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September 5

Formula for a series

Is there a formula for the series 1, 8, 18, 32, 50? Sandbh (talk) 12:30, 5 September 2022 (UTC)[reply]

Are you sure the "1" is correct? If it should be 2, then the formula is (2 x n^2), for n = positive integers starting at 1. 41.23.55.195 (talk) 14:20, 5 September 2022 (UTC)[reply]

Also true for negative integers since n is squared. 41.246.128.27 (talk) 18:36, 5 September 2022 (UTC)[reply]

For questions like this, ask the On-Line Encyclopedia of Integer Sequences. It does not know anything about the original series, but, if, as the previous answer suggests, the first term is the number 2, then it matches A001105, about which a lot is known. --Stephan Schulz (talk) 17:28, 5 September 2022 (UTC)[reply]

in short 2*n^2. Dhrm77 (talk) 17:44, 5 September 2022 (UTC)[reply]

There is a formula for every finite sequence (what you have here is a sequence, not a series; series implies summation of terms). But the formula may be complicated and messy, depending on the sequence. In general, for a sequence of N terms, there is a polynomial of degree N-1 or lower that produces that sequence. For example, the five terms of the OP's sequence (assuming the initial 1 was not a typo) are generated by

f(n)={\frac {1}{24}}(-n^{4}+10n^{3}+13n^{2}+146n+24)

, for n = 0 to 4. CodeTalker (talk) 00:21, 6 September 2022 (UTC)[reply]

Thank you for the responses. The 1 is correct. Google gives some hits to this sequence but the suggested "formulae" are incorrect in that they're right for 8, 18, 32, 50 but not for the 1, at the start.

The 2, 8, 18, 32, 50 sequence discussed earlier corresponds to the period lengths in a periodic table, where the first period has H-He, the next two periods are Li-Be-B-C-N-O-F-Ne and (8) then Na-Mg-Al-Si-P-S-Cl-Ar (8) and so on. The period lengths become 1, 8, 18, 32, 50 etc if H is instead place over He. This may seem odd but there are chemistry-based reasons for its plausibility. Among other things it results in a regularity of period doubling that the traditional sequence of 2, 8, 8, 18, 18, 32, 32 lacks.

CodeTalker, that impressive formula works for n = 0 to 4, but for n = 5 the result is 71, whereas the expected result in theoretical periodic table terms is 72.

Thanks again. Sandbh (talk) 02:19, 6 September 2022 (UTC)[reply]

If you think about the periodic table, you only have a finite number of elements and hence distances to contend with. So you can always find a polynomial of a sufficiently high degree to produce all the desired numbers - but that may not have much explanatory power... --Stephan Schulz (talk) 14:23, 6 September 2022 (UTC)[reply]

If you want to add a sixth number, 72, to the sequence, then it would possible to construct a fifth-degree polynomial that generates all six numbers, although I don't feel like doing the algebra right now. However it would probably be more, not less, complicated than the fourth degree polynomial I gave above for the first 5 terms. I think the fact that there isn't a simple clean formula for this sequence starting with 1, while there IS a simple clean formula (

2n^{2}

) for the sequence starting with 2, which works for all electron shells, is a strong indication that the latter sequence is a more appropriate representation of the physics. I've always heard that the capacity of the n-th shell is

2n^{2}

, which also is easily derived from the possible values of the four quantum numbers work. CodeTalker (talk) 18:50, 6 September 2022 (UTC)[reply]

Mostly out of curiosity, I checked, and the fifth-degree polynomial you were talking about is

{\frac {x^{5}-15x^{4}+85x^{3}+15x^{2}+754x+120}{120}}

. It certainly is more complicated. GalacticShoe (talk) 20:13, 6 September 2022 (UTC)[reply]

Since it's off-by-one from a simpler polynomial, a simpler formula exists:

2x^{2}+{\frac {(x-2)(x-3)(x-4)(x-5)(x-6)}{120}}.

The reason for the difference is that this one assumes you start at x=1 and the one you have assumes you start at x=0; it's never really specified what the starting value of x (a.k.a. offset) is.

By the way, there's no reason in chemistry that the periodic table couldn't continue for more rows. The reason that the elements that we have only go up to a little over a hundred is that heavier nuclei would be too unstable to exist. But it is possible to predict some of the chemistry of these hypothetical elements. I'm not sure it would be science any longer because it would be impossible to verify the predictions, but the models do, at least, make the predictions. --RDBury (talk) 22:30, 6 September 2022 (UTC)[reply]

