A306090 G.f. A(x) satisfies: (1 + A(x))^A(x) = (1 + x)^x ; this sequence gives the numerators of the coefficients of x^n in g.f. A(x).
-1, 1, -1, 1, -1, 143, -79, 8483, -2953, 391753, -77983, 20963473, -182269, 192178874539, -355629691849, 248105704337, -206101262483, 253628381647657, -222936799599583, 37078279922025269, -43439069697425189, 1498102421014867632661, -951127545430874789837, 375811649512893381826067, -18430176119809328448967
Offset: 1
Examples
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 + ... such that (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.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.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! + ... FUNCTIONAL EQUATION. The series A(x) satisfies: (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 + ... GENERATING METHOD. 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. If we start with A = -x, and iterate (E.5) A = (A + x*log(1 + x)/log(1 + A))/2 then A will converge to g.f. A(x). SPECIFIC VALUES. The series A(x) diverges at x = -1. Here we evaluate some specific values. A(t) = 1 at t = A(1) = -0.653676637721419077935143447819113227... A(t) = 1/2 at t = A(1/2) = -0.398639649051906807220717042823223882... A(t) = 1/3 at t = A(1/3) = -0.285386940618446074866834432180324876... A(t) = 1/4 at t = A(1/4) = -0.222107177448605275724475246853117193... A(t) = -1/2 at t = A(-1/2) = 0.673256694764839886028076283033520406... A(t) = -1/3 at t = A(-1/3) = 0.400909109269336524244889643832206510... A(t) = -1/4 at t = A(-1/4) = 0.285959998501428938843181474481790362...
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..300
Programs
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Mathematica
nmax = 25; sol = {a[1] -> -1}; 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}]; sol /. Rule -> Set; a /@ Range[1, nmax] // Numerator (* Jean-François Alcover, Nov 02 2019 *) -
PARI
/* From Functional Equation (1 + A(x))^A(x) = (1 + x)^x */ {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 ); numerator( polcoeff(A,n) )} for(n=1,40, print1(a(n),", "))
Formula
(1) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*A(x) = 1.
(2) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + (k - p)*A(x) = (1 + x)^p.
(3) Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k - m)*x + k*A(x) = (1 + A(x))^m.
(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.
(5) A(A(x)) = x.
(6) (1 + A(x))^A(x) = (1 + x)^x.
(7) Sum_{n>=1} (-A(x))^(n+1) / n = x*log(1+x).
(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).
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