A362371 -id:A362371 - cp's OEIS frontend

A362551 a(0)=0. For each digit d in the sequence, append the smallest unused integer such that its last digit equals d.

0, 10, 1, 20, 11, 2, 30, 21, 31, 12, 3, 40, 22, 41, 13, 51, 61, 32, 23, 4, 50, 42, 52, 14, 71, 81, 33, 5, 91, 6, 101, 43, 62, 72, 53, 24, 15, 60, 34, 82, 25, 92, 111, 44, 7, 121, 8, 131, 63, 73, 35, 9, 141, 16, 151, 70, 161, 54, 83, 26, 102, 17, 112, 45, 93

Offset: 0

Gavin Lupo, Apr 24 2023

This sequence is a permutation of the nonnegative integers.

			a(0) =  0
a(1) = 10 (d=0 from a(0)=0, smallest integer other than 0 that ends with 0).
a(2) =  1 (d=1 from a(1)=10, smallest integer that ends with a 1).
a(3) = 20 (d=0 from a(1)=10, smallest integer other than 0 and 10 that ends with 0).
a(4) = 11 (d=1 from a(2)=1, smallest integer other than 1 that ends with 1).
a(5) =  2 (d=2 from a(3)=20, smallest integer that ends with 2).
a(6) = 30 (d=0 from a(3)=20, smallest integer other than 0, 10, and 20 that ends with 0).

Gavin Lupo, Table of n, a(n) for n = 0..10000
Gavin Lupo, Image of the first 10000 terms.
Index entries for sequences that are permutations of the nonnegative integers

Cf. A106001, A362371.

from itertools import count, islice
def agen(): # generator of terms
    s, lastd = "0", [(k for k in count(i, 10)) for i in range(10)]
    for n in count(0):
        an = next(lastd[int(s[0])])
        s = s[1:] + str(an)
        yield an
print(list(islice(agen(), 65))) # Michael S. Branicky, Apr 25 2023

A362896 a(0)=2. For n>0, let d = n-th digit in the sequence thus far. a(n) = a(n-1) + d if d is even. Otherwise, a(n) = a(n-1) - d.

Original entry on oeis.org

2, 4, 8, 16, 15, 21, 20, 15, 17, 16, 18, 18, 17, 12, 11, 4, 3, 9, 8, 16, 15, 23, 22, 15, 14, 16, 15, 14, 18, 15, 6, 14, 13, 19, 18, 13, 15, 12, 14, 16, 15, 10, 9, 13, 12, 18, 17, 12, 11, 15, 14, 22, 21, 16, 22, 21, 25, 24, 21, 20, 11, 10, 18, 17, 14, 13, 8, 7

Offset: 0

Gavin Lupo, May 09 2023

"-" signs on negative values are ignored when determining the n-th digit.

			a(0) =  2.
a(1) =  4. 1st digit is 2, which is even. 2 + 2 = 4.
a(2) =  8. 2nd digit is 4, which is even. 4 + 4 = 8.
a(3) = 16. 3rd digit is 8, which is even. 8 + 8 = 16.
a(4) = 15. 4th digit is 1, which is odd.  16 - 1 = 15.
a(5) = 21. 5th digit is 6, which is even. 15 + 6 = 21.

Gavin Lupo, Table of n, a(n) for n = 0..10000

Cf. A362371, A362551.

a = [2]
split = []
for i in range(100):
    split += [int(j) for j in str(abs(a[i]))]
    if split[i] % 2 == 0:
        a.append((a[i] + split[i]))
    else:
        a.append((a[i] - split[i]))
print(a)

A384632 a(0)=0. For each digit d in the sequence, let a(n) equal the smallest unused integer which has at least d divisors.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 12, 5, 7, 16, 24, 8, 18, 9, 10, 30, 11, 36, 48, 13, 14, 15, 17, 19, 20, 21, 28, 22, 40, 23, 25, 26, 27, 29, 32, 31, 42, 33, 60, 34, 35, 37, 38, 39, 54, 41, 43, 44, 45, 46, 49, 47, 50, 51, 52, 53, 56, 55, 72, 57, 58, 62, 59, 63, 61, 64, 65

Offset: 0

Gavin Lupo, Jun 05 2025

			a(0) = 0.
a(1) = Smallest unused integer with at least 0 divisors = 1.
a(2) = Smallest unused integer with at least 1 divisor = 2.
a(3) = Smallest unused integer with at least 2 divisors = 3.
a(4) = Smallest unused integer with at least 3 divisors = 4.
a(5) = Smallest unused integer with at least 4 divisors = 6.
a(6) = Smallest unused integer with at least 6 divisors = 12.
a(7) = Smallest unused integer with at least 1 divisor = 5.
a(8) = Smallest unused integer with at least 2 divisors = 7.

Cf. A000005, A362371, A362551.

from sympy import divisor_count
a = [0]
for n in range(100):
    if a[n] >= 10:
        split = [int(d) for d in str(a[n])]
    else:
        split = [a[n]]
    for s in split:
        num = 1
        while True:
            if divisor_count(num) >= s and num not in a:
                a.append(num)
                break
            num += 1
print(a)

cp's OEIS Frontend

A362551 a(0)=0. For each digit d in the sequence, append the smallest unused integer such that its last digit equals d.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A362896 a(0)=2. For n>0, let d = n-th digit in the sequence thus far. a(n) = a(n-1) + d if d is even. Otherwise, a(n) = a(n-1) - d.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Python

A384632 a(0)=0. For each digit d in the sequence, let a(n) equal the smallest unused integer which has at least d divisors.

Views

Author

Keywords

Examples

Crossrefs

Programs

Python