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A356782
Expansion of g.f. A(x) satisfies A(x) = x * Product_{n>=0} (1 + 2*A(x)^(2^n)).
6
1, 2, 6, 24, 106, 496, 2428, 12288, 63762, 337392, 1813628, 9876096, 54365876, 302037408, 1691327224, 9536234496, 54093070626, 308474110000, 1767481876540, 10170367611008, 58746459504884, 340513035730944, 1979964903739992, 11546094361266176, 67509252360531940
OFFSET
1,2
COMMENTS
From Paul D. Hanna, May 29 2024: (Start)
Compare to the following identities of the Catalan function C(x) = x + C(x)^2 (A000108):
(1) C(x)^2 = C( x*C(x)*(1 + C(x)) ),
(2) C(x)^4 = C( x*C(x)^3*(1 + C(x))*(1 + C(x)^2) ),
(3) C(x)^(2^n) = C( x*C(x)^(2^n-1)*Product_{k=0..n-1} (1 + C(x)^(2^k)) ) for n > 0. (End)
LINKS
FORMULA
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
(1) A(x)^2 = A( x*A(x) + 2*x*A(x)^2 ).
(2) A(x)^4 = A( x*A(x)^3*(1 + 2*A(x))*(1 + 2*A(x)^2) ).
(3) A(x) = x * Product_{n>=0} (1 + 2*A(x)^(2^n)).
(4) A(x) = x * Sum_{n>=0} 2^A000120(n) * A(x)^n, where A000120(n) = number of 1's in binary expansion of n.
(5) A(x) = x * Sum_{n>=0} A001316(n) * A(x)^n, where A001316(n) = number of odd entries in row n of Pascal's triangle.
(6) A(x) = Series_Reversion( x / Product_{n>=0} (1 + 2*x^(2^n)) ).
(7) A(x)^(2^n) = A( x*A(x)^(2^n-1)*Product_{k=0..n-1} (1 + 2*A(x)^(2^k)) ) for n > 0. - Paul D. Hanna, May 29 2024
a(n) ~ c * d^n / n^(3/2), where d = 6.21914802593104186425760427502624741814921... and c = 0.11969150889542130656096151211011193088986347... - Vaclav Kotesovec, Apr 05 2024
The radius of convergence r of g.f. A(x) and A(r) satisfy 1 = Sum_{n>=0} 2^(n+1) * A(r)^(2^n) / (1 + 2*A(r)^(2^n)) and r = A(r) / Product_{n>=0} (1 + 2*A(r)^(2^n)), where r = 0.160793728631389877... = 1/d (d is given above) and A(r) = 0.368704226036154413721793821323276205873691236089885... - Paul D. Hanna, May 04 2024
EXAMPLE
G.f.: A(x) = x + 2*x^2 + 6*x^3 + 24*x^4 + 106*x^5 + 496*x^6 + 2428*x^7 + 12288*x^8 + 63762*x^9 + 337392*x^10 + 1813628*x^11 + 9876096*x^12 + ...
such that
A(x) = x * (1 + 2*A(x)) * (1 + 2*A(x)^2) * (1 + 2*A(x)^4) * (1 + 2*A(x)^8) * (1 + 2*A(x)^16) * ... * (1 + 2*A(x)^(2^n)) * ...
which is equivalent to
A(x) = x * (1 + 2*A(x) + 2*A(x)^2 + 4*A(x)^3 + 2*A(x)^4 + 4*A(x)^5 + 4*A(x)^6 + 8*A(x)^7 + 2*A(x)^8 + 4*A(x)^9 + 4*A(x)^10 + 8*A(x)^11 + 4*A(x)^12 + 8*A(x)^13 + 8*A(x)^14 + 16*A(x)^15 + ... + 2^A000120(n)*A(x)^n + ...).
Also, g.f. A(x) satisfies
A(x)^2 = A( x*A(x) + 2*x*A(x)^2 )
where
A(x)^2 = x^2 + 4*x^3 + 16*x^4 + 72*x^5 + 344*x^6 + 1704*x^7 + 8688*x^8 + 45328*x^9 + 240856*x^10 + 1298984*x^11 + 7092544*x^12 + ...
PROG
(PARI) {a(n) = my(A = serreverse( x/prod(k=0, #binary(n), 1 + 2*x^(2^k) +x*O(x^n)) ));
polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 01 2022
STATUS
approved