A271935 G.f. A(x) satisfies: A(x) = A( x^2 + 8*x*A(x)^2 )^(1/2), with A(0)=0, A'(0)=1.
1, 4, 26, 200, 1691, 15204, 142710, 1382568, 13721765, 138802136, 1425785270, 14832383488, 155947271878, 1654494195340, 17690004381000, 190426309700616, 2062071992480208, 22447191471665160, 245501068961175090, 2696300196714320520, 29725402250477117175, 328835072363241763920
Offset: 1
Keywords
Examples
G..f.: A(x) = x + 4*x^2 + 26*x^3 + 200*x^4 + 1691*x^5 + 15204*x^6 + 142710*x^7 + 1382568*x^8 + 13721765*x^9 + 138802136*x^10 + 1425785270*x^11 + ... where A(x)^2 = A( x^2 + 8*x*A(x)^2 ). RELATED SERIES. A(x)^2 = x^2 + 8*x^3 + 68*x^4 + 608*x^5 + 5658*x^6 + 54336*x^7 + 534984*x^8 + 5373824*x^9 + 54866075*x^10 + 567775856*x^11 + 5942353444*x^12 + ... Let B(x) be the series reversion of the g.f. A(x), A(B(x)) = x, then: B(x) = x - 4*x^2 + 6*x^3 - 15*x^5 + 90*x^7 - 660*x^9 + 5310*x^11 - 45765*x^13 + 413640*x^15 - 3864345*x^17 + 37014120*x^19 + ... + A264413(n)*x^(2*n+1) + ... such that B(x) = x*G(x^2) - 4*x^2 where G(x)^2 = G(x^2) + 12*x, and G(x) is the g.f. of A264413. From _Paul D. Hanna_, May 20 2024: (Start) The series (A(x)/x)^(1/4) seems to consist solely of integer coefficients (A(x)/x)^(1/4) = 1 + x + 5*x^2 + 34*x^3 + 268*x^4 + 2305*x^5 + 20988*x^6 + 198891*x^7 + 1941111*x^8 + 19377707*x^9 + 196936775*x^10 + ... and continues to be integral for at least the initial 400 coefficients. (End)
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..300
Programs
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PARI
{a(n) = my(A=x+x^2,X=x+x*O(x^n)); for(i=1,n, A = subst(A,x, x^2 + 8*X*A^2)^(1/2) ); polcoeff(A,n)} for(n=1,30,print1(a(n),", "))
Formula
G.f. A(x) satisfies: A( x*G(x^2) - 4*x^2 ) = x, where G(x)^2 = G(x^2) + 12*x, and G(x) is the g.f. of A264413.
Comments