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Devil's Sequence

Problem



    Robodevil likes to do some mathematics between rehearsals of his orchestra. Today he invented devilish sequence No. 1729:
    

x0 = 0,
x1 = 1,
xn = (xn - 1 + xn - 2) / 2.


    For example, x10 = 0.666015625. Robodevil became interested at once how many `sixes` there were at the beginning of an arbitrary xn. In 6 nanoseconds, he had a formula. Can you do the same?
    Input
    You are given an integer n; `2 ≤ n ≤ 100000`.
    Output
    Output the number of sixes at the beginning of xn in decimal notation.
    Example



Input
Output


10
3

Difficulty: EASY
Tags: Mathematics

Starter code:
def count_sixes(n):
	

Solutions:
from math import floor, log

def count_sixes(n):
	return floor((n - n % 2) * log(2, 10))


count_sixes = lambda n: len(str(1 << n - n % 2)) - 1


from math import log10, floor, log

def count_sixes(n):
	k = floor(log10(3) + (n - 1) * log10(2))
	return k - (k * log(10) > (n - (n & 1)) * log(2))


import math
count_sixes = lambda n: int((n - n % 2) * math.log10(2))


def count_sixes(n):
	res = str(10 ** n // 15 - (-5) ** n // 15)
	return len(res) - len(res.lstrip('6'))


def definer(left, nL, right, nR, name):
	((inpL, outL), (inpR, outR)) = (left[:2], right[:2])
	return (inpL * nL + inpR * nR, outL * nL + outR * nR, left, nL, right, nR, name)
INF = 10 ** 6
C00 = (4, 0, None, None, None, None, 'C00')
C2 = (2, 1, None, None, None, None, 'C2')
C4 = (4, 1, None, None, None, None, 'C4')
C6 = definer(C2, 1, C4, 1, 'C6')
C10 = definer(C4, 1, C6, 1, 'C10')
C19_6 = definer(C10, 19, C6, 1, 'C19_6')
C18_6 = definer(C10, 18, C6, 1, 'C18_6')
C19x5_18x1 = definer(C19_6, 5, C18_6, 1, 'C19x5_18x1')
C19x4_18x1 = definer(C19_6, 4, C18_6, 1, 'C19x4_18x1')
C_x5andx4 = definer(C19x5_18x1, 1, C19x4_18x1, 1, 'C_x5andx4')
C_x13andx4 = definer(C_x5andx4, 13, C19x4_18x1, 1, 'C_x13andx4')
C_TOP = definer(C_x13andx4, INF, C2, 0, 'C_TOP')
START = definer(C00, 1, C_TOP, 1, 'START')

def count_sixes(n, c=START):
	(_, out, left, nL, right, _, name) = c
	if left:
		(inpL, outL) = left[:2]
	if not left:
		return out
	elif n < inpL * nL:
		(complete, r) = divmod(n, inpL)
		return outL * complete + (left and count_sixes(r, left))
	else:
		return outL * nL + (right and n and count_sixes(n - inpL * nL, right))


import math

def count_sixes(n):
	n0 = int(math.log10(2) * (n - 1))
	n1 = int(math.log10(2) * n)
	n2 = int(math.log10(2) * (n + 1))
	len = n1
	if n1 == n2 and n1 == n0 + 1 and (n % 2 != 0):
		len = n0
	return len


from fractions import Fraction
from decimal import *

def count_sixes(n):
	a = 6 << n - 2
	b = Fraction(2, 3)
	c = Fraction(1, a)
	d = 0
	if n % 2 == 0:
		d = b - c
	else:
		d = b + c
	getcontext().prec = 100000
	e = str(Decimal(d.numerator) / Decimal(d.denominator))
	f = 0
	for i in range(2, len(e)):
		if e[i] == '6':
			f += 1
		else:
			break
	return f


def count_sixes(n):
	num = 3 * 2 ** (n - 1)
	c = 0
	if n % 2 == 0 and int(str(num)[:2]) < 15:
		c = 1
	elif n % 2 == 1 and int(str(num)[0]) < 3:
		c = 1
	return len(str(num)) - 1 - c