1s are located at (i-th Tri-angular number + 1)th position

Number of zeros in bracket: 0 1 2 3 4

Sequence : (1) (10) (100) (1000) (10000)

1 is found at : T0+1 T1+1 T3+1 T4+1 T5+1

Value of Ti+1 : 0+1 1+1 3+1 6+1 10+1

The i-th triangular number is the sum of the i natural numbers from 1 to i.

Ti = i(i+1)/2

i-th '1' is located at (Ti+1)-th position.

Let, z = (Ti + 1) = (i(i+1)/2) + 1

-> z = (i^2 + i + 2)/2

-> 2z = i^2 + i + 2

-> (1)x(i^2) + (1)x(i^1) + (-2(z-1)x(i^0) = 0

So, now if we solve for i using quadratic formula,

i = (-1 +- squareRoot(8z-7))/2 [do calculation on your own]

z = {1,2,4,7,11,...}

plugin these values and you will find that squareRoot(8z-7) = an integersquare number for any z. So, a number belongs to z only and if only squareRoot(8z-7) can produce an integer. And here z is the position number where the digits are 1.

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*Edited by author 24.12.2020 19:00*

*Edited by moderator 14.02.2021 18:16*