A341909 a(0) = 0; for n > 0, a(n) is the smallest positive integer not yet in the sequence such that the first digit of a(n) differs by 1 from the last digit of a(n-1).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 80, 10, 11, 20, 12, 13, 21, 22, 14, 30, 15, 40, 16, 50, 17, 60, 18, 70, 19, 81, 23, 24, 31, 25, 41, 26, 51, 27, 61, 28, 71, 29, 82, 32, 33, 42, 34, 35, 43, 44, 36, 52, 37, 62, 38, 72, 39, 83, 45, 46, 53, 47, 63, 48, 73, 49, 84, 54, 55, 64, 56, 57, 65, 66, 58, 74, 59

Offset: 0

Author

Scott R. Shannon, Feb 23 2021

Keywords

• nonn
• base

Examples

			a(10) = 80 as the last digit of a(9) = 9 is 9, thus the first digit of a(10) must be 8. As 8 has already been used the next smallest number starting with 8 is 80.
a(16) = 21 as the last digit of a(15) = 13 is 3, thus the first digit of a(16) must be 2 or 4. As 2, 4 and 20 have already been used the next smallest number starting with 2 is 21.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A001477, A000027, A341692, A341002, A331163, A331375.

Cf. A131835, A217394, A217395, A217397, A217398, A217399, A217400, A217401, A217402.

Programs

Mathematica

Block[{a = {0}, k}, Do[k = 1; While[Nand[FreeQ[a, k], Abs[First@ IntegerDigits[k] - Mod[a[[-1]], 10]] == 1], k++]; AppendTo[a, k], {i, 76}]; a] (* Michael De Vlieger, Feb 23 2021 *)

Python

def nextd(strn, d):
  n = int(strn) if strn != "" else 0
  return n+1 if str(n+1)[0] == str(d) else int(str(d)+'0'*len(strn))
def aupton(term):
  alst, aset = [0], {0}
  lastdstr = ["" for d in range(10)]
  for n in range(1, term+1):
    lastdig = alst[-1]%10
    firstdigs = set([max(lastdig-1, 0), min(lastdig+1, 9)]) - {0}
    cands = [nextd(lastdstr[d], d) for d in firstdigs]
    m = min(cands)
    argmin = cands.index(m)
    alst.append(m)
    strm = str(m)
    lastdstr[int(strm[0])] = strm
  return alst
print(aupton(76)) # Michael S. Branicky, Feb 23 2021