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 A249785 E.g.f. A(x) satisfies: (A(x)^3 - 6*x)^2 = (2 - A(x)^2)^3. 5
 1, 1, -1, 4, -19, 160, -1405, 17920, -231175, 3942400, -67260025, 1435033600, -30485899675, 780658278400, -19861319603125, 593300291584000, -17590824363487375, 600419895083008000, -20332162643728560625, 780545863607910400000, -29725015420426543091875, 1267606482499246489600000 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 LINKS Vaclav Kotesovec, Table of n, a(n) for n = 0..200 FORMULA E.g.f.: (1 + 2*Series_Reversion(G(x)))^(1/2), where G(x) = ((1+2*x)^(3/2) - (1-2*x)^(3/2))/6  = x + Sum_{n>=1} x^(2*n+1)/(2*n+1)! * Product_{k=0..n-1} (4*k-1)*(4*k+1). E.g.f. A = A(x) satisfies: (1) A(x)^2 + A(-x)^2 = 2. (2) A(x)^3 - A(-x)^3 = 6*x. (3) (A(x) - A(-x))/2 = Series_Reversion(x - 2*x^3/3). (4) x = (A(x)^3 - (2 - A(x)^2)^(3/2))/6. Recurrence: 4*a(n) = 8*(9*n^2 - 45*n + 55)*a(n-2) - (3*n-11)*(135*n^3 - 1449*n^2 + 5037*n - 5555)*a(n-4) + 9*(n-5)^2*(3*n-19)*(3*n-17)*(3*n-11)*(3*n-7)*a(n-6). - Vaclav Kotesovec, Nov 15 2014 a(n) ~ (-1)^(n+1) * 3^(n-1/2) * n^(n-1) / (2^((n-1)/2) * exp(n)). - Vaclav Kotesovec, Nov 15 2014 EXAMPLE E.g.f.: A(x) = 1 + x - x^2/2! + 4*x^3/3! - 19*x^4/4! + 160*x^5/5! - 1405*x^6/6! + 17920*x^7/7! - 231175*x^8/8! + 3942400*x^9/9! - 67260025*x^10/10! +... where A(x)^2 = 1 + 2*x + 2*x^3/3! + 50*x^5/5! + 4130*x^7/7! + 723170*x^9/9! + 219099650*x^11/11! + 102195343250*x^13/13! + 68022601897250*x^15/15! +... A(x)^3 = 1 + 3*x + 3*x^2/2! + 21*x^4/4! + 975*x^6/6! + 117705*x^8/8! + 27118875*x^10/10! + 10188133725*x^12/12! + 5671841550375*x^14/14! +... Thus the coefficients of even powers of x in A(x)^2 equal zero: [1, 2, 0, 2, 0, 50, 0, 4130, 0, 723170, 0, 219099650, 0, 102195343250, 0, ...], while the coefficients of odd powers of x in A(x)^3 equal zero: [1, 3, 3, 0, 21, 0, 975, 0, 117705, 0, 27118875, 0, 10188133725, 0, ...], after a few initial terms. EXPLICIT FORMULA. Let G(x) = ((1+2*x)^(3/2) - (1-2*x)^(3/2))/6, which begins G(x) = x - 1*x^3/3! - 1*15*x^5/5! - 1*15*63*x^7/7! - 1*15*63*143*x^9/9! - 1*15*63*143*255*x^11/11! +...+ [Product_{k=0..n-1} (16*k^2 - 1)]*x^(2*n+1)/(2*n+1)! +... then (A(x)^2 - 1)/2 = Series_Reversion(G(x)). The coefficients in G(x) form double factorials (A001147) that begin: [1, 0, -1, 0, -15, 0, -945, 0, -135135, 0, -34459425, 0, -13749310575, 0, ...]. PROG (PARI) /* Explicit formula: */ {a(n)=local(A, X=x+x^2*O(x^n), G=((1+2*X)^(3/2) - (1-2*X)^(3/2))/6); A=(1 + 2*serreverse(G))^(1/2); n!*polcoeff(A, n)} for(n=0, 25, print1(a(n), ", ")) (PARI) /* Formula using series expansion: */ {a(n)=local(A, G=x + sum(m=1, n\2+1, x^(2*m+1)/(2*m+1)!*prod(k=0, m-1, (4*k-1)*(4*k+1)) +x^2*O(x^n))); A=sqrt(1 + 2*serreverse(G)); n!*polcoeff(A, n)} for(n=0, 25, print1(a(n), ", ")) (PARI) /* Alternating zero coefficients in A(x)^2 and A(x)^3: */ {a(n)=local(A=[1, 1], E=1, M); for(i=1, n, A=concat(A, 0); M=#A; E=sum(m=0, M-1, A[m+1]*x^m/m!)+x*O(x^M); A[M]=if(M%2==0, -(M-1)!*Vec(E^3/3)[M], -(M-1)!*Vec(E^2/2)[M])); A[n+1]} for(n=0, 25, print1(a(n), ", ")) CROSSREFS Cf. A249786 (dual), A249787, A249788. Sequence in context: A067577 A225904 A203010 * A155804 A292167 A126147 Adjacent sequences:  A249782 A249783 A249784 * A249786 A249787 A249788 KEYWORD sign AUTHOR Paul D. Hanna, Nov 13 2014 STATUS approved

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Last modified January 25 19:09 EST 2020. Contains 331249 sequences. (Running on oeis4.)