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A073941 a(n) = ceiling((Sum_{k=1..n-1} a(k)) / 2) for n >= 2 starting with a(1) = 1.

%I A073941 #102 Sep 21 2022 01:42:26
%S A073941 1,1,1,2,3,4,6,9,14,21,31,47,70,105,158,237,355,533,799,1199,1798,
%T A073941 2697,4046,6069,9103,13655,20482,30723,46085,69127,103691,155536,
%U A073941 233304,349956,524934,787401,1181102,1771653,2657479,3986219,5979328,8968992
%N A073941 a(n) = ceiling((Sum_{k=1..n-1} a(k)) / 2) for n >= 2 starting with a(1) = 1.
%C A073941 a(n) is the number of even integers that have n-1 digits when written in base 3/2. For example, there are 2 even integers that use three digits in base 3/2: 6 and 8: they are written as 210 and 212, respectively. - _Tanya Khovanova_ and PRIMES STEP Senior group, Jun 03 2018
%C A073941 From _Petros Hadjicostas_, Jul 20 2020: (Start)
%C A073941 We describe Schuh's counting-off game (pp. 373-375 and 377-379). Assume m people are standing on a circle and they are labeled 1 through m (say clockwise). We start with the person labeled 1 and every 3rd person drops out (in a variation of the famous Josephus problem). The process is repeated until only one person is left.
%C A073941 This sequence describes those numbers m for which either the person labeled 1 or the person labeled 2 is the last survivor.
%C A073941 From a(4) = 2 to a(53) = 775795914 (see _T. D. Noe_'s b-file), the values agree with those in Schuh (1968, p. 374) and Burde (1987, p. 207). a(54) = 1163693871 while both Schuh and Burde have 1063693871 (a difference in the 2nd digit starting on the left). a(55) = 1745540806 while both Schuch and Burde have 1595540806.
%C A073941 Schuh (1968) obtains the numbers in the following way. Suppose we know a(n) and the corresponding number i(n) of the last survivor (i(n) = 1 or 2). We multiply a(n) by 3/2 (cf. Burde's use of fractional bases).
%C A073941 If the product is an integer, that is a(n+1) and the corresponding last survivor is the same.
%C A073941 If the product is not an integer, then a(n+1) = floor(a(n)*3/2) if the last survivor i(n) = 2 (and the new last survivor is i(n+1) = 1), and a(n+1) = ceiling(a(n)*3/2) if the last survivor is i(n) = 1 (and the new last survivor is i(n+1) = 2).
%C A073941 Note that a(53) = 775795914 and a(54) = (3/2)*a(53) = 1163693871 (not 1063693871), so it seems Schuh did a mistake and Burde copied it. Also (3/2)*1163693871 = 1745540806.5. Since a(53) = 775795914 corresponds to number 2, we round down, i.e., a(54) = 1745540806 (and move to number 1). If, however, we multiply the incorrect 1063693871 by 3/2 and round down, we get Schuh and Burde's incorrect value 1595540806 for a(54).
%C A073941 Numbers a(n) that correspond to last survivors being number 1 are tabulated in A081614 while numbers a(n) that correspond to last survivors being number 2 are tabulated in A081615. (End)
%C A073941 a(n) is the number of times (n-1) appears in A061420. - _Chinmaya Dash_, Aug 19 2020
%D A073941 Fred Schuh, The Master Book of Mathematical Recreations, Dover, New York, 1968. [See Table 18, p. 374. Only the terms from a(6) = 4 forward are shown in the table. The table is definitely related to this sequence.]
%H A073941 T. D. Noe, <a href="/A073941/b073941.txt">Table of n, a(n) for n = 1..500</a>
%H A073941 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. Here k = 3. See the table on p. 207. See also the review in MathSciNet (MR0889384) by R. G. Stoneham.]
%H A073941 B. Chen, R. Chen, J. Guo, S. Lee et al, <a href="http://arxiv.org/abs/1808.04304">On Base 3/2 and its sequences</a>, arXiv:1808.04304 [math.NT], 2018.
%H A073941 Tom Edgar, Hailey Olafson, and James Van Alstine, <a href="https://www.emis.de/journals/INTEGERS/papers/q63/q63.Abstract.html">Approximating the Fibonacci Sequence</a>, Integers 16 (2016), #A63.
%F A073941 a(n) = ceiling(c*(3/2)^n-1/2) where c = 0.3605045561966149591015446628665... - _Benoit Cloitre_, Nov 22 2002
%F A073941 If 2^m divides a(i) then 2^(m-1)*3^1 divides a(i+1) and so on... until finally, 3^m divides a(i+m). - _Ralf Stephan_, Apr 20 2003
%F A073941 a(n) = A081848(n)/3. - _Tom Edgar_, Jul 21 2014
%F A073941 a(n) = A005428(n-2). - _Tanya Khovanova_ and PRIMES STEP Senior group, Jun 03 2018
%t A073941 f[s_] := Append[s, Ceiling[Plus @@ s/2]]; Nest[f, {1}, 41] (* _Robert G. Wilson v_, Jul 07 2006 *)
%o A073941 (PARI) v=vector(100);s=v[1]=1;for(i=2,#v,s+=(v[i]=(s+1)\2));v \\ _Charles R Greathouse IV_, Feb 11 2011
%o A073941 (Haskell)
%o A073941 a073941 n = a073941_list !! (n-1)
%o A073941 a073941_list = 1 : f [1] where
%o A073941    f xs = x' : f (x':xs) where x' = (1 + sum xs) `div` 2
%o A073941 -- _Reinhard Zumkeller_, Oct 26 2011
%o A073941 (Python)
%o A073941 from itertools import islice
%o A073941 def A073941_gen(): # generator of terms
%o A073941     a, c = 1, 0
%o A073941     yield 1
%o A073941     while True:
%o A073941         yield (a:=(c:=c+a)+1>>1)
%o A073941 A073941_list = list(islice(A073941_gen(),70)) # _Chai Wah Wu_, Sep 20 2022
%Y A073941 Same as log_2(A082125(n)), for n > 2. - _Ralf Stephan_, Apr 16 2002
%Y A073941 Apart from initial term, same as A005428, which has further information.
%Y A073941 a(n+4) = A079719(n)+2. Cf. A082416.
%Y A073941 Partial sums for various start indices are in A006999, A061419, A061418. - _Ralf Stephan_, Apr 17 2003
%Y A073941 Is this the same as A081848/3?
%Y A073941 The constant c is (2/9)*K(3) (see A083286). - _Ralf Stephan_, May 29 2003
%Y A073941 Cf. A081614, A081615.
%K A073941 nonn,nice
%O A073941 1,4
%A A073941 _Reinhard Zumkeller_, Nov 20 2002