Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #52 Jan 25 2023 11:51:48
%S 1,1,2,3,5,8,14,22,36,58,94,152,247,399,646,1045,1691,2736,4428,7164,
%T 11592,18756,30348,49104,79453,128557,208010,336567,544577,881144,
%U 1425722,2306866,3732588,6039454,9772042,15811496,25583539,41395035,66978574,108373609
%N a(n) = a(n-1) + a(n-2) + 1 if n is a multiple of 6, otherwise a(n) = a(n-1) + a(n-2).
%C Also: convolution of A000045 with the period-6 sequence (0,0,0,0,0,0, 1,...). - _R. J. Mathar_, May 30 2008
%C Sequences of the form s(0)=a, s(1)= b, s(n) = s(n-1) + s(n-2) + k if n mod m = p, else s(n) = s(n-1) + s(n-2) have a form s(n) = fibonacci(n-1)*a + fibonacci(n)*b + P(x)*k. a(n) is the P(x) sequence for m=6. s(n) = fib(n)*a + fib(n-1)*b + a(n-6-p)*k. - _Gary Detlefs_, Dec 05 2010
%C a(n) is the number of compositions of n where the order of the 2 and the 3 does not matter. - _Gregory L. Simay_, May 18 2017
%H H. Matsui et al., <a href="http://www.fq.math.ca/Problems/elementary45-2.pdf">Problem B-1035</a>, Fibonacci Quarterly, Vol. 45, Number 2; 2007; p. 182.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,0,0,0,1,-1,-1).
%F O.g.f.: 1/((1-x^6)(1 - x - x^2)). - _R. J. Mathar_, May 30 2008
%F a(n) = ((-1)^n-1)/6 + A099837(n+3)/12 + A000045(n+4)/4 + A057079(n)/12. - _R. J. Mathar_, Dec 07 2010
%F a(n) = floor(A066983(n+4)/3). - _Gary Detlefs_, Dec 19 2010
%F a(n) = round((1 + sqrt(5))/8 A000045(n+3)). - _John M. Campbell_, Jul 06 2016
%F a(n) = (number of compositions of n consisting of only 1 or 2 or 6) - (number of compositions with only 7 or ((1 or 2) and 7)) - (number of compositions with only 8 or ((1 or 2) and 8)). The "or" is inclusive. - _Gregory L. Simay_, May 25 2017
%e Since 5 is not a multiple of 6, a(5) = a(4) + a(3) = 5 + 3 = 8. Since 6 is a multiple of 6, a(6) = a(5) + a(4) + 1 = 8 + 5 + 1 = 14. - _Michael B. Porter_, Jul 26 2016
%p A131132:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 6 = 1 then t1:=1 else t1:=0; fi: procname(n-1)+procname(n-2)+t1; end; [seq(A131132(n), n=1..100)]; # _N. J. A. Sloane_, May 25 2008; Typo corrected by _R. J. Mathar_, May 30 2008
%t Print[Table[Round[(1 + Sqrt[5])/8 Fibonacci[n + 3]], {n, 0, 50}]] ;
%t Print[RecurrenceTable[{c[n] == c[n - 1] + c[n - 2] + c[n - 6] - c[n - 7] - c[n - 8], c[0] == 1, c[1] == 1, c[2] == 2, c[3] == 3, c[4] == 5, c[5] == 8, c[6] == 14, c[7] == 22}, c, {n, 0, 50}]] ; (* _John M. Campbell_, Jul 07 2016 *)
%t LinearRecurrence[{1, 1, 0, 0, 0, 1, -1, -1}, {1, 1, 2, 3, 5, 8, 14, 22}, 40] (* _Vincenzo Librandi_, Jul 07 2016 *)
%Y Cf. A052952, A004695, A080239, A124502, A066983.
%K nonn
%O 0,3
%A _N. J. A. Sloane_, May 25 2008
%E More specific name from _R. J. Mathar_, Dec 09 2009