by Michael Thomas De Vlieger, St. Louis, Missouri, 2020 0730.
Consider a(n) = A330170(n) = 2n + 3n + 6n − 1. For every prime p, there exists n ≥ 1 such that p divides a(n). Further, for every prime p, p | a(n) for n ≡ −1 (mod p − 1), as a consequence of Fermat’s Little Theorem.
All terms of a(n) are even, since we have as addends three perfect powers, two odd, one even, and we offset by 1.
The sequence is a linear recurrence (12,−47,72,−36), with generating function 2x(5 − 36x + 72x² − 36x³) / ((1 − x)(1 − 2x)(1 − 3x)(1 − 6x)), starting {10, 48, 250, 1392, 8050, 47448, 282250, 1686432, …}. See Codes 1.1, 1.2, and 1.3 to generate a(n).
We examine residues n (mod p – 1) such that p | a(n), knowing that we have p | a(n) for n ≡ −1 (mod p − 1).
We know that for n ≡ 0 (mod 1), 2 | a(n), i.e., all terms of a(n) are even.
For n ≡ 0 (mod 2), 3 | a(n), i.e., all terms a(k) are divisible by 3 for even k.
For n ≡ ±1 (mod 4), 5 | a(n), i.e., all terms a(k) are divisible by 5 for odd k.
For n ≡ −1 (mod 6), 7 | a(n), e.g., 7 | a(5) = 8050 = 2 × 5² × 7 × 23, 7 | a(11) = 2 × 54 × 7 × 13 × 3191, etc.
For n ≡ −2 or −1 (mod 10), 11 | a(n), 11 | a(8) = 1686432, 11 | a(9) = 10097890, etc.
…
For n ≡ {5, 10, −6, −1} (mod 22), 23 | a(n); 23 | a(5) = 8050 = 2 × 5² × 7 × 23, …
For n ≡ {12, 25, 36, 47, −36, −23, −12, −1} (mod 96), 97 | a(n), e.g., 97 | a(12) = 24 × 3² × 97² × 1607.
Therefore we observe, that though p | a(n) for n ≡ −1 (mod p − 1), for certain primes we have additional residues r (mod p − 1) that have a(r) divisible by p. We might write a sequence b(n) = number of residues r such that p | A330170(n) with n ≡ r (mod p − 1) (see code 2.1). The first terms are:
1, 1, 2, 1, 2, 1, 1, 2, 4, 4, 1, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 1, 1, 8, 4, 1, ...
Further, we might write a sequence c(n) containing primes q = π(n) for which b(n) = 1 (see code 3.1):
2, 3, 7, 13, 17, 31, 37, 61, 79, 83, 89, 103, 109, 131, 151, 179, 223, 233, 251, ...
In other words, these are the n-th primes q for indices n of 1s in b(n). For these primes q, q | a(n) ONLY for n ≡ −1 (mod q − 1), and for no other residue.
We define a sequence s(n) that lists the least prime p with n residues r such that p | a(n) for n ≡ r (mod p − 1).
Let’s start by examining the record transform of b(n), finding records m at index n, also listing π(n) in Table 4.1 (See codes 4.1, 4.2):
m n π(n)
1 1 2
2 3 5
4 9 23
8 25 97
10 83 431
12 117 643
16 197 1201
20 456 3221
56 847 6553
72 3358 31177
...
The sparsity of Table 4.1, given 10000 terms of b(n), suggests examining the least indices n where a given value m first appears in b(n) in Table 4.2 (See code 4.3):
m n π(n)
-------------------
1 1 2
2 3 5
3 37 157
4 9 23
5 164 971
6 35 149
7 732 5531
8 25 97
9 85 439
10 83 431
11 7774 79333
12 117 643
13 8113 83071
14 1766 15121
15 1876 16111
16 197 1201
17 848 6563
18 1255 10243
19 8884 91961
...
