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If $p,q$ are a pair of twin primes, then $x = \dfrac{p+ q}{2} = q-1 = p+1$ is their twin prime average.

Conjecture. Every twin prime average $x \gt 6$ is the sum of two smaller twin prime averages, $x = x' + x''$.

This doesn't require infinitude. So wondering if we can prove it.

12 = 6 + 6
18 = 6 + 12
30 = 12 + 18
42 = 12 + 30
60 = 18 + 42
72 = 12 + 60
102 = 30 + 72
108 = 6 + 102
138 = 30 + 108
150 = 12 + 138
180 = 30 + 150
192 = 12 + 180
198 = 6 + 192
228 = 30 + 198
240 = 12 + 228
270 = 30 + 240
282 = 12 + 270
312 = 30 + 282
348 = 108 + 240
420 = 72 + 348
432 = 12 + 420
462 = 30 + 432
522 = 60 + 462
570 = 108 + 462
600 = 30 + 570
618 = 18 + 600
642 = 42 + 600
660 = 18 + 642
810 = 150 + 660
822 = 12 + 810
828 = 6 + 822
858 = 30 + 828
882 = 60 + 822
1020 = 138 + 882
1032 = 12 + 1020
1050 = 18 + 1032
1062 = 12 + 1050
1092 = 30 + 1062
1152 = 60 + 1092
1230 = 138 + 1092
1278 = 228 + 1050
1290 = 12 + 1278
1302 = 12 + 1290
1320 = 18 + 1302
1428 = 108 + 1320
1452 = 150 + 1302
1482 = 30 + 1452
1488 = 6 + 1482
1608 = 180 + 1428
1620 = 12 + 1608
1668 = 60 + 1608
1698 = 30 + 1668
1722 = 102 + 1620
1788 = 180 + 1608
1872 = 150 + 1722
1878 = 6 + 1872
1932 = 60 + 1872
1950 = 18 + 1932
1998 = 570 + 1428
2028 = 30 + 1998
2082 = 150 + 1932
2088 = 6 + 2082
2112 = 30 + 2082
2130 = 18 + 2112
2142 = 12 + 2130
2238 = 108 + 2130
2268 = 30 + 2238
2310 = 42 + 2268
2340 = 30 + 2310
2382 = 42 + 2340
2550 = 240 + 2310
2592 = 42 + 2550
2658 = 108 + 2550
2688 = 30 + 2658
2712 = 570 + 2142
2730 = 18 + 2712
2790 = 60 + 2730
2802 = 12 + 2790
2970 = 180 + 2790
3000 = 30 + 2970
3120 = 150 + 2970
3168 = 198 + 2970
3252 = 282 + 2970
3258 = 6 + 3252
3300 = 42 + 3258
3330 = 30 + 3300
3360 = 30 + 3330
3372 = 12 + 3360
3390 = 18 + 3372
3462 = 72 + 3390
3468 = 6 + 3462
3528 = 60 + 3468
3540 = 12 + 3528
3558 = 18 + 3540
3582 = 42 + 3540
