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A283394
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a(n) = 3*n*(3*n + 7)/2 + 4.
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5
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4, 19, 43, 76, 118, 169, 229, 298, 376, 463, 559, 664, 778, 901, 1033, 1174, 1324, 1483, 1651, 1828, 2014, 2209, 2413, 2626, 2848, 3079, 3319, 3568, 3826, 4093, 4369, 4654, 4948, 5251, 5563, 5884, 6214, 6553, 6901, 7258, 7624, 7999, 8383, 8776, 9178, 9589, 10009
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OFFSET
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0,1
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COMMENTS
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Sum_{k = 0..n} (3*k + r)^3 is divisible by 3*n*(3*n + 2*r + 3)/2 + r^2: the sequence corresponds to the case r = 2 of this formula (other cases are listed in Crossrefs section).
Also, Sum_{k = 0..n} (3*k + 2)^3 / a(n) gives 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, ... (A005449).
a(n) is even if n belongs to A014601. No term is divisible by 3, 5, 7 and 11.
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LINKS
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FORMULA
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O.g.f.: (4 + 7*x - 2*x^2)/(1 - x)^3.
E.g.f.: (8 + 30*x + 9*x^2)*exp(x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
The inverse binomial transform yields 4, 15, 9, 0 (0 continued), therefore:
a(n) = 4*binomial(n,0) + 15*binomial(n,1) + 9*binomial(n,2).
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MATHEMATICA
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Table[3 n (3 n + 7)/2 + 4, {n, 0, 50}]
LinearRecurrence[{3, -3, 1}, {4, 19, 43}, 50] (* Harvey P. Dale, Mar 02 2019 *)
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PROG
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(Python) [3*n*(3*n+7)/2+4 for n in range(50)]
(Sage) [3*n*(3*n+7)/2+4 for n in range(50)]
(Maxima) makelist(3*n*(3*n+7)/2+4, n, 0, 50);
(Magma) [3*n*(3*n+7)/2+4: n in [0..50]];
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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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