A321604 G.f. A(x) satisfies: 1 = Sum_{n>=0} ((1+x)^(4*n) - A(x))^n.

1, 4, 22, 564, 25157, 1499576, 109904860, 9480509576, 937113401201, 104240673195936, 12878161899791760, 1749261564410844864, 259132759251207789056, 41585481940418457992816, 7188476201158569394613976, 1331880173688346226092103696, 263358773243148578509342224153, 55363099822436514905885084770968, 12330972024423209530808891225876436, 2900976547500300324930009436969260936

Offset: 0

Author

Paul D. Hanna, Nov 14 2018

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + 4*x + 22*x^2 + 564*x^3 + 25157*x^4 + 1499576*x^5 + 109904860*x^6 + 9480509576*x^7 + 937113401201*x^8 + 104240673195936*x^9 + ...
such that
1 = 1  +  ((1+x)^4 - A(x))  +  ((1+x)^8 - A(x))^2  +  ((1+x)^12 - A(x))^3  +  ((1+x)^16 - A(x))^4  +  ((1+x)^20 - A(x))^5  +  ((1+x)^24 - A(x))^6  +  ((1+x)^28 - A(x))^7 + ...
Also,
1 = 1/(1 + A(x))  +  (1+x)^4/(1 + (1+x)^4*A(x))^2  +  (1+x)^16/(1 + (1+x)^8*A(x))^3  +  (1+x)^36/(1 + (1+x)^12*A(x))^4  +  (1+x)^64/(1 + (1+x)^16*A(x))^5  +  (1+x)^100/(1 + (1+x)^20*A(x))^6  + ...
RELATED SERIES.
The logarithmic derivative of the g.f. begins
A'(x)/A(x) = 4 + 28*x + 1492*x^2 + 91788*x^3 + 6981484*x^4 + 621939700*x^5 + 63151305340*x^6 + 7181135905380*x^7 + 903210250234696*x^8 + ...
the coefficients of which are all divisible by 4:
(1/4) * A'(x)/A(x) = 1 + 7*x + 373*x^2 + 22947*x^3 + 1745371*x^4 + 155484925*x^5 + 15787826335*x^6 + 1795283976345*x^7 + 225802562558674*x^8 + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A303056, A321602, A321603, A321605.

Cf. A326264.

Programs

PARI

{a(n) = my(A=[1]); for(i=0, n, A=concat(A, 0); A[#A] = Vec( sum(m=0, #A, ((1+x)^(4*m) - Ser(A))^m ) )[#A] );H=A; A[n+1]}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) 1 = Sum_{n>=0} ((1+x)^(4*n) - A(x))^n.

(2) 1 = Sum_{n>=0} (1+x)^(4*n^2) / (1 + (1+x)^(4*n)*A(x))^(n+1).