3
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Define $\sum_{q=1}^{u}q^{u-1}= S(u)$

Problem 1, show that

Let $p$ be a odd prime, if $S(3p)\equiv t \pmod {2p}$ then $t\in\{0,p\}$


Update

Can we show that $\frac{t}{p}=1-$ parity of $\frac{p(p+1)}{2}$

Link for $t/p$ sequence https://oeis.org/A100672

Check solution for above


Now $p_1,p_2$ and $(p_2>p_1)$ are consecutive odd prime gives smallest non negative integer $t_1,t_2$ as

$S(3p_1)\equiv t_1\pmod {2p_1}$ and $S(3p_2)\equiv t_2\pmod {2p_2} $

Problem 2, show that

If $(\frac{p_2-p_1}{2})\equiv0\pmod2$ then $ \mid \frac{t_1}{p_1}-\frac{t_2}{p_2}\mid =0$

If $(\frac{p_2-p_1}{2})\equiv1\pmod2$ then $ \mid \frac{t_1}{p_1}-\frac{t_2}{p_2}\mid =1$

For example

$p_1$ and $p_2 $ are twin prime so there are two possibile cases

Case(1): $S(3p_1)\equiv 0 \pmod {2p_1}$ and $S(3p_2)\equiv 0 \pmod {2p_2}$

Case(2): $S(3p_1)\equiv p_1 \pmod {2p_1}$ and $S(3p_2)\equiv p_2 \pmod {2p_2}$

Means $(t_1,t_2)\in\{(0,0),(p_1,p_2)\}$


I check upto $3\le p \le 1500 $

Source code PARI GP

forprime(p=3, 1500, print ([p,sum(q=0,3*p,q^(3*p-1))%(2*p),nextprime(p+1)-p]))

