This can't be new--it's essentially the coefficient of x^n in coth(x) tanh(x): Sum[((-1 + 2^j)*BernoulliB[i - j]*BernoulliB[j])/((i - j)!*j!), {j, 2, -2 + i}]== -((-1 + 2^i)*BernoulliB[i])/i! but I don't see it in GKP's Concrete Math, nor DLMF. But that's probably because this is a poorly phrased special case of something. In particular, it has Bernoulli #s instead of Bernoulli polynomials. Anyone know better? Also, everybody knows the row sums of Pascal' triangle are 2^n, and the alternating row sums are 0^n, and the row sums of the Eulerian triangle are n!, but did you know the alternating row sums, 1, 1, 0, -2, 0, 16, 0, -272, 0, 7936,... A155585, are 2^n*EulerE[n, 1] = 2^(n + 1)*(2^(n + 1) - 1)*BernoulliB[n + 1, 1]/(n + 1) ? (http://dlmf.nist.gov/26.14#E11) --rwg