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Initial terms were calculated by Peter J.C. Moses; see comment in A129995.
Comment from Max Alekseyev: To check whether 14! divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 +14) it is enough to check whether every prime power q from prime factorization of 14! divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for n=0,1,...,q-1.
Note that 14! = 2^11 * 3^5 * 5^2 * 7^2 * 11 * 13.
It is easy to verify that:
i) 2^11 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every n=0,1,...,2^11-1;
ii) 3^5 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every n=0,1,...,3^5-1;
iii) 7^2 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every n=0,1,...,7^2-1,
except for n=2, 9, 16, 23, 30, 37 (i.e., n of the form 7m+2 but not 49m+44);
iv) 11 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every n=0,1,...,11-1;
v) 13 divides (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^14 + 14) for every n=0,1,...,13-1.
This proves that for k=14, a(n) is nonintegral only for n in the set difference { 7m+2} \ { 49m+44 } .
In simple cases like { 7m+2 } \ { 49m+44 }, one can get an explicit formula.
Note that { 7m+2 } \ { 49m+44 } can be viewed as the arithmetic sequence 7m + 2 where m cannot be 6 modulo 7.
The number of nonnegative integers equal 6 modulo 7 not exceeding m is equal [(m+1)/7] (here [x] stands for the integer part of x).
Therefore in our sequence the (m-[(m+1)/7])-th term equals 7m+2 when m is minimum possible.
Let us find the minimum m satisfying m-[(m+1)/7]=n.
Let t = (m+1) mod 7. Then m-((m+1)-t)/7=n or 6m+t-1=7n, implying that m=(7n-t+1)/6.
Depending on the value of t we have m = [(7n+1)/6] or m = [(7n+1)/6] - 1.
Since m must be minimum possible, we have the second case when [(7n+1)/6] mod 7 > (7n+1) mod 6; and the first case otherwise.
In other words, the explicit formula for m is m = [(7n+1)/6] - 1, if [(7n+1)/6] mod 7 > (7n+1) mod 6; m = [(7n+1)/6], otherwise.
We can also consider all possible residues of n modulo 6*7 and find out that the inequality [(7n+1)/6] mod 7 > (7n+1) mod 6 holds for n mod 42 in { 5, 11, 17, 23, 29, 35, 41 } or alternatively n mod 6 = 5.
Therefore m = [(7n+1)/6] - 1, if n mod 6 = 5; m = [(7n+1)/6], otherwise.
So for the sequence {2,9,16,23,30,37,51,58,65,72,79,86,100,...} we can give formula a(n) = 7*[(7n+1)/6] - 5, if n mod 6 = 5; a(n) = 7*[(7n+1)/6] + 2, otherwise.
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