A371911 -id:A371911 - cp's OEIS frontend

1, 1, 1, 1, 4, 7, 13, 25, 49, 94, 181, 349, 673, 1297, 25, 2344, 4339, 85, 6793, 13561, 24778, 45217, 9349, 9295, 88639, 1525, 1888, 11347, 13399, 28159, 54793, 17698, 11449, 11299, 95239, 135685, 253672, 495895, 98491, 983743, 183181, 176131, 1441546, 278461, 279319, 2175457

Offset: 0

Bryle Morga, Apr 12 2024

It is not known whether this sequence cycles, but it is conjectured to cycle just like A243063 and A371911 (have periods of 912 and 300056874, respectively) because the expected growth factor in the number of digits of successive terms is 0.9.

It's been computationally verified that if the sequence does cycle, then s+p > 10^10, where s and p are the starting index and period of the cycle, respectively.

			a(14) = Zr(a(13)+a(12)+a(11)+a(10)) = Zr(1297+673+349+181) = Zr(2500) = 25.

Cf. A000045, A000213, A000288, A004719, A242350, A243657, A243658, A243063.

a[0]=a[1]=a[2]=a[3]=1; a[n_]:=FromDigits[DeleteCases[IntegerDigits[a[n-1]+a[n-2]+a[n-3]+a[n-4]], 0]]; Array[a, 46, 0] (* Stefano Spezia, Apr 12 2024 *)

def a(n):
    a, b, c, d = 1, 1, 1, 1
    for _ in range(n):
        a, b, c, d = b, c, d, int(str(a+b+c+d).replace('0', ''))
    return a

# faster for initial segment of sequence
from itertools import islice
def agen(): # generator of terms
    a, b, c, d = 1, 1, 1, 1
    while True:
        yield a
        a, b, c, d = b, c, d, int(str(a+b+c+d).replace("0", ""))
print(list(islice(agen(), 45))) # Michael S. Branicky, Apr 13 2024

a(n) = Zr(a(n-1)+a(n-2)+a(n-3)+a(n-4)), where the function Zr(k) removes zero digits from k.

cp's OEIS Frontend

A371916 Zeroless analog of tetranacci numbers.

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