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March 4

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Formula for a sequence

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Is there a formula for the sequence 2,2,8,8,18,18,32,32? It is the number of chemical elements in each period of a left-step or Janet periodic table.

In contrast, the formula for the number of elements in each period of a conventional periodic table (2,8,8,18,18,32,32) is:

a(n) = (2*n+3+(−1)^n)^2/8

thank you, Sandbh (talk) 02:28, 4 March 2023 (UTC)[reply]

You can use b(n) = a(n−1) = (2*(n−1)+3+(−1)^(n−1))^2/8 = (2*n+1−(−1)^n)^2/8.  --Lambiam 03:20, 4 March 2023 (UTC)[reply]
Thank you. Sandbh (talk) 06:15, 6 March 2023 (UTC)[reply]
A simple formula that generates these particular 8 numbers is . CodeTalker (talk) 07:31, 4 March 2023 (UTC)[reply]
Do the square brackets mean round up? Sandbh (talk) 06:15, 6 March 2023 (UTC)[reply]
Yes, they denote the ceiling function. But note that their shapes have one less corner than square brackets.  --Lambiam 09:42, 6 March 2023 (UTC)[reply]

Why does the units digit of an integer equal the units digit of its fifth power?

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Fifth_power_(algebra)#Properties notes

Is it a coincidence, or is there a fundamental reason it must be so? Thanks, cmɢʟeeτaʟκ 07:22, 4 March 2023 (UTC)[reply]

Let n be written as , where b is the low order digit (between 0 and 9 inclusive).
Then expanding by the binomial theorem,
Every term except the last is a multiple of 10, so to show that , we just need to show that for every value of b between 0 and 9. This can be done by enumeration, and observing that the last digit of is equal to in all cases:
. CodeTalker (talk) 07:52, 4 March 2023 (UTC)[reply]
Another, rather different proof. The decimal representations of and share their last digit when is a multiple of or, equivalently, when it is both a multiple of and of . From the factorization it is obvious that the difference is even. It remains to show that it is a multiple of . For this, we use modular arithmetic modulo In what follows, I write as for the sake of concision. Using the fact that modular congruence is a congruence relation with respect to the ring structure of arithmetic, we have:
For all , one of these five factors is congruent to  --Lambiam 13:12, 4 March 2023 (UTC)[reply]
by Fermat's little theorem; trivially; therefore . 116.86.4.41 (talk) 13:44, 4 March 2023 (UTC)[reply]
Thank you very much, everyone. So it's a bit of both: our base, 10 happens to be 2 × 5. Cheers, cmɢʟeeτaʟκ 15:17, 4 March 2023 (UTC)[reply]
Indeed, it is a bit of a coincidence, not a truly fundamental reason. For example, it does not work in bases 12 and 16: (212)5 = 25 = 32 = 2812 and (216)5 = 25 = 32 = 2016. Next to base 10, the property holds for bases 2, 3, 5, 6, 15 and 30.  --Lambiam 16:41, 4 March 2023 (UTC)[reply]
More generally, I think this holds: There exist a power m > 1 such that nm always ends in the same base b digit as n iff b is square-free. PrimeHunter (talk) 21:30, 4 March 2023 (UTC)[reply]
Moreover, it appears that m has this property iff m − 1 is an integral multiple of p − 1 for each prime factor p of b.  --Lambiam 23:43, 4 March 2023 (UTC)[reply]
Thanks very much for the general formula. Cheers, cmɢʟeeτaʟκ 12:38, 6 March 2023 (UTC)[reply]
The OEIS sequence A197658 being listed as one plus the reduced totient function A002322 (at least, when squarefree) seems to corroborate this. GalacticShoe (talk) 19:41, 6 March 2023 (UTC)[reply]