A131189 Numbers n >= 0 such that d(n) = (n^1 + 1)*(n^2 + 2) ... (n^14 + 14) / 14!, e(n) = (n^1 + 1)*(n^2 + 2) ... (n^15 + 15) / 15!, and f(n) = (n^1 + 1)*(n^2 + 2) ... (n^16 + 16) / 16! take nonintegral values.

2, 9, 16, 23, 30, 37, 51, 58, 65, 72, 79, 86, 100, 107, 114, 121, 128, 135, 149, 156, 163, 170, 177, 184, 198, 205, 212, 219, 226, 233, 247, 254, 261, 268, 275, 282, 296, 303, 310, 317, 324, 331, 345, 352, 359, 366, 373, 380, 394, 401, 408, 415, 422, 429

Offset: 1

Author

Alexander R. Povolotsky, Sep 25 2007

Keywords

• nonn
• easy

Comments

Initial terms were calculated by Peter J. C. Moses; see comment in A129995.

From Max Alekseyev: (Start)

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 the 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 to 6 modulo 7 not exceeding m is equal to floor((m+1)/7).

Therefore in our sequence the (m-floor((m+1)/7))-th term equals 7m+2 when m is minimum possible.

Let us find the minimum m satisfying m - floor((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 = floor((7n+1)/6) or m = floor((7n+1)/6) - 1.

Since m must be the minimum possible, we have the second case when floor((7n+1)/6) mod 7 > (7n+1) mod 6; and the first case otherwise.

In other words, the explicit formula for m is m = floor((7n+1)/6) - 1, if floor((7n+1)/6) mod 7 > (7n+1) mod 6; m = floor((7n+1)/6), otherwise.

We can also consider all possible residues of n modulo 6*7 and find out that the inequality floor((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 = floor((7n+1)/6) - 1, if n mod 6 = 5; m = floor((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*floor((7n+1)/6) - 5, if n mod 6 = 5; a(n) = 7*floor((7n+1)/6) + 2, otherwise.

(End)

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).

Programs

Maple

a:= n-> `if`((n-1) mod 6 = 5, 7*floor((7*n-6)/6)-5, 7*floor((7*n-6)/6)+2):
seq(a(n), n=1..80);  # Alois P. Heinz, Aug 08 2013

Mathematica

LinearRecurrence[{1, 0, 0, 0, 0, 1, -1}, {2, 9, 16, 23, 30, 37, 51}, 50] (* G. C. Greubel, Feb 19 2017 *)

PARI

x='x+O('x^50); Vec(x*(12*x^6+7*x^5+7*x^4+7*x^3+7*x^2+7*x+2) / ((x-1)^2*(x+1)*(x^2-x+1)*(x^2+x+1))) \\ G. C. Greubel, Feb 19 2017

Formula

a(n) = 7*floor((7n+1)/6) - 5, if n mod 6 = 5; a(n) = 7*floor((7n+1)/6) + 2, otherwise. For e(n) and g(n), proof is similar; see also the comments. - Max Alekseyev

G.f.: x*(12*x^6+7*x^5+7*x^4+7*x^3+7*x^2+7*x+2) / ((x-1)^2*(x+1)*(x^2-x+1)*(x^2+x+1)). - Colin Barker, Aug 08 2013

Extensions

More terms from Colin Barker, Aug 08 2013