OFFSET
1,3
COMMENTS
Also the sum of consecutive Lucas numbers because the difference L(i) - L(j) equals the sum L(j+1) + ... + L(i+2).
Conjecture: L(m) - L(n) with m > 1 and m > n >= 0 is a perfect power but not a square only for (m,n) = (7,0), (5,2). This has been verified for n < m <= 500. Note that L(7) - L(0) = 29 - 2 = 3^3 and L(5) - L(2) = 11 - 3 = 2^3. - Zhi-Wei Sun, Jan 02 2025
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
MATHEMATICA
Lucas[n_] := Fibonacci[n+1]+Fibonacci[n-1]; Union[Flatten[Table[Lucas[n]-Lucas[i], {n, 13}, {i, 0, n-2}]]]
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Oct 17 2005
STATUS
approved