1,1
Table of n, a(n) for n=1..7.
a(n) = floor(2^((5/2)^(n-1))).
Floor[RecurrenceTable[{a[1]==2, a[n]==a[n-1]^(5/2)}, a, {n, 1, 10}]]
Table[Floor[2^((5/2)^(n-1))], {n, 1, 10}]
Cf. A254406.
Sequence in context: A301993 A128297 A183291 * A216847 A102983 A038583
Adjacent sequences: A254402 A254403 A254404 * A254406 A254407 A254408
nonn,easy
Vaclav Kotesovec, Jan 30 2015
approved