

A108116


Base 10 weak SkolemLangford numbers.


6



2002, 131003, 231213, 300131, 312132, 420024, 12132003, 14130043, 15120025, 23121300, 23421314, 25121005, 25320035, 30023121, 31213200, 31413004, 34003141, 40031413, 41312432, 45001415, 45121425, 45300435, 50012152, 51410054, 52002151, 52412154, 53002352, 53400354, 61310036
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OFFSET

1,1


COMMENTS

Selfdescribing numbers: between two digits "d" there are d digits.
a(n) has either 0 or 2 instances of any digit, hence even number of digits.
Largest element is a(20120) = 978416154798652002.
Named after the Norwegian mathematician Thoralf Albert Skolem (18871963) and the British chemist and mathematics teacher Charles Dudley Langford (19051969).  Amiram Eldar, Jun 17 2021


REFERENCES

E. Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 3235, Volume 59 (Jeux math'), April/June 2008, Paris.


LINKS



EXAMPLE

In "2002" there are 2 digits between the two 2's and 0 digits between the two 0's.
In "131003" there is 1 digit between the two 1's, 3 digits between the two 3's and 0 digit between the two 0's.


PROG

(Python)
def SL(d, s):
for i1 in range(int(d[0]=="0"), len(s)int(d[0])1):
i2 = i1 + int(d[0]) + 1
if not (s[i1] or s[i2]):
s[i1] = s[i2] = d[0]
r = d[1:]
if r: yield from SL(r, s)
else: yield int("".join(s))
s[i1] = s[i2] = 0
from itertools import chain, combinations as C
def A108116gen():
for numd in range(1, 11):
dset, s = "0123456789", [0 for _ in range(2*numd)]
for an in sorted(
chain.from_iterable(SL("".join(c), s) for c in C(dset, numd))):
yield an
for n, an in enumerate(A108116gen(), start=1):


CROSSREFS

Base 10 strong SkolemLangford numbers are in A132291.
Base 10 weaker SkolemLangford numbers are in A357826.


KEYWORD

base,easy,fini,full,nonn


AUTHOR



EXTENSIONS



STATUS

approved



