A335972 The Locomotive Pushing the Wagons to the Right sequence (see Comments lines for definition).

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

Offset: 1

Author

Eric Angelini and Carole Dubois, Jul 03 2020

Keywords

• base
• nonn

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A335971 (locomotive pulling to the left) and A335973 (two locomotives).

Programs

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[n-1] for a(n)
print(aupto(66))  # Michael S. Branicky, Dec 14 2020