[p, t,p2-p1]
[3, 3, 2]
[5, 0, 2]
[7, 7, 4]
[11, 11, 2]
[13, 0, 4]
[17, 0, 2]
[19, 19, 4]
[23, 23, 6]
[29, 0, 2]
[31, 31, 6]
[37, 0, 4]
[41, 0, 2]
[43, 43, 4]
[47, 47, 6]
[53, 0, 6]
[59, 59, 2]
[61, 0, 6]
[67, 67, 4]
[71, 71, 2]
[73, 0, 6]
[79, 79, 4]
[83, 83, 6]
[89, 0, 8]
[97, 0, 4]
[101, 0, 2]
[103, 103, 4]
[107, 107, 2]
[109, 0, 4]
[113, 0, 14]
[127, 127, 4]
[131, 131, 6]
[137, 0, 2]
[139, 139, 10]
[149, 0, 2]
[151, 151, 6]
[157, 0, 6]
[163, 163, 4]
[167, 167, 6]
[173, 0, 6]
[179, 179, 2]
[181, 0, 10]
[191, 191, 2]
[193, 0, 4]
[197, 0, 2]
[199, 199, 12]
[211, 211, 12]
[223, 223, 4]
[227, 227, 2]
[229, 0, 4]
[233, 0, 6]
[239, 239, 2]
[241, 0, 10]
[251, 251, 6]
[257, 0, 6]
[263, 263, 6]
[269, 0, 2]
[271, 271, 6]
[277, 0, 4]
[281, 0, 2]
[283, 283, 10]
[293, 0, 14]
[307, 307, 4]
[311, 311, 2]
[313, 0, 4]
[317, 0, 14]
[331, 331, 6]
[337, 0, 10]
[347, 347, 2]
[349, 0, 4]
[353, 0, 6]
[359, 359, 8]
[367, 367, 6]
[373, 0, 6]
[379, 379, 4]
[383, 383, 6]
[389, 0, 8]
[397, 0, 4]
[401, 0, 8]
[409, 0, 10]
[419, 419, 2]
[421, 0, 10]
[431, 431, 2]
[433, 0, 6]
[439, 439, 4]
[443, 443, 6]
[449, 0, 8]
[457, 0, 4]
[461, 0, 2]
[463, 463, 4]
[467, 467, 12]
[479, 479, 8]
[487, 487, 4]
[491, 491, 8]
[499, 499, 4]
[503, 503, 6]
[509, 0, 12]
[521, 0, 2]
[523, 523, 18]
[541, 0, 6]
[547, 547, 10]
[557, 0, 6]
[563, 563, 6]
[569, 0, 2]
[571, 571, 6]
[577, 0, 10]
[587, 587, 6]
[593, 0, 6]
[599, 599, 2]
[601, 0, 6]
[607, 607, 6]
[613, 0, 4]
[617, 0, 2]
[619, 619, 12]
[631, 631, 10]
[641, 0, 2]
[643, 643, 4]
[647, 647, 6]
[653, 0, 6]
[659, 659, 2]
[661, 0, 12]
[673, 0, 4]
[677, 0, 6]
[683, 683, 8]
[691, 691, 10]
[701, 0, 8]
[709, 0, 10]
[719, 719, 8]
[727, 727, 6]
[733, 0, 6]
[739, 739, 4]
[743, 743, 8]
[751, 751, 6]
[757, 0, 4]
[761, 0, 8]
[769, 0, 4]
[773, 0, 14]
[787, 787, 10]
[797, 0, 12]
[809, 0, 2]
[811, 811, 10]
[821, 0, 2]
[823, 823, 4]
[827, 827, 2]
[829, 0, 10]
[839, 839, 14]
[853, 0, 4]
[857, 0, 2]
[859, 859, 4]
[863, 863, 14]
[877, 0, 4]
[881, 0, 2]
[883, 883, 4]
[887, 887, 20]
[907, 907, 4]
[911, 911, 8]
[919, 919, 10]
[929, 0, 8]
[937, 0, 4]
[941, 0, 6]
[947, 947, 6]
[953, 0, 14]
[967, 967, 4]
[971, 971, 6]
[977, 0, 6]
[983, 983, 8]
[991, 991, 6]
[997, 0, 12]
[1009, 0, 4]
[1013, 0, 6]
[1019, 1019, 2]
[1021, 0, 10]
[1031, 1031, 2]
[1033, 0, 6]
[1039, 1039, 10]
[1049, 0, 2]
[1051, 1051, 10]
[1061, 0, 2]
[1063, 1063, 6]
[1069, 0, 18]
[1087, 1087, 4]
[1091, 1091, 2]
[1093, 0, 4]
[1097, 0, 6]
[1103, 1103, 6]
[1109, 0, 8]
[1117, 0, 6]
[1123, 1123, 6]
[1129, 0, 22]
[1151, 1151, 2]
[1153, 0, 10]
[1163, 1163, 8]
[1171, 1171, 10]
[1181, 0, 6]
[1187, 1187, 6]
[1193, 0, 8]
[1201, 0, 12]
[1213, 0, 4]
[1217, 0, 6]
[1223, 1223, 6]
[1229, 0, 2]
[1231, 1231, 6]
[1237, 0, 12]
[1249, 0, 10]
[1259, 1259, 18]
[1277, 0, 2]
[1279, 1279, 4]
[1283, 1283, 6]
[1289, 0, 2]
[1291, 1291, 6]
[1297, 0, 4]
[1301, 0, 2]
[1303, 1303, 4]
[1307, 1307, 12]
[1319, 1319, 2]
[1321, 0, 6]
[1327, 1327, 34]
[1361, 0, 6]
[1367, 1367, 6]
[1373, 0, 8]
[1381, 0, 18]
[1399, 1399, 10]
[1409, 0, 14]
[1423, 1423, 4]
[1427, 1427, 2]
[1429, 0, 4]
[1433, 0, 6]
[1439, 1439, 8]
[1447, 1447, 4]
[1451, 1451, 2]
[1453, 0, 6]
[1459, 1459, 12]
[1471, 1471, 10]
[1481, 0, 2]
[1483, 1483, 4]
[1487, 1487, 2]
[1489, 0, 4]
[1493, 0, 6]
[1499, 1499, 12]

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We can be more specific in the statement of problem 1 (update: this is equivalent to your expression for $t/p$):

Let $p$ be an odd prime. Modulo $2p$, we have $$S(3p)\equiv \begin{cases} 0, & \text{if } p \equiv 1 \pmod 4 \\ p, & \text{if } p \equiv 3 \pmod 4. \end{cases} $$

To prove this statement, first note that $$S(3p) \equiv \frac{3p +1}{2} \pmod 2,$$ so $S(3p)$ is even if and only if $p \equiv 1 \pmod 4$.

Now we just need to show that $p$ divides $S(3p)$. If we write $$S(3p)\equiv 3\sum_{k =1}^{p - 1} k^{3p - 1} \pmod p,$$ then it is clear that $p$ divides $S(3p)$ when $p = 3$. For $p > 3$, note that $p - 1$ does not divide $3p -1$, so we can use a standard argument using primitive roots to conclude that $$\sum_{k =1}^{p - 1} k^{3p - 1} \equiv 0 \pmod p,$$ so $p$ divides $S(3p)$.


With this more specific formula for $S(3p)$ mod $2p$, your second problem becomes trivial. Can you fill in with the details?

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