

A072493


a(1) = 1 and a(n) = ceiling((Sum_{k=1..n1} a(k))/3) for n >= 2.


82



1, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 22, 29, 39, 52, 69, 92, 123, 164, 218, 291, 388, 517, 690, 920, 1226, 1635, 2180, 2907, 3876, 5168, 6890, 9187, 12249, 16332, 21776, 29035, 38713, 51618, 68824, 91765, 122353, 163138, 217517, 290023, 386697
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,5


COMMENTS

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
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
From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture 1: a(n) equals the number of multiples of 3 whose representation in base 4/3 (see A024631) has n1 digits. For example, a(8) = 4 because there are four multiples of 3 with n1 = 7 digits in their representation in base 4/3: 33 = 3210201, 36 = 3210230, 39 = 3210233, and 42 = 3213122.
Conjecture 2: a(n) equals 1/4 times the number of nonnegative integers with the property that their 4/3expansion has n digits (assuming that the 4/3expansion 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)


LINKS

Michel Marcus, Table of n, a(n) for n = 1..1000
K. Burde, Das Problem der Abzählreime und Zahlentwicklungen mit gebrochenen Basen [The problem of counting rhymes and number expansions with fractional bases], J. Number Theory 26(2) (1987), 192209. [The author deals with the representation of n in fractional bases k/(k1) and its relation to countingoff games (variations of Josephus problem). Here k = 4. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33(2) (1991), 235240.
Index entries for sequences related to the Josephus Problem


FORMULA

a(n) = ceiling(c*(4/3)^n  1/2) where c = 0.389324199524937508840138455...
From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture: The constant c above equals (3/16)*K(4), where K(q) = C(q/(q1)) (q > 1) is described in Odlyzko and Wilf (1991).
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.
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)


MATHEMATICA

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


PROG

(PARI) lista(nn) = {va = vector(nn); va[1] = 1; for (n=2, nn, va[n] = ceil(sum(k=1, n1, va[k])/3); ); va; } \\ Michel Marcus, Apr 16 2015


CROSSREFS

Cf. A005427, A005428, A024631, A073941, A083286.
Sequence in context: A182097 A290697 A290821 * A064324 A173090 A032277
Adjacent sequences: A072490 A072491 A072492 * A072494 A072495 A072496


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Nov 22 2002


STATUS

approved



