A094766 Trajectory of 11 under repeated application of the map n -> n + 2*square excess of n (see A094765).

11, 15, 27, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, 273, 307, 343, 381, 421, 463, 507, 553, 601, 651, 703, 757, 813, 871, 931, 993, 1057, 1123, 1191, 1261, 1333, 1407, 1483, 1561, 1641, 1723, 1807, 1893, 1981, 2071, 2163, 2257, 2353, 2451, 2551, 2653

Offset: 0

Author

N. J. A. Sloane, Jun 10 2004

Keywords

• nonn

Comments

The trajectory of 3 gives A002061 and 5 gives essentially the same trajectory as 3.

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• S. H. Weintraub, An interesting recursion, Amer. Math. Monthly, 111 (No. 6, 2004), 528-530.
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Crossrefs

Cf. A002061, A053186, A094765.

Programs

Maple

f:= n -> 3*n - 2*floor(sqrt(n))^2:
g:= proc(n) option remember; f(procname(n-1)) end proc:
g(0):= 11:
seq(g(n),n=0..100); # Robert Israel, Oct 23 2015

Mathematica

NestList[3*#-2*Floor[Sqrt[#]]^2&,11,50] (* Harvey P. Dale, Feb 26 2022 *)

PARI

lista(nn) = {print1(n=11, ", "); for (k=2, nn, m = 3*n - 2*sqrtint(n)^2; print1(m, ", "); n = m;);} \\ Michel Marcus, Oct 23 2015

PARI

Vec(4+2*x+6*x^2+(7-8*x+3*x^2)/(1-x)^3 + O(x^100)) \\ Altug Alkan, Oct 23 2015

Formula

Numbers given satisfy a(n) = n^2 + 5n + 7, for n>2. - Ralf Stephan, Dec 04 2004

From Robert Israel, Oct 23 2015: (Start)

This is because for x = m^2 + 5*m + 7, (m+2)^2 < x < (m+3)^2 so A094765(x) = x + 2*(x-(m+2)^2) = m^2 + 7*m + 13 = (m+1)^2 + 5*(m+1) + 7.

Similarly, for any positive integer k, the trajectory of k^2 + k + 1 is n^2 + (2k+1) n + k^2 + k + 1 for n >= 0.

G.f.: 4 + 2*x + 6*x^2 + (7-8*x+3*x^2)/(1-x)^3. (End)