

A076024


a(n)=(2^n+4)*(2^n1)/6.


2



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

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.
Second binomial transform of A078008(n1)+0^n/2.  Paul Barry, Apr 27 2004


REFERENCES

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)


LINKS

Ivan Panchenko, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Folding
Index to sequences with linear recurrences with constant coefficients, signature (7,14,8).


FORMULA

a(n) = Sum_{k <= n} A007582(k).
G.f.: x(13x)/((1x)(12x)(14x)); E.g.f.: exp(2x)/2+exp(4x)/62exp(x)/3; E.g.f.: exp(2x)(2cosh(x)/3sinh(x)/3)2/3; a(n)=sum{k=0..n, C(n, k)(3^(k1)+14*0^k/3)/2}; a(n)=sum{k=0..n, C(n, k+1)(3^k+1)}.


EXAMPLE

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.


PROG

(PARI) th(n)=if(n<1, y, th(n1)*(th(n1)+1)/2) and for(n=2, 30, print1(numerator(polcoeff(th(n), 2^n3))", "))


CROSSREFS

Cf. A007582.
Sequence in context: A211308 A087945 A051924 * A062807 A117421 A034743
Adjacent sequences: A076021 A076022 A076023 * A076025 A076026 A076027


KEYWORD

easy,nonn


AUTHOR

Britney C. Gallivan (ogallivan(AT)verizon.net), Sep 30 2002


EXTENSIONS

Example corrected by Rick L. Shepherd, May 08 2003


STATUS

approved



