login
Triangle read by rows T(n, k) = Sum_{h>=0} Bernoulli(h)*binomial(k+h-1, h)*abs(Stirling1(n, h+k))*n^h for n>=0 and 0<=k<=n.
3

%I #19 Jul 21 2022 01:55:15

%S 1,0,1,0,0,1,0,-1,0,1,0,0,-5,0,1,0,24,0,-15,0,1,0,0,238,0,-35,0,1,0,

%T -3396,0,1281,0,-70,0,1,0,0,-51508,0,4977,0,-126,0,1,0,1706112,0,

%U -408700,0,15645,0,-210,0,1,0,0,35028576,0,-2267320,0,42273,0,-330,0,1

%N Triangle read by rows T(n, k) = Sum_{h>=0} Bernoulli(h)*binomial(k+h-1, h)*abs(Stirling1(n, h+k))*n^h for n>=0 and 0<=k<=n.

%H G. C. Greubel, <a href="/A339207/b339207.txt">Rows n = 0..50 of the triangle, flattened</a>

%H René Gy, <a href="http://math.colgate.edu/~integers/u67/u67.mail.html">Bernoulli-Stirling Numbers</a>, INTEGERS 20 (2020), #A67. See Table 1 p. 9.

%F T(n, k) = Sum_{h>=0} Bernoulli(h)*binomial(k+h-1, h)*abs(Stirling1(n, h+k))*n^h.

%e Triangle begins

%e 1;

%e 0 1;

%e 0 0 1;

%e 0 -1 0 1;

%e 0 0 -5 0 1;

%e 0 24 0 -15 0 1;

%e 0 0 238 0 -35 0 1;

%e ...

%t T[0, 0] = 1; T[n_, k_] := Sum[BernoulliB[j] * Binomial[k + j - 1, j] * Abs[StirlingS1[n, k + j]] * n^j, {j, 0, n - k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Amiram Eldar_, Nov 28 2020 *)

%o (PARI) T(n, k) = sum(h=0, n-k, bernfrac(h)*binomial(k+h-1,h)*abs(stirling(n, h+k, 1))*n^h);

%o (Magma)

%o function T(n, k)

%o if k eq n then return 1;

%o else return (&+[Binomial(k+j-1,j)*Bernoulli(j)*(-1)^j*StirlingFirst(n,k+j)*n^j: j in [0..n-k]]);

%o end if; return T; end function;

%o [T(n, k): k in [0..n], n in [0..15]]; // _G. C. Greubel_, Jul 21 2022

%o (SageMath)

%o def A339207(n,k):

%o if (k==n): return 1

%o else: return sum( binomial(k+j-1,j)*bernoulli(j)*stirling_number1(n,k+j)*n^j for j in (0..n-k) )

%o flatten([[A339207(n,k) for k in (0..n)] for n in (0..15)]) # _G. C. Greubel_, Jul 21 2022

%Y Cf. A339208, A339209.

%Y Cf. A027641/A027642 (Bernoulli), A007318 (binomial), A000254 (unsigned Stirling1).

%K sign,tabl

%O 0,13

%A _Michel Marcus_, Nov 27 2020