3672 = 282 + 3390
3768 = 228 + 3540
3822 = 150 + 3672
3852 = 30 + 3822
3918 = 150 + 3768
3930 = 12 + 3918
4002 = 72 + 3930
4020 = 18 + 4002
4050 = 30 + 4020
4092 = 42 + 4050
4128 = 108 + 4020
4158 = 30 + 4128
4218 = 60 + 4158
4230 = 12 + 4218
4242 = 12 + 4230
4260 = 18 + 4242
4272 = 12 + 4260
4338 = 108 + 4230
4422 = 150 + 4272
4482 = 60 + 4422
4518 = 180 + 4338
4548 = 30 + 4518
4638 = 420 + 4218
4650 = 12 + 4638
4722 = 72 + 4650
4788 = 138 + 4650
4800 = 12 + 4788
4932 = 282 + 4650
4968 = 180 + 4788
5010 = 42 + 4968
5022 = 12 + 5010
5100 = 312 + 4788
5232 = 432 + 4800
5280 = 180 + 5100
5418 = 138 + 5280
5442 = 420 + 5022
5478 = 60 + 5418
5502 = 60 + 5442
5520 = 18 + 5502
5640 = 138 + 5502
5652 = 12 + 5640
5658 = 6 + 5652
5742 = 102 + 5640
5850 = 108 + 5742
5868 = 18 + 5850
5880 = 12 + 5868
6090 = 240 + 5850
6132 = 42 + 6090
6198 = 108 + 6090
6270 = 72 + 6198
6300 = 30 + 6270
6360 = 60 + 6300
6450 = 150 + 6300
6552 = 102 + 6450
6570 = 18 + 6552
6660 = 108 + 6552
6690 = 30 + 6660
6702 = 12 + 6690
6762 = 60 + 6702
6780 = 18 + 6762
6792 = 12 + 6780
6828 = 138 + 6690
6870 = 42 + 6828
6948 = 858 + 6090
6960 = 12 + 6948
7128 = 180 + 6948
7212 = 420 + 6792
7308 = 180 + 7128
7332 = 462 + 6870
7350 = 18 + 7332
7458 = 108 + 7350
7488 = 30 + 7458
7548 = 60 + 7488
7560 = 12 + 7548
7590 = 30 + 7560
7758 = 198 + 7560
7878 = 420 + 7458
7950 = 72 + 7878
8010 = 60 + 7950
8088 = 138 + 7950
8220 = 270 + 7950
8232 = 12 + 8220
8292 = 60 + 8232
8388 = 828 + 7560
8430 = 42 + 8388
8538 = 108 + 8430
8598 = 60 + 8538
8628 = 30 + 8598
8820 = 192 + 8628
8838 = 18 + 8820
8862 = 42 + 8820
8970 = 108 + 8862
9000 = 30 + 8970
9012 = 12 + 9000
9042 = 30 + 9012
9240 = 198 + 9042
9282 = 42 + 9240
9342 = 60 + 9282
9420 = 138 + 9282
9432 = 12 + 9420
9438 = 6 + 9432
9462 = 30 + 9432
9630 = 192 + 9438
9678 = 240 + 9438
9720 = 42 + 9678
9768 = 138 + 9630
9858 = 138 + 9720
9930 = 72 + 9858

