A352613 Multiplicative, with a(p^k) determined using the same rule as Van Eck's sequence for any prime p and k > 0.

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, 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, 0, 0, 1, 0, 2, 0, 0, 3, 3, 0, 3, 3, 1, 0, 4, 0, 15, 0, 0

Offset

1,5

Comments

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.

Links

Table of n, a(n) for n=1..87.

Example

a(1) = 1 (as this sequence is multiplicative).
a(2) = 0 as there is no m < 1 such that a(m) = a(1) (by Van Eck's rule).
a(3) = 0 as there is no m < 2 such that a(m) = a(2) (by Van Eck's rule).
a(4) = 3-2 = 1 as a(2) = a(3) (by Van Eck's rule).
a(5) = 4-1 = 3 as a(1) = a(4) (by Van Eck's rule).
a(6) = a(2)*a(3) = 0 (as this sequence is multiplicative).

Prog

(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]", ")) }

Crossrefs

Cf. A181391, A350228.

Sequence in context: A374139 A298082 A085919 * A105824 A171911 A180193

Adjacent sequences: A352610 A352611 A352612 * A352614 A352615 A352616

Keyword

nonn,mult

Author

Rémy Sigrist, Mar 23 2022

Status

approved