perplexus.info :: Just Math : Odd divisor summation

Odd divisor summation (Posted on 2024-03-21)

For each positive integer n, let D_n be the greatest odd divisor of n. (For example, D₁₆₈ = 21.) Find D₁ + D₂ + D₃ + ⋅ ⋅ ⋅ + D₂₀₄₈.

No Solution Yet Submitted by Danish Ahmed Khan

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Computer solution

| Comment 2 of 4 |

1,398,102

-------------

def largestOddDivisor(n):

""" input integer, output the largest odd divisor """

if n%2 == 1:

return n

for i in range(int(n/2)+1-int(n/2)%2 , 0, -2):

if n%i == 0:

return i

return None

mysum = 0

for n in range(1,2049):

mysum += largestOddDivisor(n)

print(mysum)

Posted by Larry on 2024-03-21 11:30:55

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