login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A277179
G.f. A(x) satisfies: Series_Reversion( A(x)/(1 - i*x + A(x)^2) ) = A(x)/(1 + i*x + A(x)^2), where odd function A(x) = Sum_{n>=1} a(n)*x^(2*n-1) and i^2 = -1.
0
1, 1, 3, 12, 59, 365, 3088, 38996, 740467, 19758435, 690334227, 30150545092, 1599014847188, 100918669227260, 7463532377496184, 638749818745548988, 62599568958563222499, 6962493257384838286351, 872012649206880910980233, 122142967347570710975572644, 19017873674952442001496543299, 3273737565549680748476297065053, 619999424134642103881677814415308, 128612434698134902683744004202176480
OFFSET
1,3
COMMENTS
Compare to the following related identities.
Let F(x) = x*C(x^2) such that C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108), then F(x) satisfies:
(1) Series_Reversion( F(x)/(1 - i*F(x) + F(x)^2) ) = F(x)/(1 + i*F(x) + F(x)^2),
(2) Series_Reversion( F(x)/(1 - i*x + x*F(x)) ) = F(x)/(1 + i*x + x*F(x)),
in which the inverse of the composite function equals the conjugate of that function.
FORMULA
G.f. A(x) satisfies: A(x)^2 = x*B1(x)/B2(x) - 1, where B1(x) + i*B2(x) = A(x)/(1 - i*x + A(x)^2).
EXAMPLE
G.f.: A(x) = x + x^3 + 3*x^5 + 12*x^7 + 59*x^9 + 365*x^11 + 3088*x^13 + 38996*x^15 + 740467*x^17 + 19758435*x^19 +...
such that the inverse of function A(x)/(1-i*x + A(x)^2) equals its conjugate.
RELATED SERIES.
A(x)^2 = x^2 + 2*x^4 + 7*x^6 + 30*x^8 + 151*x^10 + 920*x^12 + 7404*x^14 + 87774*x^16 + 1589695*x^18 + 41348962*x^20 +...
Let A(x)/(1 - i*x + A(x)^2) = B1(x) + i*B2(x),
A(x)/(1 - i*x + A(x)^2) = x + i*x^2 - x^3 - 2*i*x^4 + 3*x^5 + 3*i*x^6 - 4*x^7 - 8*i*x^8 + 11*x^9 + 9*i*x^10 - 13*x^11 - 35*i*x^12 + 46*x^13 + 13*i*x^14 - 26*x^15 - 193*i*x^16 + 239*x^17 - 177*i*x^18 + 151*x^19 - 1472*i*x^20 + 1711*x^21 - 3328*i*x^22 + 3479*x^23 - 14789*i*x^24 + 16500*x^25 - 45614*i*x^26 + 49093*x^27 - 173708*i*x^28 + 190208*x^29 - 598306*i*x^30 +...
then the series reversion of B1(x) + i*B2(x) equals the conjugate B1(x) - i*B2(x).
Note that A(x)^2 = x*B1(x)/B2(x) - 1, which holds because
B1(x) = A(x)*(1 + A(x)^2)/((1 + A(x)^2)^2 + x^2), and
B2(x) = A(x)*x/((1 + A(x)^2)^2 + x^2).
PROG
(PARI) {a(n) = my(Oxn=x*O(x^(2*n)), A = x +Oxn); for(i=1, 2*n, A = A + (x - subst(A/(1+I*x + A^2), x, A/(1-I*x + A^2) ))/2); polcoeff(A, 2*n-1)}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
Sequence in context: A101054 A122752 A020102 * A201013 A379616 A379514
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 10 2016
STATUS
approved