A208130 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, ...

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

Offset: 1

Author

Francis J. McDonnell, Mar 29 2012

Keywords

• nonn
• base
• fini
• full

Comments

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

Examples

			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.
		

Links

• Francis J. McDonnell, Java program

Crossrefs

Cf. A032799 (does not sort the digits prior to raising to powers).

Programs

Java

// See McDonnell link.

Python

from itertools import combinations_with_replacement
A208130_list = []
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:
            A208130_list.append(x)
A208130_list = sorted(A208130_list)  # Chai Wah Wu, May 20 2017

Extensions

More terms added by Francis J. McDonnell, Apr 12 2012

Faster program used to obtain more terms included by Francis J. McDonnell, Apr 16 2012