A371695 The smallest composite number that divides the reverse of the concatenation of its ascending ordered prime factors, with repetition, when written in base n.

623, 4, 114, 4, 57, 4, 9, 4, 26, 4, 185, 4, 9, 4, 1718, 4, 343, 4, 9, 4, 70, 4, 25, 4, 9, 4, 195, 4, 226, 4, 9, 4, 25, 4, 123, 4, 9, 4, 654, 4, 862, 4, 9, 4, 42, 4, 49, 4, 9, 4, 3385, 4, 25, 4, 9, 4, 238, 4, 202, 4, 9, 4, 25, 4, 453, 4, 9, 4, 2435, 4, 721, 4, 9, 4, 49, 4, 70, 4, 9, 4, 186

Offset: 2

Author

Scott R. Shannon, Apr 03 2024

Keywords

• nonn
• base

Comments

See A371641 for an explanation of multiple terms being 4 and 9. The largest number in the first 10000 terms is a(5980) = 1030778.

Examples

			a(2) = 623 as 623 = 7_10 * 89_10 = 111_2 * 1011001_2 = "1111011001"_2 which when reversed is "1001101111"_2 = 623_10 which is divisible by 623.
a(4) = 114 as 114 = 2_10 * 3_10 * 19_10 = 2_4 * 3_4 * 103_4 = "23103"_4 which when reversed is "30132"_4 = 798_10 which is divisible by 114.
		

Links

• Scott R. Shannon, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A020639, A371641, A027746, A259047, A322843, A248915, A371948.

Programs

Python

from itertools import count
from sympy.ntheory import digits
from sympy import factorint, isprime
def fromdigits(d, b):
    n = 0
    for di in d: n *= b; n += di
    return n
def a(n):
    for k in count(4):
        if isprime(k): continue
        sf = []
        for p, e in list(factorint(k).items())[::-1]:
            sf.extend(e*digits(p, n)[1:][::-1])
        if fromdigits(sf, n)%k == 0:
            return k
print([a(n) for n in range(2, 83)]) # Michael S. Branicky, Apr 16 2024

Formula

If n+1 is composite, then a(n) <= A020639(n+1)^2. The numbers n where n+1 is composite and a(n) < A020639(n+1)^2 are 288, 298, 340, 360, 376, 516, 526, 550, 582, 736, ... and appear to be identical to A371948. - Chai Wah Wu, Apr 16 2024