

A212161


Numbers 6 or 10 modulo 17.


5



6, 10, 23, 27, 40, 44, 57, 61, 74, 78, 91, 95, 108, 112, 125, 129, 142, 146, 159, 163, 176, 180, 193, 197, 210, 214, 227, 231, 244, 248, 261, 265, 278, 282, 295, 299, 312, 316, 329, 333, 346, 350
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OFFSET

0,1


COMMENTS

A001844(N) = N^2 + (N+1)^2 = 4*A000217(N) + 1 is divisible by 17 if and only if N=a(n), n>=0. For the proof it suffices to show that only N=6 and N=10 from {0,1,..,16} satisfy A001844(N)== 0 (mod 17). Note that only primes of the form p= 4*k+1 (A002144) can be divisors of A001844 (see a Wolfdieter Lang comment there giving the reference). Note also that if N^2 + (N+1)^2 == 0 (mod p), with any prime p (necessarily from A002144), then also p1N satisfies this congruence. This explains why 10 = 1716 is the (incongruent) companion of 6.
Partial sums of the sequence 6,4,13,4,13,4,13,4,13,4,13,... (see the o.g.f., and subtract 6 to see the remaining 4, 13=174 periodicity).


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000


FORMULA

Bisection: a(2*n) = 17*n + 6, a(2*n+1) = 17*n + 10, n>=0.
O.g.f.: (6 + 4*x + 7*x^2)/((1x)*(1x^2)).
a(n) = (1/4)*(34*n + 9*(1)^n + 15).  Paolo P. Lava, May 16 2012


EXAMPLE

Divisibility of A001844 by 17:
n=0: A001844(6) = 85 = 5*17 == 0 (mod 17).
n=2: A001844(23) = 1105 = 5*13*17 == 0 (mod 17).
However, 8^2 + 9^2 = 145 == 9 (mod 17) is not divisible by 17 because 8 is not a member of the present sequence.


MATHEMATICA

Table[1/4*(34*n+9*(1)^n+15), {n, 0, 60}] (* Vincenzo Librandi, May 24 2012 *)


PROG

(MAGMA) [1/4*(34*n+9*(1)^n+15): n in [0..60]]; // Vincenzo Librandi, May 24 2012


CROSSREFS

Cf. A047219 (p=5), A212160 (p=13).
Sequence in context: A166305 A047651 A079426 * A132994 A199885 A024499
Adjacent sequences: A212158 A212159 A212160 * A212162 A212163 A212164


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, May 09 2012


STATUS

approved



