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a(0) = 5, a(1) = 7; for n>1, a(n) = a(n-1)+a(n-2)-(2n-2).
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%I #8 Apr 23 2018 11:37:00

%S 5,7,10,13,17,22,29,39,54,77,113,170,261,407,642,1021,1633,2622,4221,

%T 6807,10990,17757,28705,46418,75077,121447,196474,317869,514289,

%U 832102,1346333,2178375,3524646,5702957,9227537,14930426,24157893

%N a(0) = 5, a(1) = 7; for n>1, a(n) = a(n-1)+a(n-2)-(2n-2).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-1,1).

%F a(n) = Fibonacci(n+1)+2n+4. - _Ralf Stephan_, Feb 24 2004

%t RecurrenceTable[{a[0]==5,a[1]==7,a[n]==a[n-1]+a[n-2]-(2n-2)},a,{n,40}] (* or *) LinearRecurrence[{3,-2,-1,1},{5,7,10,13},40] (* _Harvey P. Dale_, Apr 23 2018 *)

%Y Cf. A088981, A000045.

%K nonn

%O 0,1

%A _Kurmang. Aziz. Rashid_, Dec 02 2003