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A004116
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a(n) = floor((n^2 + 6n - 3)/4).
(Formerly M2524)
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10
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1, 3, 6, 9, 13, 17, 22, 27, 33, 39, 46, 53, 61, 69, 78, 87, 97, 107, 118, 129, 141, 153, 166, 179, 193, 207, 222, 237, 253, 269, 286, 303, 321, 339, 358, 377, 397, 417, 438, 459
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OFFSET
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1,2
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COMMENTS
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a(n)-3 is the maximal size of a regular triangulation of a prism over a regular n-gon.
Solution to a postage stamp problem with 2 denominations.
This sequence is half the degree of the denominator of a certain sequence of rational polynomials defined in the referenced paper by G. Alkauskas. Although this fact is not documented in the paper it can be verified by running the author's code and evaluating degree(denom(...)). - Stephen Crowley, Sep 18 2011
Consider quadratic functions x^2+ax+b. Then a(n) is the number of these functions with 0 <= a+b < n, modulo changing x to x+c for a constant c.
For a(6)=17, four functions are excluded, because:
x^2 + 2x + 1 = (x+1)^2 + 0(x+1) + 0
x^2 + 2x + 2 = (x+1)^2 + 0(x+1) + 1
x^2 + 2x + 3 = (x+1)^2 + 0(x+1) + 2
x^2 + 3x + 2 = (x+1)^2 + 1(x+1) + 0 (End)
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(n) = floor((1/4)*n^2 + (3/2)*n + 1/4) - 1.
a(n) = (1/8)*(-1)^(n+1) - 7/8 + (3/2)*n + (1/4)*n^2.
O.g.f.: x*(1 + x - x^3)/((1 - x)^3*(1 + x)).
E.g.f.: (8 + sinh(x) - cosh(x) + (2*x^2 + 14*x - 7)*exp(x))/8.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
a(n) = Sum_{k=0..n-1} A266977(k). (End)
Sum_{n>=1} 1/a(n) = 2 + tan(sqrt(13)*Pi/2)*Pi/sqrt(13) - cot(sqrt(3)*Pi)*Pi/(2*sqrt(3)). - Amiram Eldar, Aug 13 2022
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MAPLE
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MATHEMATICA
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Table[Floor[(n^2 + 6 n - 3)/4], {n, 40}] (* or *)
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PROG
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(PARI) a(n)=(n^2+6*n-3)>>2
(Magma) [Floor( (n^2 + 6*n - 3)/4 ) : n in [1..50]]; // Vincenzo Librandi, Sep 19 2011
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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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