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
Keywords
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).
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)
Comments