Absolutely they do. The models and equations used to describe atomic orbitals do not have an upper bound, and you can just plug in quantum numbers into the relevant equations and get solutions. This web page here summarizes the solutions for g, h, and i orbitals, which if atoms existed that utilized these as ground state configurations, would slot into the periodic table with the same pattern we have already; just as we have an s block (2 columns), p block (6 columns) d block (10 columns) and f block (14 columns) the same pattern would continue with the g block (18 columns), h block (22 columns) and and i block (26 columns). This continues the sequence perfectly. The reason why we don't use those orbitals in our modelling of atomic electron configuration is that no atom yet created would even hypothetically need them for their ground state electron configuration. There are some practical quantum effect problems as well, as the level of resolution between the energy states of neighboring orbitals that high up the ladder starts to fall well below the uncertainty principle limit; and at that point who the heck knows what will happen. But, if we ignore that, the first atom to hypothetically need a 5g electron in its ground state would be element number 121, which is actually only 3 higher than the current largest atom, Oganesson, so we may very well make such an atom in our lifetimes; whether we will get enough of it to get an accurate atomic spectrum and be able to resolve the atomic orbitals that way to see if it really does have a 5g electron remains to be seen. --Jayron 32 15:30, 7 September 2022 (UTC)[reply]

@Jayron32: There's lots of predictions at Extended periodic table#Predicted properties of eighth-period elements, all the way into the ninth period. I don't think we'll actually be able to get that far, though, unless udQM is stable over ordinary matter at high mass numbers. Probably ground-state element 121 as a single atom does not fill the 5g shell (though it should be within reach for chemistry): La doesn't fill 4f, and relativistic destabilisation should be even stronger here. BTW, I haven't read computational chemistry stuff in a while, but I imagine g, h, and higher orbitals would not be unusual to put in basis sets. Double sharp (talk) 09:53, 12 September 2022 (UTC)[reply]

Putting H over He certainly does seem odd to me - it would imply H was a noble gas. What period doublin are you referring to? And wouldn't your sequence then be 1 1 8 18 32 50...? NadVolum (talk) 16:04, 7 September 2022 (UTC)[reply]

f(n) = 2n² - δ_n1. --Amble (talk) 17:57, 7 September 2022 (UTC)[reply]

Stephan Schulz, CodeTalker, GalacticShoe, RDBury, Jayron32, Amble: Thanks very much for your contributions.

I agree about the practical quantum effect problems; in periodic table terms it looks like things start becoming moot from, probably at least 119 onwards. I say moot in that 119 is expected to show oxidation states exceeding the +1 normally seen for the alkali metals in group 1. So the approximate recurrence of properties for elements 1 to 118 is expected to start going feral.

The simple formula works up to element 218, which is most elegantly more than adequate.

On H not being a noble gas bear in mind that while H is normally placed above Li in group 1, this does not imply that H is an alkali metal, whereas Li and its heavier congeners certainly are.

The period doubling I'm referring to is the 8, 8, 18, 18 32, 32 occurring in the traditional periodic table from row two onwards:

H                      He
Li Be...B  C  N  O  F  Ne
Na Mg...Al Si P  S  Cl Ar

The sum of the period lengths is 2+8+8+18+18+32+32 = 118.

OTOH, in philosophy of chemistry there is a fair appreciation of the left step form of periodic table, which has regular periods of length 2, 2, 8, 8, 18, 18, 32, 32. This is achieved by moving He over Be, so that the periods look like this:

                  H  He
                  Li Be
B  C  N  O  F  Ne Na Mg
Al Si P  S  Cl Ar K  Ca

The sum of the period lengths is 2+2+8+8+18+18+32+32 = 120.

While the regularity of period lengths is considered by some to be quite significant, "bangs and stinks" chemists tend to balk at He over Be, since Be is not a noble gas. As well, the left step form scrambles the familiar pattern of metals on the left and nonmetals on the right seen in the traditional periodic table. That said, there are philosophical arguments supporting the left step table, which I won't go into here.

Getting back to H over He, hydrogen is relatively unreactive at room temperature. This is due to the heat of dissociation of the H₂ molecule being extremely high, the H-H bond energy being larger than for almost all other single bonds.

Of course H is more reactive than He which is expected to be more reactive than Ne, with neon being thought to be the most noble of the noble gases. Thus H >> He > Ne < Ar < Kr < Xe < Rn. Equally, on the left side of the periodic table, Cs is thought to be the most reactive of the metals rather than Fr. Thus Li < Na < K < Rb < Cs > Fr.

There are other arguments in support of H over He, which I'm not inclined to go into, at least not in this forum.

In terms of precedents:

Rodorf (1904, pp. 87, 269) was of the opinion that H belonged to the noble gases but, according to van Spronsen (1969 p. 303), his reasoning was not clear.
Cuthbertson & Parr (1907) showed H almost over He.
LeRoy (1927) showed H, with links to F and Li, floating above He..

Sources

Rudorf G 1904, Das periodische System, seine Geschichte und Bedeutung für chemische Systematik ("The periodic table, its history and importance for chemical systematics"), L. Voss, Hamburg-Leipzig.
van Spronsen JW 1969, The Periodic System of Chemical Elements: A History of the First Hundred Years, Elsevier, Amsterdam.

--- Sandbh (talk)