A088713 G.f. A(x) satisfies A(x/A(x)) = 1/(1-x).
1, 1, 2, 6, 24, 118, 674, 4308, 30062, 225266, 1791964, 15009118, 131566314, 1201452248, 11389283418, 111761444078, 1132680800640, 11834071103246, 127261591139010, 1406778021294220, 15967144849210158, 185897394076705298
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 24*x^4 + 118*x^5 + 674*x^6 +... Illustration of logarithmic derivation. If we form an array of coefficients of x^k in A(x)^n, n>=1, like so: A^1: [1],1, 2, 6, 24, 118, 674, 4308, ...; A^2: [1, 2], 5, 16, 64, 308, 1716, 10724, ...; A^3: [1, 3, 9], 31, 126, 600, 3278, 20070, ...; A^4: [1, 4, 14, 52], 217, 1032, 5560, 33440, ...; A^5: [1, 5, 20, 80, 345], 1651, 8820, 52270, ...; A^6: [1, 6, 27, 116, 519, 2514],13385, 78420, ...; A^7: [1, 7, 35, 161, 749, 3689, 19663], 114269, ...; ... then the sums of the coefficients of x^k, k=0..n-1, in A(x)^n (shown above in brackets) begin: 1 = 1; 1 + 2 = 3; 1 + 3 + 9 = 13; 1 + 4 + 14 + 52 = 71; 1 + 5 + 20 + 80 + 345 = 451; 1 + 6 + 27 + 116 + 519 + 2514 = 3183; 1 + 7 + 35 + 161 + 749 + 3689 + 19663 = 24305; ... and equal the coefficients in log(A(x)): log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 71*x^4/4 + 451*x^5/5 + 3183*x^6/6 + 24305*x^7/7 + 197551*x^8/8 +... The main diagonal in the above table forms the g.f. G(x) of A088714: [1/1, 2/2, 9/3, 52/4, 345/5, 2514/6, 19663/7, ...] where G(x) = 1 + x + 3*x^2 + 13*x^3 + 69*x^4 + 419*x^5 + 2809*x^6 +... satisfies A'(x)/A(x) = (G(x) + x*G'(x)) / (1 - x*G(x)).
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..300
Programs
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Mathematica
terms = 22; A[] = 1; Do[A[x] = 1 + x*A[x]*A[1 - 1/A[x]] + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)
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PARI
a(n)=local(A=1+x);for(i=1,n,A=(1+A*serreverse(x/(A+x*O(x^n))))^1);polcoeff(A,n) for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Dec 06 2009
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PARI
{a(n)=local(A=1+x);if(n==0,1,for(i=1,n, A=1+x*exp(sum(k=1,n-1,sum(j=0,k,polcoeff(A^k+x*O(x^j),j))*x^k/k)+x*O(x^n)))); polcoeff(A+x*O(x^n),n)} for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Dec 09 2013
Formula
G.f. satisfies: A(x) = 1 + x*A(x)*A(1-1/A(x)).
G.f.: A(x*g(x)) = g(x) = (1-1/A(x))/x where g(x) is the g.f. of A088714.
From Paul D. Hanna, Dec 06 2009: (Start)
G.f. satisfies: A(x) = 1 + A(x)*Series_Reversion(x/A(x)).
G.f. satisfies: A( (x/(1+x)) / A(x/(1+x)) ) = 1 + x.
(End)
Logarithmic derivative: given g.f. A(x), let G(x) = A(x*G(x)) be the g.f. of A088714, then A'(x)/A(x) = (G(x) + x*G'(x)) / (1 - x*G(x)).