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a(1) = 1 and a(n) = ceiling((Sum_{k=1..n-1} a(k))/3) for n >= 2.
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%I #47 Jul 22 2020 05:23:45

%S 1,1,1,1,2,2,3,4,5,7,9,12,16,22,29,39,52,69,92,123,164,218,291,388,

%T 517,690,920,1226,1635,2180,2907,3876,5168,6890,9187,12249,16332,

%U 21776,29035,38713,51618,68824,91765,122353,163138,217517,290023,386697

%N a(1) = 1 and a(n) = ceiling((Sum_{k=1..n-1} a(k))/3) for n >= 2.

%C Is this sequence, with its first 8 terms removed, the same as A005427? See also the similar conjecture with A005428/A073941. - _Ralf Stephan_, Apr 04 2003

%C Yes; the first 8 terms sum to 15, so upon dividing by 3 they are equivalent to the +5 in the formula for A005427. - _Charlie Neder_, Jan 10 2019

%C From _Petros Hadjicostas_, Jul 21 2020: (Start)

%C Conjecture 1: a(n) equals the number of multiples of 3 whose representation in base 4/3 (see A024631) has n-1 digits. For example, a(8) = 4 because there are four multiples of 3 with n-1 = 7 digits in their representation in base 4/3: 33 = 3210201, 36 = 3210230, 39 = 3210233, and 42 = 3213122.

%C Conjecture 2: a(n) equals 1/4 times the number of nonnegative integers with the property that their 4/3-expansion has n digits (assuming that the 4/3-expansion of 0 has 1 digit). For example, a(7)*4 = 12 because the following 12 numbers have 4/3 expansions with n = 7 digits: 32 = 3210200, 33 = 3210201, 34 = 3210202, ..., 42 = 3213122, 43 = 3213123. (End)

%H Michel Marcus, <a href="/A072493/b072493.txt">Table of n, a(n) for n = 1..1000</a>

%H K. Burde, <a href="http://dx.doi.org/10.1016/0022-314X(87)90078-3">Das Problem der Abzählreime und Zahlentwicklungen mit gebrochenen Basen [The problem of counting rhymes and number expansions with fractional bases]</a>, J. Number Theory 26(2) (1987), 192-209. [The author deals with the representation of n in fractional bases k/(k-1) and its relation to counting-off games (variations of Josephus problem). Here k = 4. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]

%H A. M. Odlyzko and H. S. Wilf, <a href="https://doi.org/10.1017/S0017089500008272">Functional iteration and the Josephus problem</a>, Glasgow Math. J. 33(2) (1991), 235-240.

%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>

%F a(n) = ceiling(c*(4/3)^n - 1/2) where c = 0.389324199524937508840138455...

%F From _Petros Hadjicostas_, Jul 21 2020: (Start)

%F Conjecture: The constant c above equals (3/16)*K(4), where K(q) = C(q/(q-1)) (q > 1) is described in Odlyzko and Wilf (1991).

%F For a > 1, the constant C(a) = limit_{n -> infinity} f_n(a)/a^n, where f_{n+1}(a) = ceiling(a*f_n(a)) for n >= 0 and f_0(a) = 1.

%F Thus, K(4) = limit_{n -> infinity} f_n(4/3)/(4/3)^n = 2.076395730799666... We have K(2) = 1 and K(3) = A083286 = 1.622270502884767315... (End)

%t f[s_] := Append[s, Ceiling[Plus @@ s/3]]; Nest[f, {1}, 52] (* _Robert G. Wilson v_, Jul 07 2006 *)

%o (PARI) lista(nn) = {va = vector(nn); va[1] = 1; for (n=2, nn, va[n] = ceil(sum(k=1, n-1, va[k])/3);); va;} \\ _Michel Marcus_, Apr 16 2015

%Y Cf. A005427, A005428, A024631, A073941, A083286.

%K nonn

%O 1,5

%A _Benoit Cloitre_, Nov 22 2002