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A306091 G.f. A(x) satisfies: (1 + A(x))^A(x) = (1 + x)^x ; this sequence gives the denominators of the coefficients of x^n in g.f. A(x). 3

%I #22 Nov 02 2019 03:14:40

%S 1,2,4,6,8,1440,960,120960,48384,7257600,1612800,479001600,4561920,

%T 5230697472000,10461394944000,7846046208000,6974263296000,

%U 9146248151040000,8536498274304000,1502674769756160000,1857852442607616000,67440043666656460800000,44960029111104307200000,18613452051997183180800000,954536002666522214400000

%N G.f. A(x) satisfies: (1 + A(x))^A(x) = (1 + x)^x ; this sequence gives the denominators of the coefficients of x^n in g.f. A(x).

%C The numerators of the coefficients in g.f. A(x) are given by A306090.

%H Paul D. Hanna, <a href="/A306091/b306091.txt">Table of n, a(n) for n = 1..300</a>

%F G.f. A(x) = Sum_{n>=0} A306090(n)/A306091(n) * x^n satisfies:

%F (1) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*A(x) = 1.

%F (2) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + (k - p)*A(x) = (1 + x)^p.

%F (3) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k - m)*x + k*A(x) = (1 + A(x))^m.

%F (4) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k - m)*x + (k - p)*A(x) = (1+x)^p * (1 + A(x))^m.

%F (5) A(A(x)) = x.

%F (6) (1 + A(x))^A(x) = (1 + x)^x.

%F (7) Sum_{n>=1} (-A(x))^(n+1) / n = x*log(1+x).

%F (8) Let F(x,y) = Series_Reversion( (exp(-x*y) - exp(-x))/(1-y) ), where the inverse is taken wrt x, and let F'(x,y) = d/dx F(x,y), then F'(x, A(x)/x) = 1 (derived from Peter Bala's g.f. for A067948).

%e G.f.: A(x) = -x + 1/2*x^2 - 1/4*x^3 + 1/6*x^4 - 1/8*x^5 + 143/1440*x^6 - 79/960*x^7 + 8483/120960*x^8 - 2953/48384*x^9 + 391753/7257600*x^10 - 77983/1612800*x^11 + 20963473/479001600*x^12 - 182269/4561920*x^13 + 192178874539/5230697472000*x^14 - 355629691849/10461394944000*x^15 + 248105704337/7846046208000*x^16 - 206101262483/6974263296000*x^17 + 253628381647657/9146248151040000*x^18 - 222936799599583/8536498274304000*x^19 + 37078279922025269/1502674769756160000*x^20 + ... + A306090(n)/A306091(n)*x^n + ...

%e such that

%e (E.1) 1 = 1 + (x + A(x)) + (x + 2*A(x))*(2*x + A(x))/2! + (x + 3*A(x))*(2*x + 2*A(x))*(3*x + A(x))/3! + (x + 4*A(x))*(2*x + 3*A(x))*(3*x + 2*A(x))*(4*x + A(x))/4! + (x + 5*A(x))*(2*x + 4*A(x))*(3*x + 3*A(x))*(4*x + 2*A(x))*(5*x + A(x))/5! + ...

%e (E.2) (1 + x)^p = 1 + (x + (1-p)*A(x)) + (x + (2-p)*A(x))*(2*x + (1-p)*A(x))/2! + (x + (3-p)*A(x))*(2*x + (2-p)*A(x))*(3*x + (1-p)*A(x))/3! + (x + (4-p)*A(x))*(2*x + (3-p)*A(x))*(3*x + (2-p)*A(x))*(4*x + (1-p)*A(x))/4! + ...

%e (E.3) (1 + A(x))^m = 1 + ((1-m)*x + A(x)) + ((1-m)*x + 2*A(x))*((2-m)*x + A(x))/2! + ((1-m)*x + 3*A(x))*((2-m)*x + 2*A(x))*((3-m)*x + A(x))/3! + ((1-m)*x + 4*A(x))*((2-m)*x + 3*A(x))*((3-m)*x + 2*A(x))*((4-m)*x + A(x))/4! + ...

%e FUNCTIONAL EQUATIONS.

%e The series A(x) satisfies:

%e (E.4) (1 + A(x))^A(x) = (1 + x)^x = 1 + x^2 - 1/2*x^3 + 5/6*x^4 - 3/4*x^5 + 33/40*x^6 - 5/6*x^7 + 2159/2520*x^8 - 209/240*x^9 + ...

%e GENERATING METHOD.

%e Although the functional equation (1 + A(x))^A(x) = (1 + x)^x has an infinite number of solutions, one may arrive at the g.f. A(x) by the following iteration.

%e If we start with A = -x, and iterate

%e (E.5) A = (A + x*log(1 + x)/log(1 + A))/2

%e then A will converge to g.f. A(x).

%t nmax = 25; sol = {a[1] -> -1};

%t Do[A[x_] = Sum[a[k] x^k, {k, 1, n}] /. sol; eq = CoefficientList[(1 + A[x])^A[x] - (1 + x)^x + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax + 1}];

%t sol /. Rule -> Set;

%t a /@ Range[1, nmax] // Denominator (* _Jean-Fran├žois Alcover_, Nov 02 2019 *)

%o (PARI) /* From Functional Equation (1 + A(x))^A(x) = (1 + x)^x */

%o {a(n) = my(A = -x +x*O(x^n)); for(i=1,n, A = (A + x*log(1+x +x*O(x^n))/log(1+A))/2 ); denominator( polcoeff(A,n) )}

%Y Cf. A306090 (numerators).

%K nonn,frac

%O 1,2

%A _Paul D. Hanna_, Jun 21 2018

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