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A184218
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a(n) = largest k such that A000217(n+1) = A000217(n) + (A000217(n) mod k), or 0 if no such k exists.
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2
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0, 0, 0, 0, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377
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OFFSET
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1,5
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COMMENTS
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From the definition, a(n) = A000217(n) - (n + 1) if A000217(n) - (n + 1) > (n + 1), or 0 otherwise, where A000217 are the triangular numbers.
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LINKS
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FORMULA
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a(n) = (n+1)*(n-2)/2 = A000096(n-2) for n >= 5 and a(n) = 0 for n <= 4. - M. F. Hasler, Jan 10 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 7.
G.f.: x^5*(5*x^2 - 13*x + 9)/(1 - x)^3. (End)
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EXAMPLE
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For n = 3 we have A000217(3) = 6, A000217(4) = 10; there is no k such that 10 - 6 = 4 = (6 mod k), hence a(3) = 0.
For n = 5 we have A000217(5) = 15, A000217(6) = 21; 9 is the largest k such that 21 - 15 = 6 = (15 mod k), hence a(5) = 9; a(5) = A000217(5) - (5 + 1) = 15 - 6 = 9.
For n = 24 we have A000217(24) = 300, A000217(25) = 325; 275 is the largest k such that 325 - 300 = 25 = (300 mod k), hence a(24) = 275; a(24) = A000217(24) - (24 + 1) = 275.
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MATHEMATICA
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Join[{0, 0, 0, 0}, LinearRecurrence[{3, -3, 1}, {9, 14, 20}, 100]] (* G. C. Greubel, Jun 22 2016 *)
lim = 10^4; Table[SelectFirst[Reverse@ Range@ lim, Function[k, PolygonalNumber[n + 1] == # + Mod[#, k] &@ PolygonalNumber@ n]], {n, 53}] /. {k_ /; MissingQ@ k -> 0, k_ /; k == lim -> 0} (* Michael De Vlieger, Jun 30 2016, Version 10.4 *)
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PROG
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(Magma) [0, 0, 0, 0] cat [(n+1)*(n-2)/2: n in [5..60]]; // Vincenzo Librandi, Jun 22 2016
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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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