OFFSET
0,5
COMMENTS
The right-hand side of a binomial-coefficient identity.
From Steven Finch, Mar 22 2020: (Start)
a(n+2) is the total binary weight squared of all A000045(n+2) binary sequences of length n not containing any adjacent 1's.
The only three 2-bitstrings without adjacent 1's are 00, 01 and 10. The bitsums squared of these are 0, 1 and 1. Adding these give a(4)=2.
The only five 3-bitstrings without adjacent 1's are 000, 001, 010, 100 and 101. The bitsums squared of these are 0, 1, 1, 1 and 4. Adding these give a(5)=7.
The only eight 4-bitstrings without adjacent 1's are 0000, 0001, 0010, 0100, 1000, 0101, 1010 and 1001. The bitsums squared of these are 0, 1, 1, 1, 1, 4, 4, and 4. Adding these give a(6)=16. (End)
LINKS
Steven Finch, Cantor-solus and Cantor-multus Distributions, arXiv:2003.09458 [math.CO], 2020.
N. Gauthier (Proposer), Problem H-703, Fib. Quart., 50 (2012), 379-381.
Index entries for linear recurrences with constant coefficients, signature (3,0,-5,0,3,1).
FORMULA
a(n) = Sum_{k=0..n-1} k^2*binomial(n-k-1,k).
G.f.: -x^3*(x^2-x+1)/(x^2+x-1)^3. - Mark van Hoeij, Apr 10 2013
E.g.f.: 2*exp(x/2)*(sqrt(5)*(2 + 5*x^2)*sinh(sqrt(5)*x/2) - 5*x*cosh(sqrt(5)*x/2))/125. - Stefano Spezia, Mar 20 2023
MATHEMATICA
LinearRecurrence[{3, 0, -5, 0, 3, 1}, {0, 0, 0, 1, 2, 7}, 50] (* Harvey P. Dale, Jan 22 2016 *)
Table[((15 n^2 - 20 n + 4) Fibonacci[n] - (5 n - 6) n LucasL[n])/50, {n, 0, 30}] (* Vladimir Reshetnikov, Oct 10 2016 *)
PROG
(PARI) concat([0, 0, 0], Vec((x^2-x+1)/(x^2+x-1)^3+O(x^96))) \\ Charles R Greathouse IV, Mar 19 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Apr 09 2013
STATUS
approved