# Self and Earl Bird numbers: are there infinitely many natural numbers that are both?

Self-numbers (A003052 in the OEIS) are natural numbers which are not the sum of a smaller number and the sum of its digits. Early Bird numbers (A116700) are numbers that occur ahead of their natural place in the string 12345678910111213....

Are there infinitely many numbers, such as 110, that are both a Self-number (because no number smaller than 110 exists such that added to the sum of its digits equals 110) and an Early Bird number (because in the string 12345...99100101102103... 110 appears earlier than its natural place, just a few digits further)?

• The following are the first few numbers belonging to both sequences: 31, 42, 53, 64, 75, 86, 97, 110, 121, 132, 211, 222, 233, 310, 312, 323, 334, 345, 411, 413, 424, 435, 501, 512, 514, 525, 536, 547, 591, 602, 613, 615, 626, 637, 648, 659, 681, 692, 703, 714, 716, 727, 738, 749, 760, 771, 782, 793, 804, 815, 817, 828, 839, 850, 861, 872, 883, 894, 905, 916, 918, 929, 940, 951, 962, 973, 984, 995, 1021, 1109, 1111, 1122, ..., 2022, ... Nov 4, 2022 at 15:28
• And not until 2110 will the year in course be both an Early Bird number and a Self-number! Nov 4, 2022 at 15:31

Yes, there are infinite such numbers. My proof is based on this reduction test that can found here.

Take $$a = 1$$ and $$c = 10$$. Then, $$m_1 = 10 - 1 = 9$$ which is a self-number and $$m_2 = 9b - 11$$.
So let us pick any $$b$$ such that $$9b-11$$ is also a self-number.
Note the self-number generating recurrence relation (from wiki) $$C_k = 8 \cdot 10^{k - 1} + C_{k - 1} + 8, C_1 = 9$$.
This implies $$C_k = (C_{k-1} - 2) \mod 9$$, which means every value congruent to modulo $$9$$ repeats after every $$9$$ terms. Thus, there are infinite self-numbers that are congruent to $$7 \mod 9\;$$ so there will be infinite possible values for $$b$$ as well.

There is a $$10$$ at both the beginning and the end of this number so this is also an Early Bird number.

For example, take $$b=12$$ then $$m_2 = 9 \cdot 12 - 11 = 97$$ which is a self-number so $$10^{12} + 10 = 1000000000010$$ is also a self-number. This is also an Early bird number because it's a prefix of $$10000000000, 10000000001$$.

• Maybe it's there and I just didn't see it, but why are there infinitely many selfs = 7 mod 9? Nov 3, 2022 at 20:05
• @loopywalt the recurrence relation from the same wiki which says $C_k = 8 \cdot 10^{k - 1} + C_{k - 1} + 8$ implies $C_k = (C_{k-1} - 2) \mod 9$ so all values congruent to modulo 9 repeat after every 9 terms. Nov 3, 2022 at 20:17