A048112 -id:A048112 - cp's OEIS frontend

A068508 a(n) = round((a(n-1) + a(n-2))/a(n-3)) starting with a(1)=a(2)=a(3)=1.

1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1

Offset: 1

Henry Bottomley, Mar 25 2002

While this sequence has period 8, the unrounded version b(n) = (b(n-1) + b(n-2))/b(n-3) seems to have a quasi-period of about 8.7 for this particular starting point.

The unrounded version b(n) = A185332(n) / A185341(n) as given in A205303 has 8.694171... quasi-period. - Michael Somos, Oct 22 2018

			a(7) = round((a(6) + a(5))/a(4)) = round((5+3)/2) = 4.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,1).

Cf. A048112, A185332, A185341, A205303.

a(n) = a(n-8).

A048122 a(1) = 1, a(2) = 2, a(n) = a(n-1)^3 - a(n-2)^3.

Original entry on oeis.org

1, 2, 7, 335, 37595032, 53136308104856854277393, 150028625136232090492356612601560283583272160844416809510162344944689

Offset: 1

David Johnson-Davies

Next terms are 205 and 614 digits long.

Cf. A048112.

RecurrenceTable[{a[1]==1,a[2]==2,a[n]==a[n-1]^3-a[n-2]^3},a,{n,8}] (* Harvey P. Dale, Mar 20 2013 *)

More terms from Patrick De Geest, Jun 15 1999

cp's OEIS Frontend

A068508 a(n) = round((a(n-1) + a(n-2))/a(n-3)) starting with a(1)=a(2)=a(3)=1.

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A048122 a(1) = 1, a(2) = 2, a(n) = a(n-1)^3 - a(n-2)^3.

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