A371911 Zeroless analog of tribonacci numbers.

1, 1, 1, 3, 5, 9, 17, 31, 57, 15, 13, 85, 113, 211, 49, 373, 633, 155, 1161, 1949, 3265, 6375, 11589, 21229, 39193, 7211, 67633, 11437, 86281, 165351, 26369, 2781, 19451, 4861, 2793, 2715, 1369, 6877, 1961, 127, 8965, 1153, 1245, 11363, 13761, 26369, 51493, 91623, 169485, 31261

Offset

0,4

Comments

At n = 208666297 this sequence enters a cycle that has a period of 300056874 and begins: 2847, 26331, 5851, ... (only the first 3 terms of the cycle are needed to reproduce the entire cycle).

This can be compared with the sequence A243063, which enters a cycle that has a (relatively) small period of 912.

Links

Table of n, a(n) for n=0..49.

Formula

a(n) = Zr(a(n-1) + a(n-2) + a(n-3)), where the function Zr(k) removes all zero digits from k.

Example

a(9) = Zr(a(8) + a(7) + a(6)) = Zr(17 + 31 + 57) = Zr(105) = 15.

Mathematica

a[0]=a[1]=a[2]=1; a[n_]:=FromDigits[DeleteCases[IntegerDigits[a[n-1]+a[n-2]+a[n-3]], 0]]; Array[a, 50, 0] (* Stefano Spezia, Apr 12 2024 *)

Prog

(Python)
def a(n):
    a, b, c = 1, 1, 1
    for _ in range(n):
        a, b, c = b, c, int(str(a+b+c).replace('0', ''))
    return a
(Python) # faster for initial segment of sequence
from itertools import islice
def agen(): # generator of terms
    a, b, c = 1, 1, 1
    while True: yield a; a, b, c = b, c, int(str(a+b+c).replace("0", ""))
print(list(islice(agen(), 50))) # Michael S. Branicky, Apr 13 2024

Crossrefs

Cf. A000045, A000213, A004719, A242350, A243657, A243658, A243063.

Sequence in context: A102475 A066173 A114322 * A000213 A074858 A074860

Adjacent sequences: A371908 A371909 A371910 * A371912 A371913 A371914

Keyword

nonn,base

Author

Bryle Morga, Apr 11 2024

Status

approved