A329349 Number of occurrences of the largest primorial present in the greedy sum of primorials adding to A108951(n).
1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 4, 1, 2, 6, 2, 1, 2, 1, 4, 6, 2, 1, 1, 4, 2, 1, 4, 1, 1, 1, 1, 6, 2, 2, 4, 1, 2, 6, 1, 1, 1, 1, 4, 5, 2, 1, 3, 1, 8, 6, 4, 1, 2, 2, 8, 6, 2, 1, 3, 1, 2, 3, 2, 1, 12, 1, 4, 6, 5, 1, 1, 1, 2, 2, 4, 16, 12, 1, 2, 6, 2, 1, 2, 1, 2, 6, 8, 1, 10, 12, 4, 6, 2, 1, 6, 1, 2, 2, 1, 1, 12, 1, 8, 1
Offset: 1
Keywords
Examples
For n = 21 = 3 * 7, A108951(21) = A034386(3) * A034386(7) = 6 * 210, so the factor of the largest primorial present (210) in the greedy sum is 6 (as 1260 = 210 + 210 + 210 + 210 + 210 + 210), thus a(21) = 6. For n = 24 = 2^3 * 3, A108951(24) = A034386(2)^3 * A034386(3) = 2^3 * 6 = 48 = 1*30 + 3*6, and as the factor of the largest primorial in the sum is 1, we have a(24) = 1.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
- Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
- Index entries for sequences computed from indices in prime factorization
- Index entries for sequences related to primorial base
- Index entries for sequences related to primorial numbers
Crossrefs
Programs
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PARI
A034386(n) = prod(i=1, primepi(n), prime(i)); A108951(n) = { my(f=factor(n)); prod(i=1, #f~, A034386(f[i, 1])^f[i, 2]) }; \\ From A108951 A276153(n) = { my(e=0, p=2); while(n, e=n%p; n = n\p; p = nextprime(1+p)); (e); }; A329349(n) = A276153(A108951(n));
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PARI
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); }; A324886(n) = A276086(A108951(n)); A071178(n) = if(1==n,0,my(es=factor(n)[,2]); es[#es]); A329349(n) = A071178(A324886(n));
Comments