A375440 - OEIS

%I #11 Sep 12 2024 14:44:08

%S 1,1,5,65,1593,61953,3476813,265517441,26492540401,3349218907137,

%T 523572565927509,99215376614955457,22415450137196941993,

%U 5953820173628518544385,1837040977427662958973341,651657636773935012586716929,263375512326578915885862469601,120319850003020550647400856678401

%N Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*k) * ([x^k] A(x)^n) for n >= 1.

%C Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n+k, 2*k) * ([x^k] C(x)^n) for n >= 1 is satisfied by the Catalan function C(x) = 1 + x*C(x)^2 (A000108), where coefficient [x^k] C(x)^n = binomial(n+2*k-1,k)*n/(n+k).

%H Paul D. Hanna, <a href="/A375440/b375440.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) ~ c * 2^(4*n) * n^(2*n + 1/2) / (Pi^(2*n) * exp(2*n)), where c = 7.23682343848882192289996... - _Vaclav Kotesovec_, Sep 12 2024

%e G.f.: A(x) = 1 + x + 5*x^2 + 65*x^3 + 1593*x^4 + 61953*x^5 + 3476813*x^6 + 265517441*x^7 + 26492540401*x^8 + ...

%e RELATED TABLES.

%e The table of coefficients of x^k in A(x)^n begins:

%e   n=1: [1, 1,  5,  65,  1593,  61953,  3476813, ...];

%e   n=2: [1, 2, 11, 140,  3341, 127742,  7097687, ...];

%e   n=3: [1, 3, 18, 226,  5259, 197637, 10869476, ...];

%e   n=4: [1, 4, 26, 324,  7363, 271928, 14799444, ...];

%e   n=5: [1, 5, 35, 435,  9670, 350926, 18895290, ...];

%e   n=6: [1, 6, 45, 560, 12198, 434964, 23165174, ...];

%e   ...

%e from which we may illustrate the defining property given by

%e 0 = Sum_{k=0..n} (-1)^k * binomial(2*n, 2*k) * ([x^k] A(x)^n).

%e Using the coefficients in the table above, we see that

%e   n=1: 0 = 1*1 - 1*1;

%e   n=2: 0 = 1*1 - 6*2 + 1*11;

%e   n=3: 0 = 1*1 - 15*3 + 15*18 - 1*226;

%e   n=4: 0 = 1*1 - 28*4 + 70*26 - 28*324 + 1*7363;

%e   n=5: 0 = 1*1 - 45*5 + 210*35 - 210*435 + 45*9670 - 1*350926;

%e   n=6: 0 = 1*1 - 66*6 + 495*45 - 924*560 + 495*12198 - 66*434964 + 1*23165174;

%e   ...

%e The triangle A086645(n,k) = binomial(2*n, 2*k) begins:

%e   n=0: 1;

%e   n=1: 1,  1;

%e   n=2: 1,  6,   1;

%e   n=3: 1, 15,  15,   1;

%e   n=4: 1, 28,  70,  28,  1;

%e   n=5: 1, 45, 210, 210,  45,  1;

%e   n=6: 1, 66, 495, 924, 495, 66, 1;

%e   ...

%o (PARI) {a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1;

%o A[m+1] = sum(k=0, m, (-1)^(m-k+1) * binomial(2*m, 2*k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A086645, A375450, A375451.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 11 2024