A249075 - OEIS

%I #19 Feb 08 2023 07:46:03

%S 1,5,11,51,161,773,3027,15395,69881,377781,1915163,10981907,60776145,

%T 368269541,2191553891,13976179203,88489011497,591631462805,

%U 3954213899691,27619472411891,193696456198913,1408953242322117,10318990227472115,77948745858933731

%N Sum of the numbers in row n of the array at A249074.

%H Vaclav Kotesovec, <a href="/A249075/b249075.txt">Table of n, a(n) for n = 0..720</a> (terms 0..100 from Clark Kimberling)

%F a(n) = a(n-1) + 2*(n+1)*a(n-2). - _Vaclav Kotesovec_, Aug 10 2021, following a suggestion from _John M. Campbell_

%F From _Vaclav Kotesovec_, Aug 10 2021: (Start)

%F E.g.f. A(x) satisfies the differential equation 6*A(x) + (2*x + 1)*A'(x) - A''(x) = 0, A(0) = 1, A'(0) = 5.

%F E.g.f.: 1 + 2*x + sqrt(Pi) * (3 + 4*x*(1 + x)) * exp((x + 1/2)^2) * (erf(x + 1/2) - erf(1/2))/2.

%F a(n) ~ sqrt(Pi) * erfc(1/2) * 2^((n-1)/2) * n^(n/2 + 1) * exp(1/8 + sqrt(n/2) - n/2). (End)

%e First 3 rows of A249074:

%e 1

%e 4    1

%e 6    4    1

%e so that a(0) = 1, a(1) = 5, a(2) = 11.

%t z = 11; p[x_, n_] := x + 2 n/p[x, n - 1]; p[x_, 1] = 1;

%t t = Table[Factor[p[x, n]], {n, 1, z}]

%t u = Numerator[t]

%t TableForm[Table[CoefficientList[u[[n]], x], {n, 1, z}]] (* A249074 array *)

%t Flatten[CoefficientList[u, x]] (* A249074 sequence  *)

%t u /. x -> 1  (* A249075 *)

%t RecurrenceTable[{a[n] == a[n-1] + 2*(n+1)*a[n-2], a[0] == 1, a[1] == 5}, a, {n, 0, 25}] (* _Vaclav Kotesovec_, Aug 10 2021 *)

%t nmax = 25; CoefficientList[Series[1 + 2*x + Sqrt[Pi]*(3 + 4*x*(1 + x)) * E^((1/2 + x))^2 * (Erf[1/2 + x] - Erf[1/2])/2, {x, 0, nmax}], x] * Range[0, nmax]! (* _Vaclav Kotesovec_, Aug 10 2021 *)

%Y Cf. A249074.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 20 2014