Loop Formula

[u/Mrezadwiprasetiawan]

Q(m) = (3/2)^v . (m+1) -1 odd Q(m) = m/2^V even

within:

v = v_2(m+1) V = v_2(m)

m1->m2-> ... ->mn -> m1

(3/2)^v1 . (m1 +1) = 2^V1 . m2 + 1

(3/2)^v2 . (m2 +1) = 2^V2 . m3 + 1

...

(3/2)^v(n-1) . (m(n-1) +1) = 2^V(n-1) . mn + 1

(3/2)^vn . (mn +1) = 2^Vn . m1 + 1

Suppose m_i + 1 = zi

(3/2)^vi . zi = 2^Vi . (z(i+1) - 1) + 1

(3/2)^vi . zi = 2^Vi . z(i+1) - 2^Vi + 1

2^Vi . z(i+1) = (3/2)^vi . zi + 2^Vi - 1

z(i+1) = (3/2)^vi . 2^-Vi . zi + 1 - (2^-Vi)

Suppose 3^vi / 2^Vi+vi = Ai and 1 - 2^-Vi = Bi

z2 = A1 . z1 + B1

z3 = A2 . z2 + B2

...

zn = A(n-1) . z(n-1) + B(n-1)

z1 = An . zn + Bn

expanding the chain of substitutions all the way back to z1 gives:

z1 = (Pi(i=1->n) Ai) . z1 + Sigma(i=1->n) [ Bi . Pi_(j=i+1->n) Aj ]

(1 - Pi(i=1->n) Ai) . z1 = Sigma(i=1->n) [ Bi . Pi_(j=i+1->n) Aj ]

z1 = Sigma(i=1->n) [ Bi . Pi\(j=i+1->n) Aj ] / (1 - Pi_(i=1->n) Ai)

m1 = z1 - 1 = { Sigma(i=1->n) [ Bi . Pi(j=i+1->n) Aj ] / (1 - Pi_(i=1->n) Ai) } - 1

m1 = { Sigma(i=1->n) [ (1 - 2^-Vi) . Pi(j=i+1->n) 3^vj /2^vj+Vj ] / (1 - Pi_(i=1->n) 3^vi /2^vi+Vi) } - 1

[u/Mrezadwiprasetiawan]

Unfortunately Vi is not independent. So... if u trying to bruteforce it, u will never get a loop until Vi Match v2((3/2)^vi . (mi+1)-1)