A293239 Number of terms in the fully expanded n-th derivative of x^x.

1, 2, 4, 7, 11, 15, 21, 28, 35, 43, 53, 64, 76, 88, 102, 117, 133, 149, 167, 186, 206, 226, 248, 271, 295, 319, 345, 372, 400, 428, 458, 489, 521, 553, 587, 622, 658, 694, 732, 771, 811, 851, 893, 936, 980, 1024, 1070, 1117, 1165, 1213, 1263, 1314, 1366, 1418

Offset: 0

Author

Vladimir Reshetnikov, Oct 03 2017

Keywords

• nonn

Comments

Conjecture: the 2nd differences are eventually periodic: 1, 1, 1, 0, 2, 1, 0, 1, [2, 1, 1, 0].

Examples

			For n = 3, the 3rd derivative of x^x is x^x + 3*x^x*log(x) + 3*x^x*log^2(x) + x^x*log^3(x) + 3*x^(x-1) + 3*x^(x-1)*log(x) - x^(x-2), so a(3) = 7.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A281434.

Programs

Mathematica

Join[{1}, Length /@ Rest[NestList[Expand[D[#, x]] &, x^x, 53]]]

Formula

Conjecture: a(n) ~ n^2/2. - Vaclav Kotesovec, Oct 05 2017

Conjectures from Colin Barker, Oct 05 2017: (Start)

G.f.: (1 + x^2 + x^3 + x^6 - x^8 + x^9 + x^12 - x^13) / ((1 - x)^2*(1 - x^4)).

a(n) = (5 + (-1)^n + (1-i)*(-i)^n + (1+i)*i^n + 2*n + 4*n^2) / 8 for n>7 where i=sqrt(-1).

a(n) = 2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) for n>6.

(End)