A157932 Numbers k such that (3^(35*k) + 5^(21*k) + 7^(15*k)) mod 105 is prime.

0, 4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48, 52, 54, 56, 60, 64, 66, 68, 72, 76, 78, 80, 84, 88, 90, 92, 96, 100, 102, 104, 108, 112, 114, 116, 120, 124, 126, 128, 132, 136, 138, 140, 144, 148, 150, 152, 156, 160, 162, 164, 168, 172, 174

Offset: 1

Author

Kyle D. Balliet, Mar 09 2009

Keywords

• nonn
• easy

Comments

Let b(k) = (3^(35*k) + 5^(21*k) + 7^(15*k)) mod 105, then sequence {b(k)} is 3, repeat (60, 68, 75, 17, 30, 23, 60, 47, 75, 38, 30, 2), with primes 3, 17, 23, 47, 2. First differences of {a(n)} are 4, 2, 2, 4, 4, 2, 2, 4, .... - Michel Marcus, Aug 15 2013

3^(35*k) + 5^(21*k) + 7^(15*k) = (4^k)*(3^k + 5^k + 7^k) mod 105, then by the division algorithm a simple proof yields that only numbers k of the form 24*m, 24*m+4, 24*m+6, 24*m+8, 24*m+12, 24*m+16, 24*m+18, 24*m+20 will be congruent to a prime modulo 105. Thus the pattern 4, 2, 2, 4, 4, 2, 2, ... will repeat infinitely. - Kyle D. Balliet, Jan 01 2014

Even numbers that can be written as the sum of 3 of their divisors, not necessarily distinct (see A355200). Also, numbers k of the form 12*m, 12*m+4, 12*m+6, 12*m+8. - Bernard Schott, Sep 08 2023

Examples

			a(4) = 3^(35*4) + 5^(21*4) + 7^(15*4) mod 105 = 17 (prime).
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Crossrefs

Equals {0} Union (A355200 \ A016945) <=> subsequence of even terms in A355200.

Programs

Mathematica

Select[Range[0,180],PrimeQ[Mod[3^(35#)+5^(21#)+7^(15#),105]]&] (* Harvey P. Dale, Oct 10 2017 *)

PARI

isok(n) = isprime((3^(35*n)+5^(21*n)+7^(15*n)) % 105); \\ Michel Marcus, Aug 15 2013

PARI

a(n)=n\4*12+[-4,0,4,6][n%4+1] \\ Charles R Greathouse IV, Dec 27 2013

PARI

is(n)=n%=12;n==0||n==4||n==6||n==8 \\ Charles R Greathouse IV, Dec 27 2013

PARI

a(n) = (-6-(-I)^n-I^n+6*n)/2 \\ Colin Barker, Oct 19 2015

PARI

concat(0, Vec(2*x^2*(2*x^2-x+2)/((x-1)^2*(x^2+1)) + O(x^100))) \\ Colin Barker, Oct 19 2015

Formula

3n - 4 <= a(n) <= 3n - 2. - Charles R Greathouse IV, Dec 27 2013

From Colin Barker, Oct 19 2015: (Start)

a(n) = (-6 - (-i)^n - i^n + 6*n)/2, where i = sqrt(-1).

G.f.: 2*x^2*(2*x^2-x+2) / ((x-1)^2*(x^2+1)). (End)

Extensions

More terms from Michel Marcus, Aug 15 2013