A022019 Define the sequence S(a(0), a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0 . This is S(2,32).

2, 32, 513, 8224, 131841, 2113576, 33883265, 543191088, 8708032065, 139600638008, 2237972711489, 35877499765312, 575161163852417, 9220552339712072, 147816978601123073, 2369690920646861904

Offset

0,1

Comments

There is a discrepancy between terms and definition. The definition constructs 2, 32, 513, 8225, 131873, 2114346, 33899730,... - R. J. Mathar, Feb 10 2016

The data agrees with the following definition: if n is even, a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n), but if n is odd, a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n). - Robert Israel, Feb 11 2016

Links

Robert Israel, Table of n, a(n) for n = 0..828 (using my definition)

Formula

(With my definition) a(n+3)-16*a(n+2)-a(n+1)+8*a(n) = 0 holds for at least n = 0 to 20000, but this may not always be the case. - Robert Israel, Feb 11 2016

Maple

# This agrees with the given Data
g:= proc(t, n) if n::even then floor(t+1) else ceil(t-1) fi end proc:
A[0]:= 2: A[1]:= 32:
for n from 2 to 50 do A[n]:= g(A[n-1]^2/A[n-2], n) od:
seq(A[i], i=0..50); # Robert Israel, Feb 11 2016

Prog

(PARI) a=List([2, 32]); for(n=2, 50, listput(a, a[n]^2\a[n-1]+1)); Vec(a) \\ M. F. Hasler, Feb 10 2016

Crossrefs

Sequence in context: A022028 A013776 A174491 * A010045 A052151 A092844

Adjacent sequences: A022016 A022017 A022018 * A022020 A022021 A022022

Keyword

nonn

Author

R. K. Guy

Status

approved