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A108116
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Base 10 weak Skolem-Langford numbers.
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6
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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
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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Self-describing 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 (1887-1963) and the British chemist and mathematics teacher Charles Dudley Langford (1905-1969). - Amiram Eldar, Jun 17 2021
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REFERENCES
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E. Angelini, "Jeux de suites", in Dossier Pour La Science, pp. 32-35, Volume 59 (Jeux math'), April/June 2008, Paris.
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LINKS
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EXAMPLE
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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.
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PROG
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(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):
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CROSSREFS
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Base 10 strong Skolem-Langford numbers are in A132291.
Base 10 weaker Skolem-Langford numbers are in A357826.
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KEYWORD
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base,easy,fini,full,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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