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%I #9 Mar 26 2022 14:47:59
%S 1,0,0,1,3,0,3,2,0,0,1,0,2,0,0,1,5,0,3,3,0,0,1,0,2,0,2,3,8,0,4,0,0,0,
%T 9,0,2,0,0,6,0,0,1,1,0,0,1,0,2,0,0,2,3,0,3,6,0,0,1,0,2,0,0,1,6,0,3,5,
%U 0,0,1,0,2,0,0,3,3,0,3,3,1,0,4,0,15,0,0
%N Multiplicative, with a(p^k) determined using the same rule as Van Eck's sequence for any prime p and k > 0.
%C If a(m) = a(p^k-1) for some m < p^k-1, take the largest such m and set a(p^k) = p^k-1-m; otherwise a(p^k) = 0.
%e a(1) = 1 (as this sequence is multiplicative).
%e a(2) = 0 as there is no m < 1 such that a(m) = a(1) (by Van Eck's rule).
%e a(3) = 0 as there is no m < 2 such that a(m) = a(2) (by Van Eck's rule).
%e a(4) = 3-2 = 1 as a(2) = a(3) (by Van Eck's rule).
%e a(5) = 4-1 = 3 as a(1) = a(4) (by Van Eck's rule).
%e a(6) = a(2)*a(3) = 0 (as this sequence is multiplicative).
%o (PARI) { for (n=1, #a=vector(87), f=factor(n); if (#f~==1, forstep (m=n-2, 1, -1, if (a[n-1]==a[m], a[n]=n-1-m; break)), a[n]=prod (k=1, #f~, a[f[k,1]^f[k,2]])); print1 (a[n]", ")) }
%Y Cf. A181391, A350228.
%K nonn,mult
%O 1,5
%A _Rémy Sigrist_, Mar 23 2022
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