A362551 -id:A362551 - cp's OEIS frontend

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.

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

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.

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