What is the pattern rule for 1122, 1112, 1101, 1091?

Answer 1

well, the first three digits are 112,111,110,109, ...
Not sure why the last digit repeats in pairs, but I guess the next two terms could be 1080, 1070

Answer 2

No obvious pattern, but
the difference between first and second is -10
the difference between the 2nd and third is -21
the difference between the 3rd and the 2nd is -11
the difference between the 4th and the 3rd is -10

we could "force" a pattern on the numbers by assuming they are the
results of a cubic function
f(x) = ax^3 + bx^2 + cx + d , and let x = 0,1,2,3
f(0) = 0+0+0+d = 1122 , so d = 1122
f(1) = a + b + c + 1122 = 1112
or a+b+c = -10
f(2) = 8a + 4b + 2c + 1122 = 1101
8a + 4b + 2c = -21
f(3) = 27a + 9b + 3c + 1122 = 1091
27a + 9b + 3c = -31

so term(n) = (1/3)n^3 - (3/2)n^2 - (53/6)n + 1122
for n = 0, 1, 2, 3 to get your 4 terms

check: term(3) should give us 1091
term(3) = (1/3)(27) - (3/2)(9) - (53/6)(3) + 1122
= 9 - 27/2 - 53/2 + 1122
= 1091

term(2) = (1/3)(8) - (3/2)(4) - (53/6)(2) + 1122
= 1101

Looking good

I still believe you have a typo somewhere, so I was just having some fun with this.