A329344 Number of times most frequent primorial is 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, 3, 4, 2, 1, 4, 1, 5, 1, 1, 6, 2, 8, 4, 1, 2, 6, 1, 1, 1, 1, 4, 5, 2, 1, 3, 6, 8, 6, 4, 1, 2, 4, 8, 6, 2, 1, 3, 1, 2, 3, 2, 13, 12, 1, 4, 6, 5, 1, 3, 1, 2, 5, 4, 16, 12, 1, 2, 6, 2, 1, 2, 11, 2, 6, 8, 1, 10, 12, 4, 6, 2, 7, 6, 1, 12, 10, 6, 1, 12, 1, 8, 4
Offset: 1
Keywords
Examples
For n = 24 = 2^3 * 3, A108951(24) = A034386(2)^3 * A034386(3) = 2^3 * 6 = 48 = 30 + 6 + 6 + 6, and as the most frequent primorial in the sum is 6 = A002110(2), we have a(24) = 3.
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Crossrefs
Programs
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Mathematica
With[{b = Reverse@ Prime@ Range@ 120}, Array[Max@ IntegerDigits[#, MixedRadix[b]] &@ Apply[Times, Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]] &, 105] ] (* Michael De Vlieger, Nov 18 2019 *)
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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 A328114(n) = { my(s=0, p=2); while(n, s = max(s,(n%p)); n = n\p; p = nextprime(1+p)); (s); }; A329344(n) = A328114(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)); A051903(n) = if((1==n),0,vecmax(factor(n)[, 2])); A329344(n) = A051903(A324886(n));
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