On the coefficients of Psi(x,0) of the unique solution of the functional equ\ 2 ation, (Psi(x, 0) x y - y + x - y) Psi(x, y) + y - x Psi(x, 0) = 0 By Shalosh B. Ekhad Let's g(x)=Psi(x,0). The first, 50, (for the sake of the OEIS) are [1, 1, 2, 3, 7, 12, 30, 55, 143, 273, 728, 1428, 3876, 7752, 21318, 43263, 120175, 246675, 690690, 1430715, 4032015, 8414640, 23841480, 50067108, 142498692, 300830572, 859515920, 1822766520, 5225264024, 11124755664, 31983672534, 68328754959, 196947587823, 422030545335, 1219199353190, 2619631042665, 7583142491925, 16332922290300, 47365474641870, 102240109897695, 296983176369495, 642312451217745, 1868545312633440, 4048514844039120, 11793499763070480, 25594403741131680, 74650344244967400, 162250238001816900, 473770694965305348, 1031147983159782228] g=g(x) satisfies the following algebraic equation 3 2 g x - 2 g + 3 g - 1 = 0 and in Maple notation g^3*x-2*g^2+3*g-1 = 0 More generally, psi=Psi(x,y) satisisfies the algebraic equation 3 6 3 2 3 3 5 3 3 3 4 2 5 3 3 psi y - psi x y + psi y + psi x y - psi y - 3 psi y - psi x 3 2 3 2 3 3 2 2 2 2 3 + psi x y + psi x y - psi y + 3 psi x y - 3 psi x y 2 2 2 2 2 3 4 2 2 + 3 psi x y - 8 psi x y + 5 psi y + 3 psi y + 2 psi x 2 2 2 2 2 3 2 - 4 psi x y + 2 psi y - 3 psi x y + 6 psi x y - 3 psi y - 3 psi x 2 3 2 2 + 7 psi x y - 4 psi y - y + x - 3 x y + 2 y = 0 and in Maple notation psi^3*y^6-psi^3*x^2*y^3+psi^3*y^5+psi^3*x*y^3-psi^3*y^4-3*psi^2*y^5-psi^3*x^3+ psi^3*x^2*y+psi^3*x*y^2-psi^3*y^3+3*psi^2*x^2*y^2-3*psi^2*x*y^3+3*psi^2*x^2*y-8 *psi^2*x*y^2+5*psi^2*y^3+3*psi*y^4+2*psi^2*x^2-4*psi^2*x*y+2*psi^2*y^2-3*psi*x^ 2*y+6*psi*x*y^2-3*psi*y^3-3*psi*x^2+7*psi*x*y-4*psi*y^2-y^3+x^2-3*x*y+2*y^2 = 0 writing Psi(x,0)=g(x) as a Taylor series around x=0 infinity ----- \ n g(x) = ) a[n] x / ----- n = 0 The coefficients, a[n], satisfy the folllowing linear recurrence equation wi\ th polynomial coefficients of order, 2 (3 n + 2) (3 n + 1) (3 n + 8) a(n) 9 a(n + 1) -3/4 ---------------------------------- - ----------------- + a(n + 2) = 0 (3 n + 5) (n + 3) (n + 2) (n + 3) (3 n + 5) and in Maple notation -3/4*(3*n+2)*(3*n+1)*(3*n+8)/(3*n+5)/(n+3)/(n+2)*a(n)-9/(n+3)/(3*n+5)*a(n+1)+a( n+2) = 0 subject to the initial conditions a(1) = 1, a(2) = 1 Finally, just for fun here is , a(1000) a(1000) = 979126004443785412931142278503878513058808467397000810633549113420157\ 8283149523637748958186355180833072004330547196414348651307900516137271566955606\ 6750211464043783587913389858165786459115474543301980179084016196284587772296213\ 2199044811127284633412163848291265268108776927426798214055878941986697359477539\ 1532062721770795314545926837472474380661334765644319245879054765044780084512221\ 7718615069863058443201120 --------------------------------------- This ends this paper that took, 0.794, seconds to generate