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A212160
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Numbers that are congruent to {2, 10} mod 13.
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6
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2, 10, 15, 23, 28, 36, 41, 49, 54, 62, 67, 75, 80, 88, 93, 101, 106, 114, 119, 127, 132, 140, 145, 153, 158, 166, 171, 179, 184, 192, 197, 205, 210, 218, 223, 231, 236, 244, 249, 257, 262, 270, 275, 283, 288, 296, 301, 309, 314, 322, 327, 335, 340, 348
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OFFSET
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0,1
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COMMENTS
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A001844(N) = N^2 + (N+1)^2 = 4*A000217(N) + 1 is divisible by 13 if and only if N=a(n), n>=0. For the proof it suffices to show that only N=2 and N=10 from {0,1,..,12} satisfy A001844(N)== 0 (mod 13). 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).
Partial sums of the sequence [2,5,8,5,8,5,8,5,8,...] (see the o.g.f., and subtract 2 to see the 5,8 periodicity).
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LINKS
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FORMULA
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Bisection: a(2*n) = 13*n + 2, a(2*n+1) = 13*n + 10, n>=0.
O.g.f.: (2 + 8*x + 3*x^2)/((1-x)*(1-x^2)).
E.g.f.: ((26*x + 11)*exp(x) - 3*exp(-x))/4. - David Lovler, Aug 09 2022
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EXAMPLE
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n=0: A001844(2) = 13 == 0 (mod 13).
n=3: A001844(23) = 1105 = 85*13 == 0 (mod 13).
However, 8^2 + 9^2 = 145 == 2 (mod 13) is not divisible by 13 because 8 is not a member of the present sequence.
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MAPLE
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MATHEMATICA
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PROG
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(PARI) a(n) = (26*n - 3*(-1)^n + 11)/4 \\ David Lovler, Aug 09 2022
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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