A004719 -id:A004719 - cp's OEIS frontend

A371916 Zeroless analog of tetranacci numbers.

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.

A379512 Erase digits 0 and 1 from decimal expansion of n. Then keep just the coprime digits; write 0 if all digits disappear.

Original entry on oeis.org

0, 0, 2, 3, 4, 5, 6, 7, 8, 9, 0, 0, 2, 3, 4, 5, 6, 7, 8, 9, 2, 2, 0, 23, 0, 25, 0, 27, 0, 29, 3, 3, 32, 0, 34, 35, 0, 37, 38, 0, 4, 4, 0, 43, 0, 45, 0, 47, 0, 49, 5, 5, 52, 53, 54, 0, 56, 57, 58, 59, 6, 6, 0, 0, 0, 65, 0, 67, 0, 0, 7, 7, 72, 73, 74, 75, 76, 0, 78, 79, 8, 8, 0, 83, 0, 85, 0, 87

Offset: 0

Ctibor O. Zizka, Jan 21 2025

The numbers n such that a(n) = k for any fixed k are a 10-automatic sequence. - Charles R Greathouse IV, Jan 21 2025

			a(10) = 0 as we do not accept zeros and ones in n.
a(22) = 0 as gcd(2,2) = 2.
a(25) = 25 as gcd(2,5) = 1.
a(1234567890) = a(23456789) = a(3579) = a(57) = 57.
Note that numbers n with even digits and numbers n containing digits 0 and 1 only disappear immediately and we get a(n) = 0.

Cf. A004151, A004719, A059717.

a(n)=my(d=select(k->k>1, digits(n))); if(sum(i=1,#d, d[i]%2==0)>1, d=select(k->k%2,d)); if(sum(i=1,#d, d[i]%3==0)>1, d=select(k->k%3,d)); if(sum(i=1,#d, d[i]==5)>1, d=select(k->k!=5,d)); if(sum(i=1,#d, d[i]==7)>1, d=select(k->k!=7,d)); fromdigits(d) \\ Charles R Greathouse IV, Jan 21 2025

a(n) <= 9875. There are 299 distinct values in this sequence. - Charles R Greathouse IV, Jan 21 2025

A380391 Numbers k such that A343750(k) != k.

Original entry on oeis.org

10, 12, 14, 16, 18, 20, 28, 30, 31, 32, 34, 35, 36, 38, 40, 42, 50, 51, 52, 54, 56, 60, 62, 64, 68, 70, 71, 72, 73, 74, 75, 76, 78, 80, 84, 85, 90, 91, 92, 93, 94, 95, 96, 97, 98, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 112, 114, 115, 116, 118, 119, 120

Offset: 1

Ctibor O. Zizka, Jan 23 2025

			k = 13: A343750(13) = 13, thus k = 13 is not in the sequence.
k = 14: A343750(14) = 41, thus k = 14 is a term.

Cf. A000005, A004719, A010785, A272215, A343750.

q[k_] := Module[{perm = FromDigits /@ Permutations[IntegerDigits[k]], d}, d = DivisorSigma[0, perm]; Min@ perm[[Position[d, Min[d]] // Flatten]] != k]; Select[Range[120], q] (* Amiram Eldar, Jan 24 2025 *)

A380435 Erase digit 0 from decimal expansion of n. Then repeatedly apply the number of divisor function (A000005) onto each digit until a stationary value is reached. a(n) is the final stationary value (if it is reached for all digits).

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 11, 12, 12, 12, 12, 12, 12, 12, 12, 2, 21, 22, 22, 22, 22, 22, 22, 22, 22, 2, 21, 22, 22, 22, 22, 22, 22, 22, 22, 2, 21, 22, 22, 22, 22, 22, 22, 22, 22, 2, 21, 22, 22, 22, 22, 22, 22, 22, 22, 2, 21, 22, 22, 22, 22, 22, 22, 22, 22, 2, 21, 22, 22, 22, 22, 22

Offset: 1

Ctibor O. Zizka, Jan 24 2025

The number of iterations is 0, 1, 2, 3 for numbers containing the highest digits (1, 2), (3,5,7), (4, 9), (6, 8). n >= a(n) >= 1.

			For n = 21 a(21) = 21.
For n = 408 we iterate 48 --> 34 --> 23 --> 22, thus, after 3 iterations, a(408) = 22.

Cf. A002275, A002276 - A002283, A004719, A007931, A111066.

a[n_] := FromDigits[IntegerDigits[n] /. {0 -> Nothing, ?(# > 1 &) -> 2}]; Array[a, 100] (* _Amiram Eldar, Jan 24 2025 *)

a(A007931(n)) = A007931(n).
For r = 1, k >= 0:
a(10^k) = 1
a((10^k - 1)/9) = (10^k - 1)/9.
For r from [2, 9], k >= 0:
a(r*10^k) = 2.
a(r*(10^k - 1)/9) = 2*(10^k - 1)/9.

cp's OEIS Frontend

A371916 Zeroless analog of tetranacci numbers.

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A379512 Erase digits 0 and 1 from decimal expansion of n. Then keep just the coprime digits; write 0 if all digits disappear.

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A380391 Numbers k such that A343750(k) != k.

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A380435 Erase digit 0 from decimal expansion of n. Then repeatedly apply the number of divisor function (A000005) onto each digit until a stationary value is reached. a(n) is the final stationary value (if it is reached for all digits).

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