Last update: 4-30-2018

These patterns describe the first half of a number’s divisor parity set, as the second half is just a mirror image of the first half.

In the below sequences I am searching up to the number 25,000,000.

How to read the following listings

Numbers having the divisor parity of: 222: are A143928 — 2*p^2

p will always stand for a prime number and the A***** numbers are links to the corresponding sequence at The On-Line Encyclopedia of Integer Sequences. After the characters — will be a further description of the sequence.

only 2 or 1: A000040 — The prime numbers.

all 2’s: A016825 — Numbers of the form 4n+2.
22: A100484 and A001747 — 2*p^1
222: A143928 — 2*p^2
2222: 30, 42, 54, 66, 70, 78, 102, 110, 114, 130, 138, …
22222: A172191 — 2*p^4
222222: 90, 126, 150, 198, 234, 294, 306, 342, 350, 414, 486, …
2222222: 2*p^6
22222222: 210, 270, 330, 378, 390, 462, 510, 546, 570, 594, 690, …
222222222: n/2/p(n)^2=9 — 5^2*18
2222222222: 810, 1134, 1782, 2106, 2754, 3078, 3726, 3750, 4698, 5022, 5994, …
22222222222: 118098, 19531250 or also written as: 2*3^10 and 2*5^10
counting the number of 2’s in this sequence yields: A099774 — Number of divisors of 2*n-1. basically the odds

all 1’s: A005408 — The odd numbers: a(n) = 2n+1. —specific number of 1’s are…????
mixed 1’s and 2’s: A008586 — Multiples of 4. — a(n) = 4n
start 2, middle 1’s — A000079 — Powers of 2: a(n) = 2^n
for each middle 1 2^(2*middle zero count) matches two numbers the first is a square and the second is twice the first
example: 21111 = 2^8 and 2*(2^8) first number is (2^4)^2

start 2, middle 1’s end 2: nothing preexisting as a whole
starting p seems to increment at A014210 — Next prime after 2^n.
212: p*2^2 — starts at p=3
2112: p*2^3 — starts at p=5
21112: p*2^4 — starts at p=11
211112: p*2^5 — starts at p=17
2111112: p*2^6 — starts at p=37

repeating 21’s: 4,8,24,120
each one added yields the next term in: A007283 — 3*2^n.
2121: 24
21211:48
212111:96
2121111: 192
21211111: 384

first part of pair: A002063 — 9*4^n.
second part of pair: 2*(9*4^n)
21211121: 144
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2121112111: nothing?
21211121111: 576
212111211111: 1152
2121112111111: nothing?
21211121111111: 2304
212111211111111: 4608
2121112111111111: nothing?
21211121111111111: 9216
212111211111111111: 18432

each one added yields the next set of terms in: A070875 and A063920
21121: 80, 112
211211: 160, 224
2112111: 320, 448
21121111: 640, 896

each one yields more terms in: A160825 — a(2n) = 13*2^(n-2) and a(2n+1) = 11*2^(n-1) if n >= 2.
211121: 352, 416
2111211: 704, 832
21112111: 1408, 1664

2112: p*8
21121: 5*16 and 7*16 also written as 80 and 112
211211: 5*32 and 7*32 also written as 160 and 224
2112111: 5*64 and 7*64 also written as 320 and 448
21121111: 5*128 and 7*128 also written as 640 and 896
211211111: 5*256 and 7*256 also written as 1280 and 1792

212112: 72
21211212: p*24
21211221: 168, 216, 264 — 7,9,11 * 24
21221: numbers seem to always be squares and: A069262 — Numbers n such that sum(d|n,(-1)^d)=3.
……or(prime^2)*4
my finding is (2p)^2
212221: when these numbers are divided by 4 they yield: A082663
21212121:120
212212: 84, 108, 132, 156, 204, 220, 228, 260, 276, 340…
212212212: 588, 684, 828, 972, 1044…
212212212212: 3612, 3948, 4116, 4452, 4956, 5060, 5124, 5500, 5628, 5964, 6132…
2112: 8*p
211221: 8*(p^2)
21122121: 0, 760, 952, 1064, 1288, 2024, 2552, 2728… — gcd 8?
2112211121: 560
211221112112121: 3920
2121111221: 624, 816, 912, 1104 — primes: 13, 17, 19, 23, * 48
2121111212: 1392, 1488, 1776, 1968, 2064, 2256… primes from 29 on times 48
212212221212: 924, 1092, 1836, 2244, 2508, 3588, 5916, 6324, 6612…
212221212:180
21122112: divide the number by gcd: 8…then this sequence is made up of A161835 and A084968

impossible combinations?
21222