A258227 Concatenate the natural numbers, then partition into minimal strings so that adjacent terms have a common divisor greater than 1.

12, 3, 45, 6, 78, 9, 1011, 12, 1314, 15, 1617, 18, 192, 0, 2, 12, 2, 2, 32, 4, 2, 52, 6, 2, 72, 8, 2, 930, 3, 132, 3, 3, 3, 435, 3, 6, 3, 738, 3, 9, 4041, 42, 4, 34, 4, 4, 54, 6, 4, 74, 8, 4, 950, 5, 15, 25, 35, 45, 5, 5, 65, 75, 85, 960, 6, 16, 2, 6, 3, 6

Offset: 1

Author

Reinhard Zumkeller, May 23 2015

Keywords

• nonn
• base

Comments

00 -> 0 is not allowed, else all digits will not appear in the concatenation of terms. For example, a(198)..a(201) = 198, 19920, 0, 2 and not 198, 192, 0, 2. - Michael S. Branicky, Dec 03 2021

Examples

			.  a(n) | 12,3,45,6,78,9,1011,12,1314,15,1617,18,192,0,2,12,2,2,32,4,2,52
--------+----------------------------------------------------------------
.  gcd  |   3 3  3 6  3 3    3  6    3  3    3  6 192 2 2  2 2 2  4 2 2  .
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (24 terms corrected in terms 1..10000 from Reinhard Zumkeller)

Crossrefs

Cf. A002782, A007376.

Programs

Haskell

a258227 n = a258227_list !! (n-1)
a258227_list = f 12 1 (map toInteger $ tail a007376_list) where
   f x y (d:ds) | gcd x y > 1 = y : f y d ds
                | otherwise   = f x (10 * y + d) ds

Python

from math import gcd
from itertools import count
def diggen():
    for k in count(1): yield from list(map(int, str(k)))
def aupton(terms):
    g = diggen()
    alst, aset = [12], {12}
    , , nxtd, nxtnxtd = next(g), next(g), next(g), next(g)
    for n in range(2, terms+1):
        an, nxtd, nxtnxtd = nxtd, nxtnxtd, next(g)
        while gcd(an, alst[-1]) == 1 or nxtd == nxtnxtd == 0:
            an, nxtd, nxtnxtd = int(str(an) + str(nxtd)), nxtnxtd, next(g)
        alst.append(an); aset.add(an)
    return alst
print(aupton(70)) # Michael S. Branicky, Dec 03 2021

Formula

GCD(a(n), a(n+1)) > 1.