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A306367
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a(n) = numerator of (n^2 + 2)/(n + 2).
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15
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1, 1, 3, 11, 3, 27, 19, 17, 33, 83, 17, 123, 73, 57, 99, 227, 43, 291, 163, 121, 201, 443, 81, 531, 289, 209, 339, 731, 131, 843, 451, 321, 513, 1091, 193, 1227, 649, 457, 723, 1523, 267, 1683, 883, 617, 969, 2027, 353, 2211, 1153, 801, 1251, 2603, 451
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OFFSET
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0,3
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COMMENTS
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If P(x) and Q(x) are coprime integral polynomials such that Q(n) > 0 for n >= 0 then the sequence of numerators of the rational numbers P(n)/Q(n) for n >= 0 and the sequence of denominators of P(n)/Q(n) for n >= 0 are both quasi-polynomial in n. In fact, there exists a purely periodic sequence b(n) such that numerator(P(n)/Q(n)) = P(n)/b(n) and denominator(P(n)/Q(n)) = Q(n)/b(n).
Here we take P(n) = n^2 + 2 and Q(n) = n + 2. Cf. A228564 (case P(n) = n^2 + 1, Q(n) = n + 1).
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LINKS
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FORMULA
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O.g.f.: (3*x^17 + x^16 + 11*x^15 + 9*x^14 + 9*x^13 + 19*x^12 + 42*x^11 + 8*x^10 + 50*x^9 + 24*x^8 + 14*x^7 + 16*x^6 + 27*x^5 + 3*x^4 + 11*x^3 + 3*x^2 + x + 1)/(1 - x^6)^3.
a(n) = (n^2 + 2)/b(n), where (b(n))n>=0 is the purely periodic sequence [2, 3, 2, 1, 6, 1, 2, 3, 2, 1, 6, 1, ...] with period 6.
a(n) is a quasi-polynomial in n: a(6*n) = 18*n^2 + 1; a(6*n+1) = 12*n^2 + 4*n + 1; a(6*n+2) = 18*n^2 + 12*n + 3; a(6*n+3) = 36*n^2 + 36*n + 11; a(6*n+4) = 6*n^2 + 8*n + 3; a(6*n+5) = 36*n^2 + 60*n + 27.
a(n) = (n^2 + 2)/( gcd(n+2,2)*gcd(n+2,3) ).
a(n) = (n^2 + 2)/(n + 2) * A060789(n+2).
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MAPLE
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seq(numer( (n^2 + 2)/(n + 2) ), n = 0..100);
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PROG
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(GAP) List([0..100], n->NumeratorRat((n^2+2)/(n+2)); # Muniru A Asiru, Feb 25 2019
(PARI) a(n) = numerator((n^2 + 2)/(n + 2)); \\ Michel Marcus, Feb 26 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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