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A332993
a(1) = 1, for n > 1, a(n) = n + a(A032742(n)).
11
1, 3, 4, 7, 6, 10, 8, 15, 13, 16, 12, 22, 14, 22, 21, 31, 18, 31, 20, 36, 29, 34, 24, 46, 31, 40, 40, 50, 30, 51, 32, 63, 45, 52, 43, 67, 38, 58, 53, 76, 42, 71, 44, 78, 66, 70, 48, 94, 57, 81, 69, 92, 54, 94, 67, 106, 77, 88, 60, 111, 62, 94, 92, 127, 79, 111, 68, 120, 93, 113, 72, 139, 74, 112, 106, 134, 89, 131, 80, 156, 121
OFFSET
1,2
COMMENTS
Sum of those divisors of n that can be obtained by repeatedly taking the largest proper divisor (of previous such divisor, starting from n, which is included in the sum), up to and including the terminal 1.
FORMULA
a(1) = 1; and for n > 1, a(n) = n + a(A032742(n)).
a(n) = n + A006022(n).
a(n) = A332994(n) + A333791(n).
a(n) = A000203(n) - A333783(n).
It seems that for all n >= 1, a(n) <= A073934(n) <= A333794(n).
EXAMPLE
a(18) = 18 + 18/2 + 9/3 + 3/3 = 18 + 9 + 3 + 1 = 31.
PROG
(PARI) A332993(n) = if(1==n, n, n + A332993(n/vecmin(factor(n)[, 1])));
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 04 2020
STATUS
approved