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May 24

About the prime values of the sigma function and Ramanujan tau function

Let $\sigma (n)$ be the sum-of-divisors function (sequence A000203 in the OEIS) and $\tau (n)$ be the Ramanujan tau function (sequence A000594 in the OEIS).

1. If $\sigma (n)$ or/and $\tau (n)$ is prime, must $n$ be of the form $p^{q-1}$ with $p,q$ both primes?

2. Does there exist a positive integer $n$ such that $\sigma (n)$ and $\tau (n)$ are both primes? If so, do there exist infinitely many such positive integers $n$ ? 218.187.70.249 (talk) 04:05, 24 May 2023 (UTC)[reply]

The answer to the first half of the first question (i.e. for

\sigma (n)

) is yes. See OEIS A023194, particularly the last comment. GalacticShoe (talk) 04:24, 24 May 2023 (UTC)[reply]

The answer to the second half of the first question (i.e. for

\tau (n)

) is probably yes. In particular, if

\tau (n)

is an odd prime, then

n

is indeed of the specified form. Whether

|\tau (n)|=2

has any solutions is an open problem. See OEIS A135430. GalacticShoe (talk) 04:34, 24 May 2023 (UTC)[reply]

The answer to the first half of the second question is yes. The first few terms of the sequence of

n

for which

\sigma (n)

and

\tau (n)

are both primes are

458329

,

46117681

,

49182169

,

56957209

,

104509729

,

\ldots

. GalacticShoe (talk) 04:42, 24 May 2023 (UTC)[reply]

Currently how many such positive integers

n

(i.e.

\sigma (n)

and

\tau (n)

are both primes) are known? (Could you create an OEIS sequence for such

n

?) Are there any known large (> 10¹⁰⁰⁰) such

n

? (For large

n

, unproven probable primes

\sigma (n)

and

\tau (n)

are allowed, also negative primes

\tau (n)

are allowed) 218.187.70.249 (talk) 05:08, 24 May 2023 (UTC)[reply]

First question is hard to quantify, since there doesn't seem to have been any literature studying this particular sequence of integers. As for the second question, Lygeros and Rozier's paper mentioned in A135430 has a list of large

n

for which

\tau (n)

is prime or probably prime; I was not able to personally confirm whether any of those

n

have the property that

\sigma (n)

is prime or probably prime due to their sheer sizes. GalacticShoe (talk) 13:54, 28 May 2023 (UTC)[reply]

It took less than a minute to test the values in their table 3 with OpenPFGW. The only cases where

\sigma (p^{q-1})

is also prime are 677² = 458329, and 947⁴⁰.

\tau (n)

is negative in both cases but the absolute value is prime. PrimeHunter (talk) 15:19, 28 May 2023 (UTC)[reply]

I have computed nine solutions above 10¹⁰⁰⁰ with PARI/GP and OpenPFGW. PARI/GP verification:.

v=[1236727^190, 3283991^180, 3666881^180, 11534591^172, 8051963^180,\
   9224857^180, 12432997^178, 18623489^180, 7723841^192];
for(i=1, #v, n=v[i]; s=sigma(n); t=abs(ramanujantau(n));\
print([#digits(n), #digits(s), #digits(t), ispseudoprime(s), ispseudoprime(t)]))

Output:

[1158, 1158, 6366, 1, 1]
[1173, 1173, 6451, 1, 1]
[1182, 1182, 6500, 1, 1]
[1215, 1215, 6681, 1, 1]
[1244, 1244, 6837, 1, 1]
[1254, 1254, 6896, 1, 1]
[1263, 1263, 6946, 1, 1]
[1309, 1309, 7198, 1, 1]
[1323, 1323, 7274, 1, 1]

ispseudoprime makes a strong probable prime test.

\tau (n)

is positive for 1236727¹⁹⁰, 3666881¹⁸⁰, 12432997¹⁷⁸, and negative for the other six. PrimeHunter (talk) 02:02, 29 May 2023 (UTC)[reply]

OK, additional questions:

If $p$ is prime, is there always a prime $q$ such that $\sigma (p^{q-1})$ is prime? If so, are there infinitely many such primes $q$ ?
If $p$ is prime, is there always a prime $q$ such that $\tau (p^{q-1})$ is prime? If so, are there infinitely many such primes $q$ ?
If $q$ is prime, is there always a prime $p$ such that $\sigma (p^{q-1})$ is prime? If so, are there infinitely many such primes $p$ ?
If $q$ is prime, is there always a prime $p$ such that $\tau (p^{q-1})$ is prime? If so, are there infinitely many such primes $p$ ?
For each prime $p<1000$ , find the smallest prime $q$ such that $\sigma (p^{q-1})$ is prime.
For each prime $p<1000$ , find the smallest prime $q$ such that $\tau (p^{q-1})$ is prime.
For each prime $q<1000$ , find the smallest prime $p$ such that $\sigma (p^{q-1})$ is prime.
For each prime $q<1000$ , find the smallest prime $p$ such that $\tau (p^{q-1})$ is prime.

