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 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 (list; graph; refs; listen; history; text; internal format)
 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  p-1-N satisfies this congruence. This explains why 10 = 17-1-6 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=17-4 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)/((1-x)*(1-x^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

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Last modified May 5 23:41 EDT 2021. Contains 343579 sequences. (Running on oeis4.)