A156748 a(1) = a(2) = a(3) = a(4) = 0, then a(n) = abs(a(n-1) -a(n-2) +2*a(n-3) -a(n-4)) -a(n-3) -1.

0, 0, 0, 0, -1, 0, 0, 2, 2, -1, -2, -2, 4, 4, 3, 4, 0, -2, -2, 3, 2, 4, 6, 0, -5, -4, 4, 6, 4, 5, 2, -4, -6, 0, 7, 6, 4, 4, -2, -1, 0, 8, 8, 0, -1, -2, 8, 8, 4, 5, 0, 0, 0, 4, 3, 0, 0, -2, 4, 5, 4, 4, 0, -4, -5, 4, 4, 10, 14, 3, -6, -6, 4, 10, 5, 4, 4, -6, 2, 7, 16, 16, 4, -4, -9, 8, 8, 22, 30, 7, -10, -10, 8

Offset

1,8

Comments

There are 6 accumulation points for the sequence {a(n)/n} namely: -0.2, 0, 0.2, 0.4, 0.6 and 0.8.

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Mathematica

a[n_]:= a[n]= If[n<5, 0, Abs[a[n-1] -a[n-2] +2*a[n-3] -a[n-4]] -a[n-3] -1];
Table[a[n], {n, 100}] (* G. C. Greubel, Jun 16 2021 *)
nxt[{a_, b_, c_, d_}]:={b, c, d, Abs[d-c+2b-a]-b-1}; NestList[nxt, {0, 0, 0, 0}, 100][[All, 1]] (* Harvey P. Dale, Sep 01 2021 *)

Prog

(Sage)
@CachedFunction
def a(n): return 0 if (n<5) else abs(a(n-1) -a(n-2) +2*a(n-3) -a(n-4)) -a(n-3) -1
[a(n) for n in (1..100)] # a=A156748 # G. C. Greubel, Jun 16 2021

Crossrefs

Cf. A104156.

Sequence in context: A115034 A253193 A027869 * A351455 A075445 A216612

Adjacent sequences: A156745 A156746 A156747 * A156749 A156750 A156751

Keyword

sign

Author

Benoit Cloitre, Feb 14 2009

Status

approved