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a(n) = a(n-11) + a(n-12).
10

%I #43 Sep 01 2024 02:47:05

%S 1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,3,4,4,4,4,4,4,4,4,4,4,

%T 5,7,8,8,8,8,8,8,8,8,8,9,12,15,16,16,16,16,16,16,16,16,17,21,27,31,32,

%U 32,32,32,32,32,32,33,38,48,58,63,64,64,64,64,64,64,65,71,86,106,121,127

%N a(n) = a(n-11) + a(n-12).

%H <a href="/index/Rec/order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,0,1,1).

%F For n>12: a(n) = a(n-11) + a(n-12). a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = a(7) = a(8) = a(9) = a(10) = a(11) = a(12) = 1.

%F G.f.: x*(1-x^11) / ((1-x)*(1-x^11-x^12)). - _Colin Barker_, Mar 26 2013

%p A103379 := proc(n) option remember ; if n <= 12 then 1; else procname(n-11)+procname(n-12) ; fi; end: for n from 1 to 120 do printf("%d,",A103379(n)) ; od: # _R. J. Mathar_, Aug 30 2008

%t SemiprimeQ[n_]:=Plus@@FactorInteger[n][[All, 2]]?2; k11; Do[a[n]=1, {n, k+1}]; a[n_]:=a[n]=a[n-k]+a[n-k-1]; A103379=Array[a, 100] A103389=Union[Select[Array[a, 1000], PrimeQ]] A103399=Union[Select[Array[a, 300], SemiprimeQ]] N[Solve[x^12 - x - 1 == 0, x], 111][[2]] (* _Ray Chandler_ and _Robert G. Wilson v_ *)

%t LinearRecurrence[{0,0,0,0,0,0,0,0,0,0,1,1},{1,1,1,1,1,1,1,1,1,1,1,1},100] (* _Harvey P. Dale_, Jan 31 2015 *)

%Y Cf. A079398, A103372-A103378, A103380, A103389, A103399.

%K easy,nonn

%O 1,13

%A _Jonathan Vos Post_, Feb 15 2005

%E Corrected from a(11) on by _R. J. Mathar_, Aug 30 2008