OFFSET
18,1
COMMENTS
Conjecture: n = 30 is the largest number for which a(n) = -1.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 18..280
Michael De Vlieger, Plot of terms k = p^a*q^b*r^c, primes p < q < r, in row a(n) of A162306, with n = 18..58 and a(n) > -1, at (x,y,z) = (a,b,c). For a(n) there are n blocks in each diagram.
EXAMPLE
Let f = A010846.
For k >= 3 and m = 6*prime(k), f(k) = [k=3] + f(6) + Sum_{j=0..floor(log_3 m)} floor(log_2 m/3^j), with Iverson brackets. (This, since for k = 3, prime(k)^2 < 2*3*prime(k) < prime(k)^3, but for k > 3, 2*3*prime(k) < prime(k)^2.)
a(18) = 3 since 2*3*prime(3) = 30 is the smallest sphenic number k, and f(30) = A363061(3) = 18. This is to say that row 30 of A162306 = {m : rad(m) | 30} = {1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30} has cardinality 18.
a(19) = 4 since 2*3*prime(4) = 42 and f(42) = 19. This is to say that row 30 of A162306 = {m : rad(m) | 42} = {1,2,3,4,6,7,8,9,12,14,16,18,21,24,27,28,32,36,42} has cardinality 19.
a(m) = -1, m = 20..21, since for k >= 3, there exists no solution to f(6*prime(k)) = m.
a(22) = 5 since 2*3*prime(5) = 66 and f(66) = 22.
a(23) = 6 since 2*3*prime(6) = 78 and f(78) = 23.
a(24) = = -1 since for k >= 3, there exists no solution to f(6*prime(k)) = 24.
a(25) = 7 since 2*3*prime(7) = 102 and f(102) = 25.
a(26) = 8 since 2*3*prime(8) = 114 and f(114) = 26.
a(27) = 9 since 2*3*prime(9) = 138 and f(138) = 27.
a(28) = = -1 since for k >= 3, there exists no solution to f(6*prime(k)) = 28.
a(29) = 10 since 2*3*prime(10) = 174 and f(174) = 29.
a(30) = = -1 since for k >= 3, there exists no solution to f(6*prime(k)) = 30.
Note: 2*3*prime(11) = 186; f(186) = f(174) = 29.
a(31) = 12 since 2*3*prime(12) = 222 and f(222) = 31, etc.
MATHEMATICA
k = 3; Table[While[Set[t, Boole[k == 3] + 5 + Sum[Floor[1 + Log2[#/3^j]], {j, 0, Floor[Log[3, #]]}] &[6*Prime[k] ] ] < n, k++]; If[t == n, k, -1], {n, 18, 30}]
CROSSREFS
KEYWORD
sign
AUTHOR
Michael De Vlieger, Jul 08 2025
STATUS
approved