OFFSET
1,2
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
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).
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)
EXAMPLE
a(4) = 3^(35*4) + 5^(21*4) + 7^(15*4) mod 105 = 17 (prime).
MATHEMATICA
Select[Range[0, 180], PrimeQ[Mod[3^(35#)+5^(21#)+7^(15#), 105]]&] (* Harvey P. Dale, Oct 10 2017 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Kyle D. Balliet, Mar 09 2009
EXTENSIONS
More terms from Michel Marcus, Aug 15 2013
STATUS
approved