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A224227
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a(n) = (1/50)*((15*n^2-20*n+4)*Fibonacci(n) - (5*n^2-6*n)*A000032(n)).
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1
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0, 0, 0, 1, 2, 7, 16, 38, 82, 173, 352, 701, 1368, 2628, 4980, 9329, 17302, 31811, 58040, 105178, 189446, 339373, 604964, 1073593, 1897488, 3341160, 5863080, 10256065, 17888138, 31115071, 53985856, 93447278, 161397754, 278184461, 478550344, 821734901, 1408610088, 2410719084, 4119433884, 7029086705, 11977419742, 20382654971, 34643298728, 58811818210
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OFFSET
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0,5
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COMMENTS
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The right-hand side of a binomial-coefficient identity.
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)
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LINKS
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N. Gauthier (Proposer), Problem H-703, Fib. Quart., 50 (2012), 379-381.
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FORMULA
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a(n) = Sum_{k=0..n-1} k^2*binomial(n-k-1,k).
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
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MATHEMATICA
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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 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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