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A101278
Write n in base 3 as n = b_0 + b_1*3 + b_2*3^2 + b_3*3^3 + ...; then a(n) = Product_{i >= 0} prime(i+1)^b_i.
8
1, 2, 4, 3, 6, 12, 9, 18, 36, 5, 10, 20, 15, 30, 60, 45, 90, 180, 25, 50, 100, 75, 150, 300, 225, 450, 900, 7, 14, 28, 21, 42, 84, 63, 126, 252, 35, 70, 140, 105, 210, 420, 315, 630, 1260, 175, 350, 700, 525, 1050, 2100, 1575, 3150, 6300, 49, 98, 196, 147, 294, 588
OFFSET
0,2
COMMENTS
A permutation of the cubefree numbers (A004709). - Rémy Sigrist, Jul 18 2022
These are cubefree numbers organized by the highest factor. By converting to a different base, we avoid the row-by-row triangular entry used in the analogous squarefree A339195. - Gordon Hamilton, Aug 13 2025
LINKS
FORMULA
If a(bn)=x then a(bn+1)=2x, a(bn+2)=4x, ... a(bn+b-1)=2^b*x. - Robert G. Wilson v, Dec 24 2004
G.f.: (1+2x+4x^2)(1+3x^3+9x^6)(1+5x^9+25x^18)... - Paul Boddington, Jul 21 2005
a(n) = f(n, 1, 1) with f(x, y, z) = if x > 0 then f(floor(x/3), y*prime(z)^(x mod 3), z+1) else y. - Reinhard Zumkeller, Mar 13 2010
EXAMPLE
The first few terms are computed as follows:
n b2 b1 b0 a(n)
0, 0, 0, 0, 1
1, 0, 0, 1, 2
2, 0, 0, 2, 4
3, 0, 1, 0, 3
4, 0, 1, 1, 6
5, 0, 1, 2, 12
a(11) = a(102_3) and so we get prime(3)^1 * prime(2)^0 * prime(1)^2 = 5^1 * 3^0 * 2^2 = 5 * 1 * 4 = 20. - Gordon Hamilton, Aug 13 2025
MATHEMATICA
primeBase[n_Integer?Positive, base_Integer]/; base>1 := Times @@ (Table[Prime[i], {i, Floor[Log[base, n] + 1], 1, -1}]^IntegerDigits[n, base]); Table[primeBase[n, 3], {n, 59}] (* Robert G. Wilson v, Dec 24 2004 *)
PROG
(PARI) a(n) = {my(d = digits(n, 3), pr = primes(#d)); prod(i = 1, #d, pr[#d + 1 - i]^d[i])} \\ David A. Corneth, Aug 13 2025
CROSSREFS
Cf. A019565 (base 2), A101942 (base 4), A101943 (base 5), A054842 (base 10).
Sequence in context: A222600 A046793 A182940 * A329902 A366028 A091274
KEYWORD
nonn,easy,base
AUTHOR
Orges Leka, Dec 20 2004
EXTENSIONS
More terms from Robert G. Wilson v, Dec 24 2004
STATUS
approved