A176982 Sequence defined by the recursion a(n) = (1/2)*((1-signum(abs(b(n))-n))*b(n)+(1+signum(abs(b(n))-n))*a(n-1)), with a(1)=1 and b(n)=1+a(n-1-(n mod a(n-1)))-(-1)^n*a(n-1).

1, 1, 2, 1, 3, 1, 3, 1, 3, -1, -1, 1, 3, -3, -5, 3, 1, 1, 3, -1, -1, 1, 3, 1, 3, 1, 3, -1, -1, 1, 3, -3, -5, 5, 11, -13, -9, 21, 21, -21, -21, 1, 3, -23, -21, 1, 3, 1, 3, 1, 3, -1, -1, 1, 3, -3, -5, 7, 11, -9, -9, 9, 19, -21, -11, 1, 3, -13, -33, 23, 11, -21, -11, 1, 3, -1, -1, 1, 3, -3

Offset

1,3

Comments

Behaves erratically (see linked picture).

There is a similar sequence with a(1)=2.

References

G. Balzarotti and P. P. Lava, 103 curiosità matematiche, Hoepli, 2010, p. 276.

Links

Paolo P. Lava, Table of n, a(n) for n = 1..10000

Paolo P. Lava, Graph of the first 600 terms of the sequence

John A. Pelesko, Generalizing the Conway-Hofstadter $10,000 Sequence, Journal of Integer Sequences, Vol. 7 (2004), Article 04.3.5.

Klaus Pinn, A Chaotic Cousin Of Conway's Recursive Sequence, arXiv:cond-mat/9808031, 1998.

Formula

a(n) = (1/2)*((1-signum(abs(b(n))-n))*b(n)+(1+signum(abs(b(n))-n))*a(n-1)), with a(1)=1 and b(n)=1+a(n-1-(n mod a(n-1)))-(-1)^n*a(n-1).

Example

a(1)=1.
b(2)=signum(abs(1+(2-1-(2 mod 1))-(-1)^2*1)-2)=signum(abs(1+1-1)-2)=-1.
a(2)=(1/2)*(1+1)*1+(1/2)*(1-1)*1=1+0=1.

Maple

P:=proc(i) local a, n; a:=array(1..50000); a[1]:=1; print(a[1]); for n from 2 by 1 to i do a[n]:=1/2*(1-signum(abs(1+a[n-1-(n mod (a[n-1]))]-(-1)^n*a[n-1])-n))*(1+a[n-1-(n mod (a[n-1]))]-(-1)^n*a[n-1])+1/2*(1+signum(abs(1+a[n-1-(n mod (a[n-1]))]-(-1)^n*a[n-1])-n))*a[n-1]; print(a[n]); od; end: P(10000);

Crossrefs

Cf. A004001, A176075, A176076.

Sequence in context: A328049 A328050 A335620 * A079728 A181801 A029244

Adjacent sequences: A176979 A176980 A176981 * A176983 A176984 A176985

Keyword

sign,look

Author

Paolo P. Lava and Giorgio Balzarotti, Apr 30 2010

Extensions

Entries and formula corrected by Paolo P. Lava, May 04 2010

a(3) corrected by N. J. A. Sloane, Oct 02 2010

Status

approved