OFFSET
1,4
COMMENTS
LINKS
FORMULA
a(n) is, for n >= 2, the sum of all entries in the base 3 representation of the exponents of the primes in the usual prime number factorization of n.
From Antti Karttunen, Aug 12 2017: (Start)
That is, apart from the initial term, additive with a(p^e) = A053735(e).
Define b(1) = 0; and for n > 1, b(n) = A053735(A067029(n)) + b(A028234(n)). Then a(n) = b(n) for n > 1, with a(1) = 1 by convention.
(End)
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 0.38090372984518844518..., where f(x) = -x + Sum_{k>=0} (x^(3^k) + 2*x^(2*3^k))/(1 + x^(3^k) + x^(2*3^k)). - Amiram Eldar, Sep 28 2023
EXAMPLE
a(12) = 3 because the usual prime factorization is 12 = 2^2*3^1 and (2)_3 = [2] and (1)_3 = [1], hence the sum of the base-3 representations of the exponents is 3.
a(24) = 2 as 24 = 3*8, using two factors from A186285. Note also how 3*8 = 3^1 * 2^3, and ternary representations of 1 and 3 are "1" and "10", thus their digit sum is 2. - Antti Karttunen, Aug 12 2017
a(36) = 4 from 2^2*3^2, (2)_3 = [2] and 2 + 2 = 4.
MATHEMATICA
Block[{nn = 105, s}, s = Select[Select[Range@ nn, PrimePowerQ], IntegerQ@ Log[3, FactorInteger[#][[1, -1]]] &]; {1}~Join~Table[Length@ Rest@ NestWhileList[Function[{k, m}, {k/#, #} &@ SelectFirst[Reverse@ TakeWhile[s, # <= k &], Divisible[k, #] &]] @@ # &, {n, 1}, First@ # > 1 &][[All, -1]], {n, 2, nn}]] (* Michael De Vlieger, Aug 14 2017 *)
a[n_] := Total[Plus @@ IntegerDigits[#, 3] & /@ (FactorInteger[n][[;; , 2]])]; Array[a, 100] (* Amiram Eldar, May 18 2023 *)
PROG
(Scheme)
(define (A240231 n) (if (= 1 n) n (A240231with_a1_0 n)))
(definec (A240231with_a1_0 n) (if (= 1 n) 0 (+ (A053735 (A067029 n)) (A240231with_a1_0 (A028234 n)))))
;; Antti Karttunen, Aug 12 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, May 15 2014
EXTENSIONS
Description clarified and more terms added by Antti Karttunen, Aug 12 2017
STATUS
approved