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%I #16 Dec 14 2020 01:36:07
%S 2,13,14,15,16,17,18,91,201,202,31,203,24,25,26,27,28,92,41,204,35,36,
%T 37,38,93,42,51,205,46,47,48,94,52,61,206,57,58,95,62,71,207,68,96,72,
%U 81,208,97,82,131,301,302,132,133,141,303,142,134,53,143,144,63,145,64,73,146,74
%N The Locomotive Pushing the Wagons to the Right sequence (see Comments lines for definition).
%C a(1) is the (pushing) locomotive; a(2), a(3), a(4),... a(n),... are the wagons. To hook a wagon to its successor (on the right) you must be able to insert the rightmost digit of a(n) between the first two digits of a(n+1). In mathematical terms, the value of the rightmost digit of a(n) must be between (not equal to) the first and the second digit of a(n+1). This is the lexicographically earliest sequence of distinct positive terms with this property.
%C a(n) cannot end in 0 or 9. - _Michael S. Branicky_, Dec 14 2020
%H Carole Dubois, <a href="/A335972/b335972.txt">Table of n, a(n) for n = 1..5001</a>
%e The sequence starts with 2, 13, 14, 15, 16, 17, 18, 91, 201, 202,...
%e a(1) = 2 as there is no earliest possible (pushing to the right) locomotive;
%e a(2) = 13 starts with 1 and 3: now 1 < 2 < 3 [2 being the rightmost digit of a(1)];
%e a(3) = 14 starts with 1 and 4: now 1 < 3 < 4 [3 being the rightmost digit of a(2)];
%e a(4) = 15 starts with 1 and 5: now 1 < 4 < 5 [4 being the rightmost digit of a(3)];
%e (...)
%e a(8) = 91 starts with 9 and 1: now 9 > 8 > 1 [8 being the rightmost digit of a(7)];
%e a(9) = 201 starts with 2 and 0: now 2 > 1 > 0 [1 being the rightmost digit of a(8)];
%e a(10) = 202 starts with 2 and 0: now 2 > 1 > 0 [1 being the rightmost digit of a(9)]; etc.
%o (Python)
%o def dead_end(k): return k%10 in {0, 9}
%o def aupto(n, seed=2):
%o train, used = [seed], {seed}
%o for n in range(2, n+1):
%o caboose = train[-1]
%o hook = caboose % 10
%o low2, high2 = 10 + (hook + 1), 90 + (hook - 1)
%o an, pow10b = low2, 1
%o while an in used or dead_end(an): an += 1
%o first2 = an//pow10b
%o a2, b2 = divmod(first2, 10)
%o while True:
%o if a2 < hook < b2 or a2 > hook > b2:
%o train.append(an)
%o used.add(an)
%o break
%o if first2 > high2:
%o pow10b *= 10
%o an = low2*pow10b
%o else:
%o an += pow10b
%o an -= an%pow10b
%o while an in used or dead_end(an): an += 1
%o first2 = an//pow10b
%o a2, b2 = divmod(first2, 10)
%o return train # use train[n-1] for a(n)
%o print(aupto(66)) # _Michael S. Branicky_, Dec 14 2020
%Y Cf. A335971 (locomotive pulling to the left) and A335973 (two locomotives).
%K base,nonn
%O 1,1
%A _Eric Angelini_ and _Carole Dubois_, Jul 03 2020
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