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A363794
a(n) = smallest prime(n)-smooth number k such that r(k) >= r(P(n+1)), where r(n) = A010846(n) and P(n) = A002110(n).
0
16, 72, 540, 6300, 92400, 1681680, 36756720, 921470550, 27886608750, 970453984500, 37905932634570
OFFSET
1,1
COMMENTS
Let R = r(P(n)) = A010846(A002110(n)) = A363061(n).
Let S(n) be the sorted tensor product of prime power ranges {p(i)^e : i<=n, e>=0}, e.g., S(1) = A000079, S(2) = A003586, S(3) = A051037, etc.
Let T(n) = A002110(n)*S(n). Note that S(1) = T(1) since omega(A002110(1)) = 1.
Let S(n,i) be the i-th term in S(n).
Then a(n) is the smallest S(n,i), i >= R, such that S(n,i) is also in T. Equivalently, a(n) is the smallest S(n,i), i >= R, such that rad(S(n,i)) = A002110(n), where rad(n) = A007947(n).
FORMULA
a(n) >= A363061(n).
EXAMPLE
a(1) = 16 since r(2^4) = 5 and r(6) = 5; numbers in row 16 of A162306 are its divisors {1, 2, 4, 8, 16}, while row 6 of A162306 is {1, 2, 3, 4, 6}.
a(2) = 72 = A003586(18) since r(72) = r(30) = 18. 72 is the 8th term in A003586 that is not in A000961.
a(3) = 540 since r(540) = 69 which exceeds r(210) = 68.
a(4) = 6300 since r(6300) = 290 which exceeds r(2310) = 283, etc.
Table showing the relationship of a(n) to r(P(n)) = A363061(n), with p(n) = prime(n), P(n+1) = A002110(n+1), r(a(n)) = A010846(a(n)), and j the index such that S(r(a(n))) = T(j) = a(n). a(n) = m*P(n).
n p(n) P(n+1) a(n) r(P(n)) r(a(n)) j m
--------------------------------------------------------------
1 2 6 16 5 5 4 8
2 3 30 72 18 18 8 12
3 5 210 540 68 69 13 18
4 7 2310 6300 283 290 22 30
5 11 30030 92400 1161 1165 29 40
6 13 510510 1681680 4843 4848 42 56
7 17 9699690 36756720 19985 19994 53 72
8 19 223092870 921470550 83074 83435 68 95
9 23 6469693230 27886608750 349670 351047 89 125
10 29 200560490130 970453984500 1456458 1457926 107 150
MATHEMATICA
nn = 6; rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; f[x_] := FactorInteger[x][[-1, 1]]; S = Array[Product[Prime[i], {i, #}] &, nn + 1]; Table[Set[{p, q}, Prime[n + {0, 1}]]; r = Count[Range[S[[n + 1]]], _?(f[#] <= q &)]; c = k = 1; While[Or[c < r, rad[k] != S[[n]]], If[f[k] <= p, c++]; k++]; k, {n, nn}]
KEYWORD
nonn,hard,more
AUTHOR
Michael De Vlieger, Jun 22 2023
STATUS
approved