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1, 1, 3, 7, 37, 141, 1111, 5923, 62217, 426457, 5599531, 46910271, 739138093, 7318002277, 134523132927, 1536780478171, 32285551902481, 418004290062513, 9879378882159187, 142957467201379447, 3754163975220491061, 60042136224579367741, 1734423756551866870183
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The e.g.f. A(x) satisfies the differential equation (x^2 - 1)*A'(x) + (1 + 2*x - x^2)*A(x) = 0 with A(0) = 1, which leads to . Mathar's recurrence above follows from this.
From Peter Bala, Sep 05 2022: (Start)
The e.g.f. A(x) satisfies the differential equation (x^2 - 1)*A'(x) + (1 + 2*x - x^2)*A(x) = 0 with A(0) = 1, which leads to Mathar's recurrence above.
For k a positive integer, reducing the sequence modulo k produces a purely periodic sequence whose period divides k. For example, modulo 5 the sequence becomes [1, 1, 3, 2, 2, 1, 1, 3, 2, 2, ...] of period 5. (End)
nonn,easy
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a(n) = (1/(2*exp(1))) * (Integral_{t=0..2} t^n*exp(1-abs(1-t)) dt + Integral_{t=0..infinityoo} ((2+t)^n + (-t)^n) * exp(-t) dt). - Groux Roland, Jan 15 2011
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a(n) = (1/(2*exp(1))) * (Integral_{t=0..2} t^n*exp(1-abs(1-t)) dt + Integral_{t=0..infinity} ((2+t)^n + (-t)^n) * exp(-t) dt). - _Groux Roland, _, Jan 15 2011