Numbers which are Sum of Increasing Powers of Digits

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Theorem

The following integers are the sum of the increasing powers of their digits taken in order begin as follows:

$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2 \, 646 \, 798, 12 \, 157 \, 692 \, 622 \, 039 \, 623 \, 539$

This sequence is A032799 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


Proof

Single digit integers are trivial:

$d^1 = d$

for all $d \in \Z$.


Then we have:

\(\ds 8^1 + 9^2\) \(=\) \(\ds 8 + 81\) \(\ds = 89\)


\(\ds 1^1 + 3^2 + 5^3\) \(=\) \(\ds 1 + 9 + 125\) \(\ds = 135\)
\(\ds 1^1 + 7^2 + 5^3\) \(=\) \(\ds 1 + 49 + 125\) \(\ds = 175\)
\(\ds 5^1 + 1^2 + 8^3\) \(=\) \(\ds 5 + 1 + 512\) \(\ds = 518\)
\(\ds 5^1 + 9^2 + 8^3\) \(=\) \(\ds 5 + 81 + 512\) \(\ds = 598\)


\(\ds 1^1 + 3^2 + 0^3 + 6^4\) \(=\) \(\ds 1 + 9 + 0 + 1296\) \(\ds = 1306\)
\(\ds 1^1 + 6^2 + 7^3 + 6^4\) \(=\) \(\ds 1 + 36 + 343 + 1296\) \(\ds = 1676\)
\(\ds 2^1 + 4^2 + 2^3 + 7^4\) \(=\) \(\ds 2 + 16 + 8 + 2401\) \(\ds = 2427\)


Then:

\(\ds \) \(\) \(\ds 2^1 + 6^2 + 4^3 + 6^4 + 7^5 + 9^6 + 8^7\)
\(\ds \) \(=\) \(\ds 2 + 36 + 64 + 1296 + 16 \, 807 + 531 \, 441 + 1 \, 097 \, 152\)
\(\ds \) \(=\) \(\ds 2 \, 646 \, 798\)


and finally:

\(\ds \) \(\) \(\ds 1 + 2^2 + 1^3 + 5^4 + 7^5 + 6^6 + 9^7 + 2^8 + 6^9 + 2^{10} + 2^{11} + 0^{12} + 3^{13} + 9^{14} + 6^{15} + 2^{16} + 3^{17} + 5^{18} + 3^{19} + 9^{20}\)
\(\ds \) \(=\) \(\ds 1 + 4 + 1 + 625 + 16 \, 807 + 46 \, 656 + 4 \, 782 \, 969 + 256 + 10 \, 077 \, 696 + 1024 + 2048 + 0 + 1 \, 594 \, 323\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 22 \, 876 \, 792 \, 454 \, 961 + 470 \, 184 \, 984 \, 576 + 65 \, 536 + 129 \, 140 \, 163 + 3 \, 814 \, 697 \, 265 \, 625\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds 1 \, 162 \, 261 \, 467 + 12 \, 157 \, 665 \, 459 \, 056 \, 928 \, 801\)
\(\ds \) \(=\) \(\ds 12 \, 157 \, 692 \, 622 \, 039 \, 623 \, 539\)




Sources