A364835 a(1) = 1; a(n+1) = (number of times a(n)+1 has appeared) - (number of times a(n) has appeared).

1, -1, -1, -2, 1, -2, 0, 1, -3, 1, -4, 0, 2, -1, -1, -2, 1, -4, -1, -3, 1, -5, 1, -6, 0, 4, -1, -3, 0, 3, 0, 2, -1, -2, 3, -1, -3, 0, 1, -6, -1, -3, -1, -4, 2, -1, -5, 1, -6, -1, -6, -2, 7, -1, -7, 3, -2, 7, -2, 6, 1, -7, 2, -1, -8, 1, -7, 1, -8, 1, -9, 1, -10, 0, 7, -3, 1, -11, 0

Offset: 1

Author

Rok Cestnik, Aug 28 2023

Keywords

• sign
• easy
• look

Comments

Irregular until n=149695 at which point it follows a simple pattern (see formula).

Examples

			a(5) = 2 - 1 = 1 because a(4) + 1 = -1 has appeared twice before and a(4) = -2 has appeared once.
a(7) = 2 - 2 = 0 because a(6) + 1 = -1 and a(6) = -2 have both appeared twice before.
		

Links

• Rok Cestnik, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Crossrefs

Cf. A330004, A329934.

Programs

Mathematica

nmax=78; a={1}; For[n=1, n<=nmax, n++, AppendTo[a,Count[a,Part[a,n]+1]-Count[a,Part[a,n]]]]; a (* Stefano Spezia, Aug 29 2023 *)

Python

a=[1]
for n in range(1000):
    a.append(a.count(a[n]+1)-a.count(a[n]))

Python

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an = 1; c = Counter([1])
    while True: yield an; an = c[an+1] - c[an]; c[an] += 1
print(list(islice(agen(),1001))) # Michael S. Branicky, Aug 29 2023

Formula

For n >= 149695: a(n) = 49456 - n/3 if (n mod 3) = 0, otherwise a(n) = (n mod 3) - 1.