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A386883
Define a sequence of rationals by f(0) = 0, thereafter f(n) = f(n-1) - 1/n if that is >= 0, otherwise f(n) = f(n-1) + 1/n; a(n) corresponds to the number of addition steps minus the number of subtraction steps involved in calculating f(n).
1
0, 1, 0, -1, 0, -1, -2, -1, -2, -1, -2, -1, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -2, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3, -4, -3
OFFSET
0,7
COMMENTS
This sequence is unbounded below.
By contradiction:
- let M be the minimum value of the sequence and a(n) = M be the first occurrence of M in the sequence,
- as an addition step is always followed by a subtraction step, and the value M can only be followed by M+1, a(n+2*k) = M and a(n+2*k+1) = M+1 for any k >= 0,
- for any m >= n, f(m) = f(n) + Sum_{k = n+1..m} (-1)^(k-n-1) / k,
- as the alternating harmonic series converges to log(2), the sequence f will have a positive limit, say L > 0,
- hence for some m0 >= 0, f(m) > L/2 for any m >= m0,
- let k be such that n+2*k >= m0 and 1/(n+2*k+1) < L/2: f(n+2*k) > L/2, so f(n+2*k+1) = f(n+2*k) - 1/(n+2*k+1), and a(n+2*k+1) = a(n+2*k)-1 = M-1, a contradiction.
LINKS
FORMULA
a(n) = Sum_{k = 1..n} sign(A231692(n)/A231693(n) - A231692(n-1)/A231693(n-1)).
EXAMPLE
Sequence begins:
n a(n) f(n)-f(n-1)
-- ---- -----------
0 0 N/A
1 1 +1
2 0 -1/2
3 -1 -1/3
4 0 +1/4
5 -1 -1/5
6 -2 -1/6
7 -1 +1/7
8 -2 -1/8
9 -1 +1/9
10 -2 -1/10
11 -1 +1/11
12 -2 -1/12
13 -3 -1/13
14 -2 +1/14
15 -3 -1/15
MATHEMATICA
Module[{f = 0, a = 0}, Array[If[f >= 1/#, f -= 1/#; a--, f += 1/#; a++] &, 100]] (* Paolo Xausa, Aug 25 2025 *)
PROG
(PARI) { print1(0); f = 0; a = 0; for (n = 1, 65, if (f >= 1/n, f -= 1/n; a--, f += 1/n; a++); print1 (", "a); ); }
CROSSREFS
KEYWORD
sign
AUTHOR
Rémy Sigrist, Aug 06 2025
STATUS
approved