A217785 Smallest integer s>n such that 1+2*s+3*s^2+...+n*s^{n-1} is prime.

3, 12, 12, 9, 21, 12, 26, 23, 30, 24, 138, 33, 80, 32, 54, 192, 48, 40, 4500, 48, 50, 192, 30, 88, 32, 114, 178, 48, 45, 42, 356, 41, 53, 138, 174, 66, 44, 990, 120, 819, 2898, 112, 1052, 122, 164, 132, 108, 77, 540, 198, 106, 135, 237, 98, 234, 162, 83, 720, 3870, 135, 188, 1014, 94, 489, 180, 110, 204, 180, 107, 468, 1542, 508, 218, 608, 88, 102, 228, 140, 3890, 93, 361, 1848, 462, 99, 125, 390, 92, 237, 933, 172, 606, 303, 208, 924, 114, 266, 156, 410, 1330

Offset

2,1

Comments

Conjecture: For each n=2,3,... there are infinitely many primes of the form 1+2*s+...+n*s^{n-1}, where s is a positive integer; moreover, we have a(n)<12*n^2.

This is related to the following conjecture of the author: The polynomials s_n(x)=sum_{k=0}^n(k+1)x^k (n=1,2,3,...) are all irreducible over the field of rational numbers; moreover, s_n(x) is reducible modulo every prime if and only if n has the form 8k(k+1), where k is a positive integer.

Sum_{k=1..n} k*s^(k-1) = (1+n*s^(n+1)-s^n*(n+1))/(s-1)^2, see A059045. - R. J. Mathar, Mar 29 2013

Links

Zhi-Wei Sun and Charles R Greathouse IV, Table of n, a(n) for n = 2..1000 (first 450 terms from Sun)

Example

a(20)=4500<12*20^2=4800 since 4500 is the least integer s>20 with 1+2*s+3*s^2+...+20*s^{19} prime.

Mathematica

A[n_, x_]:=A[n, x]=Sum[(k+1)*x^k, {k, 0, n-1}]
Do[Do[If[PrimeQ[A[n, s]]==True, Print[n, " ", s]; Goto[aa]], {s, n+1, 12*n^2-1}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 2, 100}]

Prog

(PARI) f(n, s)=my(t); forstep(k=n, 1, -1, t=s*t+k); t
a(n)=my(s=n); while(!ispseudoprime(f(n, s++)), ); s \\ Charles R Greathouse IV, Mar 25 2013

Crossrefs

Cf. A000040.

Sequence in context: A110345 A018999 A279305 * A168410 A085272 A183508

Adjacent sequences: A217782 A217783 A217784 * A217786 A217787 A217788

Keyword

nonn

Author

Zhi-Wei Sun, Mar 24 2013

Status

approved