

A056193


Goodstein sequence with a(2)=4: to calculate a(n+1), write a(n) in the hereditary representation base n, then bump the base to n+1, then subtract 1.


7



4, 26, 41, 60, 83, 109, 139, 173, 211, 253, 299, 348, 401, 458, 519, 584, 653, 726, 803, 884, 969, 1058, 1151, 1222, 1295, 1370, 1447, 1526, 1607, 1690, 1775, 1862, 1951, 2042, 2135, 2230, 2327, 2426, 2527, 2630, 2735, 2842, 2951, 3062, 3175, 3290, 3407
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

2,1


COMMENTS

Goodstein's theorem shows that such a sequence is finite (i.e. eventually reaches 0) for any starting value [e.g. if a(2)=1 then a(3)=0; if a(2)=2 then a(5)=0; and if a(2)=3 then a(7)=0]. With a(2)=4 we have a(3*2^(3*2^27+27)1)=0, which is well beyond the 10^(10^8)th term.
The second half of such sequences is declining and the previous quarter is stable.
The resulting sequence 2,3,5,7,3*2^402653211  1, ... (see Comments in A056041) grows too rapidly to have its own entry.


REFERENCES

Goodstein, R. L., On the Restricted Ordinal Theorem, J. Symb. Logic 9, 3341, 1944.


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000
Eric Weisstein's World of Mathematics, Goodstein Sequence.
Wikipedia, Goodstein's Theorem
Reinhard Zumkeller, Haskell programs for Goodstein sequences


EXAMPLE

a(2)=4=2^2, a(3)=3^31=26=2*3^2+2*3+2, a(4)=2*4^2+2*4+21=41=2*4^2+2*4+1, a(5)=2*5^2+2*5+11=60=2*5^2+2*5, a(6)=2*6^2+2*61=83=2*6^2+6+5, a(7)=2*7^2+7+51=109 etc.


PROG

(Haskell) see Link


CROSSREFS

Cf. A056041, A056004, A059934, A057650, A059933, A059935, A059936.
Cf. A215409, A222117, A211378.
Sequence in context: A046963 A022386 A059178 * A196672 A102203 A219668
Adjacent sequences: A056190 A056191 A056192 * A056194 A056195 A056196


KEYWORD

fini,nonn


AUTHOR

Henry Bottomley, Aug 02 2000


EXTENSIONS

Edited by N. J. A. Sloane, Mar 06 2006


STATUS

approved



