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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, 33-41, 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)=3^3-1=26=2*3^2+2*3+2, a(4)=2*4^2+2*4+2-1=41=2*4^2+2*4+1, a(5)=2*5^2+2*5+1-1=60=2*5^2+2*5, a(6)=2*6^2+2*6-1=83=2*6^2+6+5, a(7)=2*7^2+7+5-1=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

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Last modified October 23 17:11 EDT 2014. Contains 248468 sequences.