Bill Gosper <rwmgosper@yahoo.com> wrote: This is my first recollection of an algebraic series for (1/3)!. inf inf j ==== ==== ==== \ \ \ s n
eulerian(s, j) = > > (- n + j + 1) (- 1) binomial(s + 1, n) = s! / / / ==== ==== ==== j = 0 j = 0 n = 0
SUM(EULERIAN(S,J),J,0,INF) = S! EULERIAN(S,J) := SUM((-N+J+1)^S*(-1)^N*BINOMIAL(S+1,N),N,0,J) This is just the interpolation of n! = nth Eulerian row sum, so is unlikely to be new. It's also the b=0 and b=1 limits of the empirical (and unpolished) result, for 0<b<1, inf ==== \
eulerian(s - 1, j) cos(%pi b (s - 2 j - 2)) / ==== j = 0
sin(%pi b) s = (----------) Gamma(s) (hurwitz_zeta(s, 1 - b) cos(%pi s) + hurwitz_zeta(s, b)) %pi SUM(EULERIAN(S-1,J)*COS(%PI*B*(S-2*J-2)),J,0,INF) = (SIN(%PI*B)/%PI)^S*GAMMA(S)*(HURWITZ_ZETA(S,1-B)*COS(%PI*S)+HURWITZ_ZETA(S,B)) This lets us multisect the Eulerian row sums, e.g., n floor(-) 2 n n + 1 ==== B (1) 2 (2 - 1) \ n! n
eulerian(n, 2 j + 1) = -- - ---------------------,
/ 2 n + 1 ==== j = 0 n floor(-) 2 n n + 1 ==== B (1) 2 (2 - 1) \ n! n
eulerian(n, 2 j) = -- + ---------------------,
/ 2 n + 1 ==== j = 0 whose difference is the alternating row sum: n n n + 1 ==== 2 B (1) 2 (2 - 1) \ j n
(- 1) eulerian(n, j) = ----------------------- / n + 1 ==== j = 0
(More evidence that Bern should have been Bernpoly(1).) These multisection formulae should be of the form integer = not obviously integer. I suspect there's an analogous StirlingS2 Gamma(s) series based on the Adamchik formula per Oleg. (Extending S2 to noninteger first arg, e.g., x x x 4 - 4 3 + 6 2 + Stirling_s2(0, x) - 4 Stirling_s2(x, 4) = ---------------------------------------- 24 .) Using the infinite versions of the Eulerian and StirlingS2 series, both Gamma(noninteger) formulas ought to have interesting imaginary parts. --rwg --------------------------------- Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now.