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A076024
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a(n) = (2^n + 4)*(2^n - 1)/6.
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9
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0, 1, 4, 14, 50, 186, 714, 2794, 11050, 43946, 175274, 700074, 2798250, 11188906, 44747434, 178973354, 715860650, 2863377066, 11453377194, 45813246634, 183252462250, 733008800426, 2932033104554, 11728128223914, 46912504507050, 187650001250986, 750599971449514
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OFFSET
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0,3
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COMMENTS
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Provides loss function for folding paper in half. It tells how much normalized paper has been lost with n folds. The sequence sets a limit on the number of times things of finite thickness can be folded in one direction.
Developed with J. R. Gallivan.
Binomial transform of A007051, with leading zero.
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REFERENCES
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Britney C. Gallivan, How to fold paper in half twelve times (an "impossible challenge" solved and explained), Historical Society of Pomona Valley, Pomona California, (2002)
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LINKS
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Eric Weisstein's World of Mathematics, Folding
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FORMULA
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G.f.: x*(1-3*x)/((1-x)*(1-2*x)*(1-4*x)).
E.g.f.: (3*exp(2*x) + exp(4*x) - 4*exp(x))/6 = (exp(2*x)*(2*cosh(x) - sinh(x)) - 2)/3.
a(n) = Sum_{k=0..n} C(n, k)*(3^(k-1) + 1 - 4*0^k/3)/2.
a(n) = Sum_{k=0..n} C(n, k+1)*(3^k + 1).
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EXAMPLE
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a(12) = 2798250 means that for the 12th folding of paper in half that 2798250 times as much material has been lost to potential folding as was lost on the first fold. [corrected by Rick L. Shepherd, May 08 2003]
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MAPLE
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MATHEMATICA
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PROG
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(PARI) a(n) = 1<<(2*n-1)\3 + 1<<(n-1); \\ Kevin Ryde, Nov 26 2022 [replacing previous incorrect code]
(Sage) [(2^n +4)*(2^n -1)/6 for n in (0..30)] # G. C. Greubel, May 04 2019
(GAP) List([0..30], n-> (2^n +4)*(2^n -1)/6) # G. C. Greubel, May 04 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Britney C. Gallivan (ogallivan(AT)verizon.net), Sep 30 2002
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STATUS
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approved
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