

A335972


The Locomotive Pushing the Wagons to the Right sequence (see Comments lines for definition).


4



2, 13, 14, 15, 16, 17, 18, 91, 201, 202, 31, 203, 24, 25, 26, 27, 28, 92, 41, 204, 35, 36, 37, 38, 93, 42, 51, 205, 46, 47, 48, 94, 52, 61, 206, 57, 58, 95, 62, 71, 207, 68, 96, 72, 81, 208, 97, 82, 131, 301, 302, 132, 133, 141, 303, 142, 134, 53, 143, 144, 63, 145, 64, 73, 146, 74
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OFFSET

1,1


COMMENTS

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.
a(n) cannot end in 0 or 9.  Michael S. Branicky, Dec 14 2020


LINKS

Carole Dubois, Table of n, a(n) for n = 1..5001


EXAMPLE

The sequence starts with 2, 13, 14, 15, 16, 17, 18, 91, 201, 202,...
a(1) = 2 as there is no earliest possible (pushing to the right) locomotive;
a(2) = 13 starts with 1 and 3: now 1 < 2 < 3 [2 being the rightmost digit of a(1)];
a(3) = 14 starts with 1 and 4: now 1 < 3 < 4 [3 being the rightmost digit of a(2)];
a(4) = 15 starts with 1 and 5: now 1 < 4 < 5 [4 being the rightmost digit of a(3)];
(...)
a(8) = 91 starts with 9 and 1: now 9 > 8 > 1 [8 being the rightmost digit of a(7)];
a(9) = 201 starts with 2 and 0: now 2 > 1 > 0 [1 being the rightmost digit of a(8)];
a(10) = 202 starts with 2 and 0: now 2 > 1 > 0 [1 being the rightmost digit of a(9)]; etc.


PROG

(Python)
def dead_end(k): return k%10 in {0, 9}
def aupto(n, seed=2):
train, used = [seed], {seed}
for n in range(2, n+1):
caboose = train[1]
hook = caboose % 10
low2, high2 = 10 + (hook + 1), 90 + (hook  1)
an, pow10b = low2, 1
while an in used or dead_end(an): an += 1
first2 = an//pow10b
a2, b2 = divmod(first2, 10)
while True:
if a2 < hook < b2 or a2 > hook > b2:
train.append(an)
used.add(an)
break
if first2 > high2:
pow10b *= 10
an = low2*pow10b
else:
an += pow10b
an = an%pow10b
while an in used or dead_end(an): an += 1
first2 = an//pow10b
a2, b2 = divmod(first2, 10)
return train # use train[n1] for a(n)
print(aupto(66)) # Michael S. Branicky, Dec 14 2020


CROSSREFS

Cf. A335971 (locomotive pulling to the left) and A335973 (two locomotives).
Sequence in context: A065584 A153651 A229908 * A320339 A075032 A032932
Adjacent sequences: A335969 A335970 A335971 * A335973 A335974 A335975


KEYWORD

base,nonn


AUTHOR

Eric Angelini and Carole Dubois, Jul 03 2020


STATUS

approved



