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A049803
a(n) = n mod 3 + n mod 9 + ... + n mod 3^k, where 3^k <= n < 3^(k+1).
5
0, 0, 0, 1, 2, 0, 1, 2, 0, 2, 4, 3, 5, 7, 6, 8, 10, 0, 2, 4, 3, 5, 7, 6, 8, 10, 0, 3, 6, 6, 9, 12, 12, 15, 18, 9, 12, 15, 15, 18, 21, 21, 24, 27, 18, 21, 24, 24, 27, 30, 30, 33, 36, 0, 3, 6, 6, 9, 12, 12, 15, 18, 9, 12, 15, 15, 18, 21, 21, 24, 27, 18
OFFSET
1,5
COMMENTS
From Petros Hadjicostas, Dec 11 2019: (Start)
Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)
Here b = 3 and a(n) = s(n,3).
We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).
If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k). (End)
LINKS
B. Dearden, J. Iiams, and J. Metzger, A Function Related to the Rumor Sequence Conjecture, J. Int. Seq. 14 (2011), #11.2.3, Example 7.
FORMULA
From Petros Hadjicostas, Dec 11 2019: (Start)
Conjecture: a(3*n+r) = 3*a(n) + r*A081604(n) = 3*a(n) + r*(floor(log_3(n)) + 1) for n >= 1 and r = 0, 1, 2.
If the conjecture above is true, the g.f. A(x) satisfies A(x) = 3*(1+x+x^2)*A(x^3) + x*(2*x+1)/(1-x^3) * Sum_{k >= 1} x^(3^k). (End)
MAPLE
a:= n-> add(irem(n, 3^j), j=1..ilog[3](n)):
seq(a(n), n=1..105); # Alois P. Heinz, Dec 13 2019
MATHEMATICA
Table[n * Floor@Log[3, n] - Sum[Floor[n*3^-k]*3^k, {k, Log[3, n]}], {n, 100}] (* after Federico Provvedi in A049802*) (* Metin Sariyar, Dec 12 2019 *)
PROG
(PARI) a(n) = sum(k=1, logint(n, 3), n % 3^k); \\ Michel Marcus, Dec 12 2019
KEYWORD
nonn
STATUS
approved