%I #45 May 20 2017 10:20:23
%S 1,2,3,4,5,6,7,8,9,89,135,2537,60409,4901732,17735872,45279768,
%T 393470463,3623008669,3893095238,229386834955666,1892713761283624,
%U 1501212693940707502,1517944702855898904,12303679765763687463,122947811178635339597,1095354314191826124704,1106509957063490820877
%N 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, ...
%C 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
%H Francis J. McDonnell, <a href="/A208130/a208130.txt">Java program</a>
%e 2537 = 2^1 + 3^2 + 5^3 + 7^4 = 2 + 9 + 125 + 2401.
%e 60409 = 0^1 + 0^2 + 4^3 + 6^4 + 9^5 = 0 + 0 + 64 + 1296 + 59049.
%o (Java) see link.
%o (Python)
%o from itertools import combinations_with_replacement
%o A208130_list = []
%o for l in range(1,23):
%o for n in combinations_with_replacement(range(10),l):
%o x = sum(b**(a+1) for a,b in enumerate(n))
%o if x > 0 and tuple(sorted(int(d) for d in str(x))) == n:
%o A208130_list.append(x)
%o A208130_list = sorted(A208130_list) # _Chai Wah Wu_, May 20 2017
%Y Cf. A032799 (does not sort the digits prior to raising to powers).
%K nonn,base,fini,full
%O 1,2
%A _Francis J. McDonnell_, Mar 29 2012
%E More terms added by _Francis J. McDonnell_, Apr 12 2012
%E Faster program used to obtain more terms included by _Francis J. McDonnell_, Apr 16 2012
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