A382122 G.f. satisfies Sum_{n>=0} x^n * abs(1/A(x)^n) = C(x), where C(x) = 1 + x*C(x)^2 and abs(F(x)) equals the series expansion formed by the unsigned coefficients in F(x).

1, 1, 3, 12, 49, 202, 838, 3486, 14575, 60820, 254406, 1061438, 4444802, 18602018, 78066384, 326985608, 1365996909, 5697914836, 23752394338, 99027785702, 413203462516, 1726164299990, 7219911692522, 30228722494504, 126658682953328, 530772842793396, 2224199143900798, 9319843329508200, 39051457052597480

Offset

0,3

Comments

Compare to Sum_{n>=0} x^n * C(x)^n = C(x), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

Conjecture: for n > 0, a(n) is odd iff n = 2^k for k >= 0.

Links

Paul D. Hanna, Table of n, a(n) for n = 0..521

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) Sum_{n>=0} x^n * abs(1/A(x)^n) = C(x), where C(x) = 1 + x*C(x)^2.

(2) Sum_{k=0..n} abs( [x^k] 1/A(x)^(n-k) ) = binomial(2*n+1,n)/(2*n+1) for n >= 0.

a(n) ~ c * d^n, where d = 4.1935797816358..., c = 0.142779... - Vaclav Kotesovec, Mar 28 2025

Example

G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 49*x^4 + 202*x^5 + 838*x^6 + 3486*x^7 + 14575*x^8 + 60820*x^9 + 254406*x^10 + 1061438*x^11 + 4444802*x^12 + ...
Below we illustrate the defining property of this sequence.
The coefficients in 1/A(x)^n begin
 1: [1,  -1, -2,  -7, -24, -84, -298, -1063, ...];
 2: [1,  -2, -3, -10, -30, -92, -283,  -858, ...];
 3: [1,  -3, -3, -10, -24, -57, -119,  -156, ...];
 4: [1,  -4, -2,  -8, -11,  -4,   82,   568, ...];
 5: [1,  -5,  0,  -5,   5,  49,  250,  1060, ...];
 6: [1,  -6,  3,  -2,  21,  90,  348,  1224, ...];
 7: [1,  -7,  7,   0,  35, 112,  364,  1070, ...];
 8: [1,  -8, 12,   0,  46, 112,  304,   672, ...];
 9: [1,  -9, 18,  -3,  54,  90,  186,   135, ...];
10: [1, -10, 25, -10,  60,  48,   35,  -430, ...];
...
The table of unsigned coefficients that form the series abs(1/A(x)^n) begins
 0: [1,  0,  0,  0,  0,   0,   0,    0,    0, ...];
 1: [1,  1,  2,  7, 24,  84, 298, 1063, 3858, ...];
 2: [1,  2,  3, 10, 30,  92, 283,  858, 2646, ...];
 3: [1,  3,  3, 10, 24,  57, 119,  156,  144, ...];
 4: [1,  4,  2,  8, 11,   4,  82,  568, 2578, ...];
 5: [1,  5,  0,  5,  5,  49, 250, 1060, 3800, ...];
 6: [1,  6,  3,  2, 21,  90, 348, 1224, 3654, ...];
 7: [1,  7,  7,  0, 35, 112, 364, 1070, 2394, ...];
 8: [1,  8, 12,  0, 46, 112, 304,  672,  469, ...];
 9: [1,  9, 18,  3, 54,  90, 186,  135, 1629, ...];
10: [1, 10, 25, 10, 60,  48,  35,  430, 3465, ...];
...
the antidiagonals of which add to the Catalan numbers (A000108):
  1 = 1;
  0 + 1 = 1;
  0 + 1 + 1 = 2;
  0 + 2 + 2 + 1 = 5;
  0 + 7 + 3 + 3 + 1 = 14;
  0 + 24 + 10 + 3 + 4 + 1 = 42;
  0 + 84 + 30 + 10 + 2 + 5 + 1 = 132;
  0 + 298 + 92 + 24 + 8 + 0 + 6 + 1 = 429;
  0 + 1063 + 283 + 57 + 11 + 5 + 3 + 7 + 1 = 1430;
  0 + 3858 + 858 + 119 + 4 + 5 + 2 + 7 + 8 + 1 = 4862;
  ...

Prog

(PARI) {a(n) = my(V=[1, 1], A, C = (1/x)*serreverse(x - x^2 +x^4*O(x^n)));
for(i=1, n, V = concat(V, 't); A = Ser(V);
V[#V] = 't + polcoef(C - sum(m=1, #V+1, x^m * Ser(abs(Vec( 1/A^m ))) ), #V) ); V[n+1]}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A000108, A382123.

Sequence in context: A037646 A012772 A012864 * A045890 A049673 A052703

Adjacent sequences: A382119 A382120 A382121 * A382123 A382124 A382125

Keyword

nonn

Author

Paul D. Hanna, Mar 16 2025

Status

approved