

A133058


a(0) = a(1) = 1; for n > 1, a(n) = a(n1) + n + 1 if a(n1) and n are coprime, otherwise a(n) = a(n1)/gcd(a(n1),n).


21



1, 1, 4, 8, 2, 8, 4, 12, 3, 1, 12, 24, 2, 16, 8, 24, 3, 21, 7, 27, 48, 16, 8, 32, 4, 30, 15, 5, 34, 64, 32, 64, 2, 36, 18, 54, 3, 41, 80, 120, 3, 45, 15, 59, 104, 150, 75, 123, 41, 91, 142, 194, 97, 151, 206, 262, 131, 189, 248, 308, 77, 139, 202, 266, 133, 199, 266, 334, 167
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

Remarkably, at n = 638 the sequence settles down and becomes quasiperiodic.  N. J. A. Sloane, Feb 22 2015. (Whenever I look at this sequence I am reminded of the great "Fly straight, dammit" scene in the movie "Avatar".  N. J. A. Sloane, Aug 06 2019)
For every choice of initial term a(0) there exist integers r and t >= 0 such that a(2*r+4*t+0) = 1, a(2*r+4*t+1) = 2*r+4*t+3, a(2*r+4*t+2) = 2*(2*r+4*t+3), a(2*r+4*t+3) = 2.  Ctibor O. Zizka, Dec 26 2007
See also the variants A255051 (which starts immediately with the same (1, x, 2x, 2) loop that the present sequence enters at n >= 641) and A255140 (which enters a different loop at n = 82).  M. F. Hasler, Feb 15 2015
With the recurrence used here (but with different starting values), if at some point we find a(2k) = 1, then from that point on the sequence looks like (1, x, 2x, 2), (1, x+4, 2(x+4), 2), (1, x+8, 2(x+8), 2), (1, x+12, 2(x+12), 2), ... where x = 2k+3. This is just a restatement of Zizka's comment above (although I have not seen a proof that this must always happen).  N. J. A. Sloane, Feb 22 2015
It is conjectured that quasiperiodic sequences exist only for R = 0, 1, 2 or 3 in a(n) = a(n1) + n + R and that for R >= 4 the recurrence is not quasiperiodic. For R = 0, 1, 2 all starting values give a quasiperiodic sequence. The respective loop is (1, x) for R = 0, (1, x, 2x, 2) for R = 1 (this sequence), (1, x, 2x, x) or (2x, x) for R = 2. For R = 3 only some starting values converge to a 6loop (4x+2, 2x+1, 3x+6, x+2, 2x+9, 3x+17).  Ctibor O. Zizka, Oct 27 2015


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 0..10000, May 26 2016 [First 1000 terms from Harvey P. Dale]
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020
N. J. A. Sloane, Graph of the first 1000 terms, showing the transition from chaos to order more dramatically.
N. J. A. Sloane and Brady Haran, Amazing Graphs, Numberphile video (2019).
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,1).


MAPLE

A[0]:= 1: A[1]:= 1:
for n from 2 to 1000 do
g:= igcd(A[n1], n);
A[n]:= A[n1]/g + `if`(g=1, n+1, 0);
od:
seq(A[i], i=0..1000); # Robert Israel, Nov 06 2015


MATHEMATICA

nxt[{n_, a_}]:={n+1, If[CoprimeQ[a, n+1], a+n+2, a/GCD[a, n+1]]}; Join[{1}, Transpose[ NestList[nxt, {1, 1}, 70]][[2]]] (* Harvey P. Dale, Feb 14 2015 *)


PROG

(PARI) (A133058_upto(N)=vector(N, n, if(gcd(N, n1)>1  n<3, N\=gcd(N, n1), N+=n)))(100) \\ M. F. Hasler, Feb 15 2015, simplified Jan 11 2020
(Magma) a:=[1]; for n in [2..70] do if Gcd(a[n1], n) eq 1 then Append(~a, a[n1] + n + 1); else Append(~a, a[n1] div Gcd(a[n1], n)); end if; end for; [1] cat a; // Marius A. Burtea, Jan 12 2020


CROSSREFS

Cf. A091508, A133579, A133580; A255051, A255140, A262922.
Quadrisections: A326991, A326992, A326993, A326994.
Sequence in context: A224536 A090488 A020848 * A011516 A088609 A163813
Adjacent sequences: A133055 A133056 A133057 * A133059 A133060 A133061


KEYWORD

nonn,look,nice,easy


AUTHOR

Ctibor O. Zizka, Dec 16 2007


EXTENSIONS

More terms from Ctibor O. Zizka, Dec 26 2007
Offset and definition corrected by N. J. A. Sloane, Feb 13 2015


STATUS

approved