How I would use this data is this. Don't try to prove this, but we can safely assume that given a complete list of twin primes $\leq x_n$ each in the list can be assumed to be the sum of two prior. I would, given such a list $X = x_0 \lt x_1 \lt \dots \lt x_n$, assume that for each $k = 1..n$ that the sublist $x_0 \lt \dots \lt x_k$ also has this summatory property. Perhaps from there we can form a contradiction if for each $x, y \in X$, we have that:

$$ (x + y)^2 - 1 = 0 \pmod q, \\ \text{ for some prime } q \leq \sqrt{x + y + 1} $$

in a Sieve-of-Eratosthenese way. We must also then note that if $y = \max X$, then $\sqrt{2y + 1} \leq y - 2$ for sufficiently large $y$, namely $y \geq 6$. That means that the primes involved in showing that for at least one prime $q$ we have $(x+y)^2 - 1= 0 \pmod q$ can be taken to be the same list $Q$ by which each $x \in X$ is such that $x \pm 1 \neq 0 \pmod q$ for some $q \in Q$, i.e. namely all $q \leq y - 2$.

Daniel Donnelly
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    What is a twin prime average? A quick search doesn't give anything. I am guessing it is the number between the two twin primes, but just to make sure. – Kamal Saleh Feb 23 '23 at 02:06
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    What evidence there is for this? – Mariano Suárez-Álvarez Feb 23 '23 at 02:07
  • @MarianoSuárez-Álvarez see evidence, no counter-example found for twin prime averages $\leq 10,000$ – Daniel Donnelly Feb 23 '23 at 02:13
  • Did you comment on your on post saying it "seems to be true?" – Clyde Kertzer Feb 23 '23 at 02:16
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    A proof seems well beyond the reach of our current abilities, particularly since the proof would have quite a different flavor if the number of twin primes were finite vs. infinite. Presumably the latter is true, and presumably we have a Goldbach-like conjecture that every proper multiple of 6 can be written as the sum of two twin-prime-averages—indeed there should be $S(n)n/(\log n)^4$ such representations when $n$ is large, where $S(n)$ is a "singular series" constant depending boundedly on $n$. – Greg Martin Feb 23 '23 at 02:22
  • @GregMartin yes there seem to be some with many possible formations of the sum. – Daniel Donnelly Feb 23 '23 at 02:44
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    @GregMartin Not every proper multiple of $6$, but maybe every one greater than $6 \times 701$: see OEIS sequence A243956. The only known members of $6 \times A243956$ that is a twin-prime average is $6$. – Robert Israel Feb 23 '23 at 03:07
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    Re: your deleted question: if you drop the (inconsistent) hypothesis that some $,p_i\mid y,$ then the system is solvable, e.g. solve $, y\equiv p_k^{-1}\pmod{\sigma(p_k)},$ for $,y,$ by CRT. More generally see here for how to generalize CRT to modular fractions. – Bill Dubuque Feb 23 '23 at 09:45
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    Also on 2nd glance I realized that you are using the version of the CRT formula that is the n-arry extension of that here. But it looks like you forget some $\sigma$'s in the formula. – Bill Dubuque Feb 23 '23 at 09:56
  • @BillDubuque I want the equation system to be insoluable. That would help prove a lemma of mine. When the equation is soluable, then you can't get to a higher twin prime average by multiplying a certain finite set of them the first few primes. That theory needs some work though; I'm writing it up. – Daniel Donnelly Feb 23 '23 at 19:00
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    Your hypothesis is equivalent to asking whether every element in OEIS A002822 other than the first can be expressed as the sum of two previous elements. – Keith Backman Feb 23 '23 at 20:46
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    @D Left Adjoint to U In elegant wording on my part; I did not intend 'previous two' to be understood as the immediately previous two, but (as you suggest), two (not necessarily different) that precede at whatever distance. In any event, the interpretation as you understood my words to mean is trivially false by example. – Keith Backman Feb 24 '23 at 15:41
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    According to my calculations, for $x\le 10^9$ there is always a solution where the smaller summand does not exceed $64578$ – Peter Feb 25 '23 at 10:06
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    Does the relationship form a tree? In other words, are none of the twin prime averages the sume of two twin prime averages two or more ways? – it's a hire car baby Jun 23 '23 at 11:29
  • @it'sahirecarbaby Not sure, good question. There seems to be many ways to sum to an average sometimes, and little ways other times. None-the-less you could just pick one of the ways, and an expression tree will be created in doing that. Or just use the full tree, but only issue is drawing it gets tangled more easily – Daniel Donnelly Jun 23 '23 at 14:44
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    @DanielDonnelly a single counterexample would be sufficient. Tbh, I assumed you would be aware of one. If you're not, then the tree provides a good way of searching for structure, possibly leading to a proof. I'll ask as a question and link to here. – it's a hire car baby Jun 23 '23 at 15:09
  • @it'sahirecarbaby That sounds like an interesting idea – Daniel Donnelly Jun 23 '23 at 15:24
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    @DanielDonnelly sorry, I misunderstood. It sounds like you already have an average for which you can find more than one pair of averages that sums to it, correct? – it's a hire car baby Jun 23 '23 at 15:36
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    @it'sahirecarbaby yes, typically its a growing number of solutions, I just printed only the first one found – Daniel Donnelly Jun 23 '23 at 16:06

0 Answers0