A295425 a(n) = smallest number > a(n-1) such that the number of preceding terms in the sequence dividing a(n) is divisible by 4; a(1) = 2.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 210, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 330, 331

Offset: 1

Author

Masahiko Shin, Feb 12 2018

Keywords

• nonn
• easy

Comments

First differs from A000040 at a(47)=210.

Examples

			3 is in the sequence because no preceding terms (i.e., 0 terms) divide it and 0 is divisible by 4.
4 is not in the sequence because there is only 1 term (i.e., a(1) = 2) that divides it and 1 is not divisible by 4.
		

Crossrefs

Subsequence of A005117.

Cf. A000040, A030059.

Programs

Mathematica

With[{k = 4}, Nest[Append[#, SelectFirst[Range[#[[-1]] + 1, #[[-1]] + 120], Function[n, Divisible[Count[#, ?(Divisible[n, #] &)], k]]]] &, {2}, 68]] (* _Michael De Vlieger, Feb 15 2018 *)

PARI

isok(k, va, nb) = (sum(j=1, nb, !(k % va[j])) % 4) == 0;
lista(nn) = {va = vector(nn); va[1] = 2; for (n=2, nn, k = va[n-1]+1; while (! isok(k, va, n-1), k++); va[n] = k;); va;} \\ Michel Marcus, Mar 01 2018

Python

import math
def getSeq(n):
    if n == 1:
        return [2]
    prev = getSeq(n-1)
    cand = max(prev)
    while True:
        cand += 1
        if len( [n for n in prev if cand % n == 0] ) % 4 == 0:
            prev.append(cand)
            return prev
print(getSeq(100))