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COMMENTS
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2, 3, 5, 7, 11, 13 are first six consecutive prime numbers.
All terms are odd. In fact, assuming m even and b(k) = 4^k + 9^k + 25^k + 49^k + 121^k + 169^k, for
. k == 0, 2, 4 (mod 6), b(k) is divisible by 5;
. k == 1, 5 (mod 6), b(k) is divisible by 377 = 13*29;
. k == 3 (mod 6), b(k) is divisible by 29. (End)
For n odd:
Let t(n) = 2^n + 3^n + 5^n + 7^n + 11^n + 13^n; then t(n) is divisible by prime p for certain pairs (p, n mod (p-1)):
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p n mod (p-1) such that p|t(n)
== ============================
2 -
3 -
5 -
7 -
11 9
13 -
17 5
19 9
23 11
29 3
31 15
37 21, 29
41 1, 19
43 11, 33, 37
47 23
53 -
59 29, 55
...
The smallest prime p that divides t(n) at more than three values of n mod (p-1) is 313: 313|t(n) when n mod 312 is any of the four values {39, 117, 195, 273}, i.e., when n mod (312/4 = 78) = 39.
The smallest prime p that divides t(n) at more than four values of n mod (p-1) is 3041: 3041|t(n) when n mod 3040 is any of the 16 values {95, 285, 475, 665, 855, 1045, 1235, 1425, 1615, 1805, 1995, 2185, 2375, 2565, 2755, 2945}, i.e., when n mod (3040/16 = 190) = 95. (End)
No other terms than the four terms cited less than 25000. - Robert G. Wilson v, Mar 07 2018
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