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Pontus von Brömssen, <a href="/A343334/b343334.txt">Table of n, a(n) for n = 1..10000</a>
nonn,look,new
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reviewed
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(Python)
def A343334_list(n_max):
a=0
a_list=[0]
count=[]
for i in range(n_max-1):
if a==len(count): count.append(0)
else: count[a]+=1
a=abs(a-count[a])
a_list.append(a)
return a_list
Every number appears, and their first occurrences are in increasing order.
allocated for Pontus von Brömssena(1) = 0; thereafter a(n+1) = abs(a(n)-y), where y is the number of numbers m < n such that a(m) = a(n).
0, 0, 1, 1, 0, 2, 2, 1, 1, 2, 0, 3, 3, 2, 1, 3, 1, 4, 4, 3, 0, 4, 2, 2, 3, 1, 5, 5, 4, 1, 6, 6, 5, 3, 2, 4, 0, 5, 2, 5, 1, 7, 7, 6, 4, 1, 8, 8, 7, 5, 0, 6, 3, 3, 4, 2, 6, 2, 7, 4, 3, 5, 1, 9, 9, 8, 6, 1, 10, 10, 9, 7, 3, 6, 0, 7, 2, 8, 5, 2, 9, 6, 1, 11, 11
1,6
Variant of A340488, with XOR(a,y) replaced by abs(a-y).
Let N_k(n) denote the number of occurrences of k among the first n terms. It appears that N_0(n) ~ sqrt(n/2) and N_k(n) ~ sqrt(2*n) for k > 0.
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nonn
Pontus von Brömssen, Apr 12 2021
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editing
allocated for Pontus von Brömssen
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