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A079272
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a(n) = ((2n+1)*3^n - 1)/2.
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4
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4, 22, 94, 364, 1336, 4738, 16402, 55768, 186988, 620014, 2037190, 6643012, 21523360, 69353050, 222408058, 710270896, 2259952852, 7167279046, 22664098606, 71479080220, 224897593864, 706073841202, 2212364702434, 6919523643784, 21605859540796, 67359444450718
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OFFSET
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1,1
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COMMENTS
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Sequence corresponds to the maximum chain length of a variant of the classical puzzle whereby, under agreed terms, a ringed golden chain asset of a(n) links, when judiciously fragmented into n opened links (through n cuts) and n pieces of lengths (2n+1), (2n+1)*3, (2n+1)*3^2, ..., (2n+1)*3^(n-1), may be used to sequentially settle for payment equivalent up to a(n)-link cost, a link-cost at a time, with swapping allowed with identical fragments owned by the creditor.
a(n) = the difference of the sum of the terms in row(n) and row(n-1) in a triangle with first column T(n-1,0) = n-1 and T(i,j) = T(i-1,j-1) + T(i,j-1) + T(i+1,j-1). - J. M. Bergot, Jul 05 2018
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LINKS
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FORMULA
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a(n) = 7*a(n-1) - 15*a(n-2) + 9*a(n-3).
G.f.: 2*x*(2-3*x)/((1-x)*(1-3*x)^2). (End)
E.g.f.: ((1+3*x)*sinh(x) + 3*x*cosh(x))*exp(2*x). - G. C. Greubel, Apr 14 2019
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EXAMPLE
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For instance, the 4 fragmented chains of original length a(4) = 364 into
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1 + 9 + 1
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243 27
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1 + 81 + 1
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when swapped with identical fragments owned by the creditor, enable the sequential payment, a link-cost at a time, for an expense up to 364 link-costs.
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MAPLE
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a:=n->sum (3^j*n^binomial(j, n), j=0..n): seq(a(n), n=1..25); # Zerinvary Lajos, Apr 18 2009
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MATHEMATICA
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Rest@ CoefficientList[Series[2x(2-3x)/((1-x)(1-3x)^2), {x, 0, 25}], x] (* Michael De Vlieger, Jul 06 2018 *)
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PROG
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(PARI) vector(25, n, ((2*n+1)*3^n - 1)/2) \\ G. C. Greubel, Apr 14 2019
(Sage) [((2*n+1)*3^n - 1)/2 for n in (1..25)] # G. C. Greubel, Apr 14 2019
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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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EXTENSIONS
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STATUS
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approved
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