A344734 a(n) is the smallest divisibility checking number for the n-th number coprime to 10.

1, 2, 2, 8, 1, 9, 5, 17, 2, 16, 8, 26, 3, 23, 11, 35, 4, 30, 14, 44, 5, 37, 17, 53, 6, 44, 20, 62, 7, 51, 23, 71, 8, 58, 26, 80, 9, 65, 29, 89, 10, 72, 32, 98, 11, 79, 35, 107, 12, 86, 38, 116, 13, 93, 41, 125, 14, 100, 44, 134, 15, 107, 47, 143, 16, 114, 50

Offset

1,2

Comments

Conjecture: the subsequence of divisibility checking numbers for numbers that end in 1 form an arithmetic sequence. This is also the case for numbers ending in 3, 7, and 9.

Links

Table of n, a(n) for n=1..67.

Formula

Conjectures from Chai Wah Wu, Dec 29 2021: (Start)

a(n) = 2*a(n-4) - a(n-8) for n > 9.

G.f.: x*(x^8 + x^7 + x^6 + 5*x^5 - x^4 + 8*x^3 + 2*x^2 + 2*x + 1)/(x^8 - 2*x^4 + 1). (End)

Example

For n = 6, the sixth number coprime to 10 is 13. The divisibility checking number for 13 is 9, as demonstrated below:
  13 divides 12961
  1296 - (1 * 9) = 1287
  13 divides 1287
  128 - (7 * 9) = 65
  13 divides 65.
This is useful in quickly checking divisibility by hand. 13 divides 65 implies that 13 divides 12961.

Prog

(Python)
def divisibility_checking_number(a):
    if a%5 == 0 or a % 2 == 0:
        return(0)
    x = 1
    while (x*10 + 1)%a != 0 :
        x += 1
    return(x)
h = []
for i in range(1, 250):
    if divisibility_checking_number(i) != 0:
        h.insert(len(h), divisibility_checking_number(i))
print(h)

Crossrefs

Cf. A045572 (numbers coprime to 10).

Sequence in context: A102645 A366324 A365747 * A037300 A368877 A029623

Adjacent sequences: A344731 A344732 A344733 * A344735 A344736 A344737

Keyword

nonn

Author

Zach Pollard, May 27 2021

Status

approved