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A126933 Quotients arising from sequence A053312.

%I A126933 #29 Mar 17 2023 02:43:15
%S A126933 1,3,14,132,691,1908,16579,47352,414301,1183713,5474669,27151397,
%T A126933 135646011,678174568,6442602909,18480090517,85533990571,424236721848,
%U A126933 4026815626549,11550150977337,53458791308981,265147974756053,1324666882885839,6622797918981982,62916043734881616,329481245744393933
%N A126933 Quotients arising from sequence A053312.
%C A126933 Take the decimal number formed by the first n digits of A023396 in reverse order and divide by 2^n.
%C A126933 The sequence A053312 gives n-digit numbers consisting entirely of 1s and 2s which are divisible by 2^n. The quotients upon division form the present sequence. The parity of the n-th term here determines the next term in A023396; if odd, it is a 1 and if even, a 2.
%C A126933 This was set as a problem in the All Union Mathematical Olympiad of 1971 and can be found in the reference cited here.
%D A126933 J. B. Tabov and P. J. Taylor, Methods of Problem Solving, Book 1, Australian Mathematics Trust, 1996.
%H A126933 David A. Corneth, <a href="/A126933/b126933.txt">Table of n, a(n) for n = 1..1431</a> (terms <= 10^1000)
%H A126933 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A126933 a(n) < 0.3 * 5^n. - _David A. Corneth_, Jun 11 2020
%e A126933 a(4) = A053312(4) / 2^4 = 2112 / 16 = 132. - _David A. Corneth_, Jun 11 2020
%o A126933 (Python)
%o A126933 from itertools import count, islice
%o A126933 def A126933_gen(): # generator of terms
%o A126933     a, b = 2, 10
%o A126933     for n in count(1):
%o A126933         a+=b if (c:=a>>n)&1 else b<<1
%o A126933         b *= 10
%o A126933         yield c
%o A126933 A126933_list = list(islice(A126933_gen(),20)) # _Chai Wah Wu_, Mar 16 2023
%Y A126933 Cf. A023396, A053312.
%K A126933 nonn,easy,base
%O A126933 1,2
%A A126933 _Gerry Leversha_, Mar 18 2007
%E A126933 Name changed and other minor edits by _Ray Chandler_, Jun 17 2020