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A208130
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Numbers that when expressed in decimal are equal to the sum of the digits sorted into nondecreasing order and raised to the powers 1, 2, 3, ...
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3
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1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 2537, 60409, 4901732, 17735872, 45279768, 393470463, 3623008669, 3893095238, 229386834955666, 1892713761283624, 1501212693940707502, 1517944702855898904, 12303679765763687463, 122947811178635339597, 1095354314191826124704, 1106509957063490820877
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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Lemma: The sequence is finite with all terms in the sequence having at most 22 digits. Proof: Let n be an m-digit natural number in the sequence for some m. Then 10^(m-1) <= n and n <= 9 + 9^2 + ... + 9^m = 9(9^m-1)/8 < (9^(m+1))/8. Thus 10^(m-1) < (9^(m+1))/8. Taking logarithms of both sides and solving yields m < 22.97. QED. The sequence listed, found by a computer program searching up to 10^22, is therefore complete. - Francis J. McDonnell, Apr 12 2012
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LINKS
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EXAMPLE
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2537 = 2^1 + 3^2 + 5^3 + 7^4 = 2 + 9 + 125 + 2401.
60409 = 0^1 + 0^2 + 4^3 + 6^4 + 9^5 = 0 + 0 + 64 + 1296 + 59049.
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PROG
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(Java) see link.
(Python)
from itertools import combinations_with_replacement
for l in range(1, 23):
for n in combinations_with_replacement(range(10), l):
x = sum(b**(a+1) for a, b in enumerate(n))
if x > 0 and tuple(sorted(int(d) for d in str(x))) == n:
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CROSSREFS
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Cf. A032799 (does not sort the digits prior to raising to powers).
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KEYWORD
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nonn,base,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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