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A264415
G.f. A(x) satisfies: A(x)^2 = A(x^2) + 30*x.
5
1, 15, -105, 1575, -29190, 603225, -13352850, 309605625, -7422255645, 182481301800, -4575894819300, 116581172754375, -3009161401332975, 78523515330379875, -2068113764887828875, 54904020923799337500, -1467692309121298737960, 39472725372798507822900, -1067296235915278105855650, 28996357915496677935088125, -791147023483262777604486675, 21669197341488265510394307750
OFFSET
0,2
LINKS
FORMULA
Given g.f. A(x), let G(x) denote the g.f. of A264227, then:
(1) G( x/(A(x)^2 - 25*x) ) = x,
(2) G( x/(A(x^2) + 5*x) ) = x,
(3) A(G(x))^2 = (1+25*x) * G(x)/x,
(4) A(G(x)^2) = (1-5*x) * G(x)/x,
where G(x)^2 = G( x^2/(1-10*x) ).
EXAMPLE
G.f.: A(x) = 1 + 15*x - 105*x^2 + 1575*x^3 - 29190*x^4 + 603225*x^5 - 13352850*x^6 + 309605625*x^7 +...
where
A(x)^2 = 1 + 30*x + 15*x^2 - 105*x^4 + 1575*x^6 - 29190*x^8 + 603225*x^10 - 13352850*x^12 + 309605625*x^14 +...
so that A(x)^2 = A(x^2) + 30*x.
Let G(x) = Series_Reversion( x / (A(x^2) + 5*x) ), then
G(x) = x + 5*x^2 + 40*x^3 + 350*x^4 + 3220*x^5 + 30500*x^6 + 294625*x^7 + 2886875*x^8 + 28598035*x^9 + 285786575*x^10 +...+ A264227(n)*x^n +...
such that G(x)^2 = G( x^2/(1-10*x) ).
PROG
(PARI) {a(n) = my(A=1); for(i=1, n, A = sqrt( subst(A, x, x^2) + 30*x +x*O(x^n))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Nov 12 2015
STATUS
approved