211.75.79.246 (talk) 06:38, 29 May 2023 (UTC)[reply]

Note: In questions 5 to 8, we allow strong probable primes > 10¹⁰⁰⁰. 211.75.79.246 (talk) 06:45, 29 May 2023 (UTC)[reply]

Questions 1 through 4 imply that there are infinitely many numbers for which

\sigma (n)

or

\tau (n)

are prime, and are thus harder than the question of whether there are infinitely many numbers for which

\sigma (n)

or

\tau (n)

are prime. As with many problems of the form "are there infinitely many numbers such that [some property regarding primality]", I expect that the answer is "probably yes, but no one has proven it yet." The best that can be done is heuristic predictions. Lygeros and Rozier's paper mentioned earlier, for example, gives an estimation on the probability that

\tau (p^{q-1})

is prime. GalacticShoe (talk) 14:22, 29 May 2023 (UTC)[reply]

For question 5, see OEIS A065854. GalacticShoe (talk) 14:29, 29 May 2023 (UTC)[reply]

For question 7, see OEIS A123487. GalacticShoe (talk) 14:39, 29 May 2023 (UTC)[reply]

Areas in projection of regular icosahedron

Assume an regular icosahedron of edge 1 unit with each side colored differently, resting on the X-Y plane. A plane at Z=10 (or any height completely above the icosahedron) is colored based on the color of the Icosahedron on the face directly below it. This will have 10 colors. The Central triangle has area root(3)/4 , the others are a set of three triangles which share an edge with the central triangle and then six even more distorted that only share a vertex. What are the areas of the other triangles? I'm thinking the three that share an edge would be the area of the central times the absolute value of the cosine of the dihedral angle between sides (which according to the article is 138.189685° = arccos(−^√5⁄₃)), but I have no idea how to figure out the area of the projection of the other six triangles.Naraht (talk) 09:17, 24 May 2023 (UTC)[reply]

If you have the

(x,y,z)

coordinates of the vertices of any of the triangles, the vertices of its orthogonal projection on any z-plane are obtained by leaving out the

z

coordinate, resulting in a set of three

(x,y)

coordinates. Several formulas for the area of a triangle can then be used. --Lambiam 16:15, 24 May 2023 (UTC)[reply]

At Regular icosahedron § Cartesian coordinates you can find the coordinates of the vertices (for edge length = 2). By ignoring the z-coords, these will give you the vertices of the projected triangles. Pick a triangle of each type, one of whose edges is parallel to an axis (eg identical x-coords on two vertices), and 1/2 base x height will give the area. To check your math, the total projected area should equal the area of a hexagon with the same circumradius as the icosahedron. -- Verbarson ^talk_edits 16:22, 24 May 2023 (UTC)[reply]

A potential alternative approach is based on the fact that the area of the orthogonal projection of any planar shape – not only polygons – is equal to the original area times the absolute value of the cosine of the angle between the two planes involved. Once you know the angles the polyhedral faces make with the z plane, you're basically done. --Lambiam 16:42, 24 May 2023 (UTC)[reply]

This was mentioned by the OP, but only for three of the triangles. The article Projected area mostly covers the cosine rule, though the article seems to be a stub and needs a lot of work. So basically this amounts to finding the cosines of the angles at the origin between the vertices of the dual, aka the dodecahedron. This can be done easily using dot products. --RDBury (talk) 16:56, 24 May 2023 (UTC)[reply]

The cosine for the "not quite adjacent" triangles is 1/3. I did this by direct calculation but in hindsight it should have been clear because of the relationship between the dodecahedron and the cube (or the relationship between the icosohedron and the octahedron). You can pick 8 vertices from a dodecahedron which form a cube, or dually, you can pick 8 faces from an icosohedron which, when extended, form an octahedron. So really the angle between the faces is the same as the angle between adjacent faces in the octahedron. The relevant coordinates involve only integers so the calculations are much easier. --RDBury (talk) 17:51, 24 May 2023 (UTC)[reply]