Testwiki:Reference desk/Archives/Mathematics/2023 March 4

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March 4

Formula for a sequence

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)

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)

Thank you. Sandbh (talk) 06:15, 6 March 2023 (UTC)

A simple formula that generates these particular 8 numbers is $2{\lceil n/2\rceil }^2$. CodeTalker (talk) 07:31, 4 March 2023 (UTC)

Do the square brackets mean round up? Sandbh (talk) 06:15, 6 March 2023 (UTC)

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)

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

Fifth_power_(algebra)#Properties notes Template:Cquote Is it a coincidence, or is there a fundamental reason it must be so? Thanks, cmɢʟee⎆τaʟκ 07:22, 4 March 2023 (UTC)

Let n be written as $10a+b$, where b is the low order digit (between 0 and 9 inclusive).
Then expanding by the binomial theorem, $n^5={(10a+b)}^5=10^5{a}^5+5\cdot 10^4{a}^{4}b+10\cdot 10^3{a}^3{b}^2+10\cdot 10^2{a}^2{b}^3+5\cdot 10a{b}^4+b^5$
Every term except the last is a multiple of 10, so to show that $n^5\equiv nmod10$, we just need to show that $b^5\equiv bmod10$ for every value of b between 0 and 9. This can be done by enumeration, and observing that the last digit of $b^5$ is equal to $b$ in all cases:
$0^5=0;1^5=1;2^5=32;3^5=243;4^5=1024;5^5=3125;6^5=7776;7^5=16807;8^5=32768;9^5=59049$. CodeTalker (talk) 07:52, 4 March 2023 (UTC)

Another, rather different proof. The decimal representations of $n^5$ and $n$ share their last digit when $n^5-n$ is a multiple of $10,$ or, equivalently, when it is both a multiple of $2$ and of $5$. From the factorization $n^5-n=(n-1)n(n+1)(n^2+1)$ it is obvious that the difference is even. It remains to show that it is a multiple of $5$. For this, we use modular arithmetic modulo $5.$ In what follows, I write $a\equiv b(\text{mod}5)$ as $a\equiv b$ 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:

$n^5-n\equiv (n-1)n(n+1)(n^2+1)\equiv$ $(n-1)n(n-4)(n^2-4)\equiv$ $(n-1)n(n-4)(n-2)(n+2)\equiv$ $(n-1)n(n-4)(n-2)(n-3)\equiv$ $n(n-1)(n-2)(n-3)(n-4).$

For all $n$, one of these five factors is congruent to $0.$  --Lambiam 13:12, 4 March 2023 (UTC)

$n^5\equiv n(\mathrm{mod}5)$ by Fermat's little theorem; $n^5\equiv n(\mathrm{mod}2)$ trivially; therefore $n^5\equiv n(\mathrm{mod}10)$. 116.86.4.41 (talk) 13:44, 4 March 2023 (UTC)

Thank you very much, everyone. So it's a bit of both: our base, 10 happens to be 2 × 5. Cheers, cmɢʟee⎆τaʟκ 15:17, 4 March 2023 (UTC)

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)

More generally, I think this holds: There exist a power m > 1 such that n^m always ends in the same base b digit as n iff b is square-free. PrimeHunter (talk) 21:30, 4 March 2023 (UTC)

Moreover, it appears that m has this property iff Template:Nowrap is an integral multiple of Template:Nowrap for each prime factor p of b.  --Lambiam 23:43, 4 March 2023 (UTC)

Thanks very much for the general formula. Cheers, cmɢʟee⎆τaʟκ 12:38, 6 March 2023 (UTC)

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)