

A073941


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


101



1, 1, 1, 2, 3, 4, 6, 9, 14, 21, 31, 47, 70, 105, 158, 237, 355, 533, 799, 1199, 1798, 2697, 4046, 6069, 9103, 13655, 20482, 30723, 46085, 69127, 103691, 155536, 233304, 349956, 524934, 787401, 1181102, 1771653, 2657479, 3986219, 5979328, 8968992
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OFFSET

1,4


COMMENTS

a(n) is the number of even integers that have n1 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
We describe Schuh's countingoff game (pp. 373375 and 377379). 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.
This sequence describes those numbers m for which either the person labeled 1 or the person labeled 2 is the last survivor.
From a(4) = 2 to a(53) = 775795914 (see T. D. Noe's bfile), 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.
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).
If the product is an integer, that is a(n+1) and the corresponding last survivor is the same.
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).
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).
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)


REFERENCES

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.]


LINKS



FORMULA

a(n) = ceiling(c*(3/2)^n1/2) where c = 0.3605045561966149591015446628665...  Benoit Cloitre, Nov 22 2002
If 2^m divides a(i) then 2^(m1)*3^1 divides a(i+1) and so on... until finally, 3^m divides a(i+m).  Ralf Stephan, Apr 20 2003


MATHEMATICA

f[s_] := Append[s, Ceiling[Plus @@ s/2]]; Nest[f, {1}, 41] (* Robert G. Wilson v, Jul 07 2006 *)


PROG

(Haskell)
a073941 n = a073941_list !! (n1)
a073941_list = 1 : f [1] where
f xs = x' : f (x':xs) where x' = (1 + sum xs) `div` 2
(Python)
from itertools import islice
def A073941_gen(): # generator of terms
a, c = 1, 0
yield 1
while True:
yield (a:=(c:=c+a)+1>>1)


CROSSREFS

Apart from initial term, same as A005428, which has further information.


KEYWORD

nonn,nice


AUTHOR



STATUS

approved



