An anonymous tipster providentially suggests: See the chapter 9 exercises of Henri Cohen's _Number Theory, Volume II: Analytic and Modern Tools_, Springer GTM 240. And then elaborates: `His exercise 4 has part (a) as follows: "By expressing e^{t(x+y)}/(e^t-1)^2 in terms of the derivative of e^{t(x+y)}/(e^t-1), prove that \sum_{0\le k\ne n}\binom nk B_k(x)B_{n-k}(y) = n(x+y-1)B_{n-1}(x+y) - (n-1)B_n(x+y)."' Assuming the \ne n is a typo, and converting to Mma, In[651]:= Sum[Binomial[n, k] BernoulliB[k, x] BernoulliB[n - k, y], {k, 0, n}] == n (x + y - 1) BernoulliB[n - 1, x + y] - (n - 1) BernoulliB[n, x + y] Out[651]= True Wait a minute. True?? This means the lhs and rhs are identical! Somebody is messing around again, and stuck this in as a special case! Proof: Divide both sides by n! : In[648]:= FunctionExpand[ Sum[BernoulliB[k, x] BernoulliB[n - k, y]/k!/(n - k)!, {k, 0, n}] -> (x + y - 1) BernoulliB[n - 1, x + y]/(n - 1)! - (n - 1)/ n! BernoulliB[n, x + y]] Now not even FunctionExpand recognizes it: Out[648]= \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(n\)] \*FractionBox[\(BernoulliB[k, x]\ BernoulliB[\(-k\) + n, y]\), \(\(k!\)\ \(\((\(-k\) + n)\)!\)\)]\) -> ((-1 + x + y) BernoulliB[-1 + n, x + y] - ((-1 + n) BernoulliB[n, x + y])/ n)/Gamma[n] Anyway, simmering down, I point out that this is an excellent example of my nag to replace Bernoulli #s with Bernoulli polynomials in a new variable. Two Bernoulli #s? Two new variables! (Well, sometimes.) Actually, since Cohen's summand lacks a 2^k-1, there probably lurks an even more general formula. --rwg Incidentally, another Cohen exercise wants proof that for odd n >1, x (x-1/2) (x-1), i.e. thrice the sum of the first x-1 squares, divides B_n(x): In[653]:= Sum[k^2, {k, x - 1}] Out[653]= 1/6 (-1 + x) x (-1 + 2 x) In[654]:= Factor[Table[BernoulliB[k, x]/%, {k, 3, 11, 2}]] Out[654]= {3, -1 - 3 x + 3 x^2, 1 + 3 x - 6 x^3 + 3 x^4, 3/5 (-3 - 9 x - x^2 + 15 x^3 + 5 x^4 - 15 x^5 + 5 x^6), (-1 - x + x^2) (-5 - 10 x + 3 x^2 + 11 x^3 + 2 x^4 - 9 x^5 + 3 x^6)} Incidentallier, Mathematica doesn't know that %653 is an integer: In[655]:= FullSimplify[Sin[\[Pi]*%%], x \[Element] Integers] Out[655]= Sin[1/6 \[Pi] (-1 + x) x (-1 + 2 x)] Macsyma has known this (c159) SIN(%PI*INTEGER*(INTEGER-1/2)*(INTEGER-1)/3); (d159) 0 (c160) BLOCK([FANCY_DISPLAY : FALSE],PLAYBACK([159,160])); "since time in memoriam" --Richard J. Daley. On Wed, Oct 28, 2015 at 12:00 PM, Bill Gosper <billgosper@gmail.com> wrote:
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