We might call column π(n) in this sequence s(n). Hence, the smallest prime p | A330170(n) ≡ r (mod p − 1) such that there is only 1 residue r is 2; and for 2 residues, 5, 3 residues 157, etc. Is s(n) defined for all n? We note that s(p) for p prime is generally greater than s(n) for adjacent composite n, but curiously not for p = 17, and that s(n) seems relatively small for n that are perfect powers of 2.
Using s(n), we can produce a triangle of residues r for which p | A330170(n) ≡ r (mod p − 1) for p in s(n) (see code 5.1):
0
1 3
89 124 155
5 10 16 21
63 280 562 621 969
37 75 83 111 139 147
789 1579 2369 3159 3949 4739 5529
12 25 36 47 60 73 84 95
91 134 145 237 280 291 383 426 437
42 85 128 171 214 257 300 343 386 429
...
Thus clearly 2 | a(n) for all n;
5 | a(n) for n ≡ {1, 3} (mod 4), in other words, odd n;
157 | a(n) for n ≡ {89, 124, 155} (mod 156);
23 | a(n) for n ≡ {5, 10, 16, 21} (mod 22); etc.
We observe 5 | {10, 250, 8050, 282250, …},
157 |
1800782593726645086383198953559179330816574049035635349624276222902690;
23 | {8050, 60526248, …}, etc.
This concludes our examination of OEIS A330170.
Code 1.1: Generate A330170 algebraically:
Array[2^# + 3^# + 6^# - 1 &, 12]
Code 1.2: Generate A330170 as a linear recurrence::
LinearRecurrence[{12, -47, 72, -36}, {10, 48, 250, 1392}, 12]
Code 1.3: Generate A330170 via generating function:
Rest@ CoefficientList[Series[2 x (5 - 36 x + 72 x^2 - 36 x^3)/((1 - x) (1 - 2 x)*(1 - 3 x)*(1 - 6 x)), {x, 0, 12}], x]
Block[{nn = 10^3, s},
s = Array[2^# + 3^# + 6^# - 1 &, Prime@ nn];
Array[Function[p, Count[s[[1 ;; p - 1]], _?(Mod[#, p] == 0 &)]]@
Prime@ # &, nn]]
Block[{nn = 10^3, s},
s = Array[2^# + 3^# + 6^# - 1 &, Prime@ nn];
Position[#, 1]&@ Array[Function[p, Count[s[[1 ;; p - 1]], _?(Mod[#, p] == 0 &)]]@
Prime@ # &, nn]
]
Code 4.1: Records transform of b(n), assuming variable b stores b(n):
Union@ FoldList[Max, b]
Code 4.2: Indices of records in b(n), assuming variable b stores b(n):
Map[FirstPosition[b, #][[1]] &, Union@ FoldList[Max, b]]
We might prepend Prime@ in front of the above code so as to yield the corresponding primes π(n).
Code 4.3: Produce Table 4.2, assuming variable s stores s(n):
Map[{#1, #2, Prime[#2]} & @@ {#, FirstPosition[s, #][[1]]} &,
#[[1 ;; LengthWhile[Differences@ #, # == 1 &]]] &@ Union@ s] // TableForm
Code 5.1: Produce triangle T(n), assuming variable a stores a(n), and variable s stores s(n):
Map[Function[p,
Position[a[[1 ;; p - 1]],
_?(Mod[#, p] == 0 &)][[All, 1]] ]@ Prime@ # &,
Map[FirstPosition[s, #][[1]] &,
Union[s][[1 ;; 10]] ] ] // TableForm
A330170: a(n) = 2n + 3n + 6n − 1.
b(n) = Number of residues r such that prime(n) = p | A330170(n) for n ≡ r (mod p − 1).
c(n) = Primes p such that p divides A330170(n) only for n ≡ −1 (mod p − 1).
s(n) = Least prime p with n residues r such that p | A330170(π(p)) for π(p) ≡ r (mod p − 1).
T(n): row n lists residues r (mod p − 1) such that prime A336685(n) = p | A330170(π(p)) for π(p) ≡ r.
2020 0730 2200 Created.
2020 0802 0745 Revised, renamed A336683 from A330170.
2020 1004 2130 Reverted to A330170.