Non-values of $a^3 + b^{c^3} - 3 a b c$ for $a,b,c\in\mathbb{N}$ like $n$ for prime $19n+17$?

Question

I stumbled on a curious case of Richard Guy's Strong Law of Small Numbers because of a typographical error. I intended to type a^3 + b^3 + c^3 - 3 a b c and look at its values (Problem A1 on the 2019 Putnam Exam), but I omitted the exponent on $b$ and it came out as a^3 + b^+c^3 - 3 a b c. Mathematica accepted this and interpreted it to mean $a^3 + b^{c^3} - 3 a b c$ and the values Mathematica came up with suggest the latter expression takes on, for nonnegative integer values of $a,b,c$, all integer values except $-6, -14, -18, -20, -30, -36, -38, -48, -50, -60, -66, -74, -78, -96 ...$

The first few terms produced one hit on OEIS:

A108977 $\ $ Numbers $n$ such that $19·n + 17$ is prime.

$0, 6, 14, 18, 20, 30, 36, 38, 44, 48, 50, 74, 78, 84, 98, 104, 108, 116, 120, 126, 140, 144, 146, 158, 168, 174, 176,...$

The first $7$ (nonzero, signless) terms agree and there are some sporadic common entries thereafter. Can anyone offer any explanation?

Answer 1

This is not at all an answer to your question but a response to Varun Vejalla's comment:

There are so many sequences (especially increasing) in the OEIS that if you type in a few small integers (with small being less than $\sim 200$), you are almost guaranteed to get a hit.

For sequences with $7$ terms this has not been my experience using the OEIS. But in any case this is a testable conjecture! We can perturb the first $7$ terms of your sequence and see how many hits we get on the OEIS each time. I will just either add or subtract $2$ from each term, since the terms are all even. This produces the following:

• $8, 14, 18, 20, 30, 36, 38$: zero hits.
• $6, 16, 18, 20, 30, 36, 38$: zero hits.
• $6, 14, 16, 20, 30, 36, 38$: zero hits.
• $6, 14, 18, 22, 30, 36, 38$: zero hits.
• $6, 14, 18, 20, 28, 36, 38$: zero hits.
• $6, 14, 18, 20, 32, 36, 38$: zero hits.
• $6, 14, 18, 20, 30, 34, 38$: zero hits.
• $6, 14, 18, 20, 28, 36, 40$: zero hits.

So it's not clear that this is entirely just a coincidence! In fact the OEIS wiki suggests that one should search the OEIS using $6$ terms.

According to Wikipedia, as of September 2020 the OEIS has about $337000$ sequences, which is really not that many: this means about $18.3$ bits worth of information uniquely pins down a sequence. The number of increasing $7$-term sequences with all terms between $1$ and, say, $50$ is ${50 \choose 7} \approx 9.99 \times 10^7$ which is about $26.5$ bits, by contrast. So $7$ terms is more than enough to get out of the realm of coincidence.

In fact $337000 < {36 \choose 5} = 376992$, so to give the OEIS a fairer test we might consider perturbing just the first $5$ terms of the sequence ($6, 14, 18, 20, 30$ already returns A108977 uniquely). This gives:

• $8, 14, 18, 20, 30$: five hits, none of which are initial.
• $6, 12, 18, 20, 30$: two hits, one of which is initial.
• $6, 16, 18, 20, 30$: zero hits.
• $6, 14, 16, 20, 30$: two hits, neither of which is initial.
• $6, 14, 18, 22, 30$: zero hits.
• $6, 14, 18, 20, 28$: zero hits.
• $6, 14, 18, 20, 32$: zero hits.

I think this justifies claiming that $5$ terms is maybe a coincidence but $7$ is maybe not.

Answer 2

As Greg Martin notes in a comment, setting $b=1$ or $c=1$ yield special cases $a^3+1-3ac$ and $a^3+b-3ab$ respectively.

In particular, when $c=1$, setting $a=1,2,4$ and letting $b$ vary rule out exactly the congruence classes mod $2$, $5$, and $11$ that would render $19n+17$ divisible by those primes. When $a=b=1$, letting $c$ vary rules out the congruence class of $1$ mod $3$*. (Mod $7$, things fail, as seen with $96$). Note that these restrictions aren't special, and the constants $19$ and $17$ in the OEIS definition are doing the work for us - in a sense, they spend the bits of not-just-a-coincidence credit we built up by finding these specific congruences and terms. So we shouldn't be surprised that this alignment happens.

This gets you most of the way there: below 100, you're left with 0, 6, 8, 14, 18, 20, 26, 30, 36, 38, 44, 48, 50, 54, 56, 60, 66, 74, 78, 80, 84, 86, 96, 98. To get alignment, they just both happen to rule out $8$ and $26$. This much I think is reasonable to chalk up to coincidence, given how often one tries out an obscure series of integers in OEIS; you're bound to get some false positives at this level of significance now and then.

*I'm letting $a,b,c\in\mathbb{Z}$ here, because I think that's what Mathematica was doing - for instance, I can't seem to get $-8$ using nonnegative values, but it's not in the list.