithm of a negative integer order</a>, Mathematics Stack Exchange question, Jun 04 2018.
%F A141697 p=12; q=14; T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n  - p*binomial(n-1, k))/2.
%F A141697 a(n) = 3*A168524(n) - 2*A154337(n). - _Thomas Baruchel_, Jun 08 2018
%e A141697 Triangle begins:
%e A141697   1;
%e A141697   1,    1;
%e A141697   1,   16,      1;
%e A141697   1,   59,     59,       1;
%e A141697   1,  158,    426,     158,       1;
%e A141697   1,  369,   2054,    2054,     369,       1;
%e A141697   1,  804,   8247,   16792,    8247,     804,       1;
%e A141697   1, 1687,  29925,  109123,  109123,   29925,    1687,      1;
%p A141697 T:= proc(n, k): 7*add((-1)^j*binomial(n+1, j)*(k-j+1)^n, j = 0..k+1) - 6*binomial(n-1, k); end proc; seq(seq(T(n,k), k=0..n-1), n=1..10); # _G. C. Greubel_, Nov 13 2019
%t A141697 i=12; l=14; Table[Table[(l*Sum[(-1)^j*Binomial[n+1, j](k+1-j)^n, {j, 0, k+1}] - i*Binomial[n-1, k])/2, {k,0,n-1}], {n,10}]//Flatten
%o A141697 (PARI) T(n,k) = 7*sum(j=0, k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n) - 6* binomial(n-1,k);
%o A141697 for(n=1,10, for(k=0,n-1, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Jun 03 2018
%o A141697 (PARI) row(n) = Vec(7*(1 - x)^(n+1)*polylog(-n,x)/x - 6*(1 + x)^(n-1)); \\ _Michel Marcus_, Jun 08 2018
%o A141697 (Magma) [ 7*(&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) - 6*Binomial(n-1,k): k in [0..n-1], n in [1..10]]; // _G. C. Greubel_, Nov 13 2019
%o A141697 (Sage) [[ 7*sum( (-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1)) - 6*binomial(n-1,k) for k in (0..n-1)] for n in (1..10)] # _G. C. Greubel_, Nov 13 2019
%Y A141697 Cf. Eulerian numbers (A008292) and Pascal's triangle (A007318).
%Y A141697 Cf. A141696.
%K A141697 nonn,tabl,easy
%O A141697 1,5
%A A141697 _Roger L. Bagula_, Sep 11 2008
%E A141697 Edited by _G. C. Greubel_, Nov 13 2019