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%I A332060 #27 Apr 18 2023 15:32:39
%S A332060 0,1,2,3,5,13,44,145,479,1582,5225,17257,56996,188245,621731,2053438,
%T A332060 6782045,22399573,73980764,244341865,807006359,2665360942,8803089185,
%U A332060 29074628497,96026974676,317155552525,1047493632251,3459636449278,11426402980085
%N A332060 a(n) = 3*a(n-1) + a(n-2) after initial values a(0..5) = (0, 1, 2, 3, 5, 13).
%C A332060 These numbers arise as borders of intervals [a(n), a(n+1)] = [b(k)=k, b(m)=m] with m := b(k) + b(k-1) after the "holes" between the borders have been filled according to b(k+1) = b(k) + b(m) and b(m-1) = b(k+1) + b(n) for any such interval of length m - k > 2, i.e., starting from k = 5, m = 13.
%C A332060 The initial terms correspond to intervals of length <= 2 with only 0 or 1 "holes" to fill: In the first case we have the same recursion rule as for the Fibonacci sequence, and when there's one hole (between 3 and 5) the next Fibonacci number b(k+1) = b(k) + b(m) = 3 + 5 = 8 gets filled in there, and the next border is m = b(k) + b(k-1) = 5 + 8 = 13. See Example for more.
%H A332060 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,1).
%F A332060 G.f.: x*(1 - x - 4*x^2 - 6*x^3 - 5*x^4)/(1 - 3*x - x^2).
%e A332060 The initial a(0) is conventional. (One could also choose a(0) = 1 to have a(2) = a(1) + a(0) as for the next two terms, but this wouldn't correspond to b(m) = b(k) + b(k-1), either.)
%e A332060 We start with [a(1), a(2)] = [1, 2].
%e A332060 No gap or "hole" here to fill, so the next interval [a(2), a(3)] has upper bound a(3) = a(2) + 1 = 3, where 1 is the element just left to the right border a(2).
%e A332060 Again, no gap or hole to fill in [2, 3], so the next interval has upper bound a(4) = a(3) + 2 = 5, where 2 is the element just left to the right border a(3).
%e A332060 Now there's a hole in (3, 5), at position 4, which is filled with 3 + 5 = 8, so the next upper bound is a(5) = a(4) + 8 = 5 + 8 = 13: here the number 8 was the element left to the right border 5.
%e A332060 Now there are several holes in (5, 13). The leftmost one (position 6) is filled with 5 + 13 = 18, and the rightmost (position 12) is filled with 18 + 13 = 31. So the next interval [a(5), a(6)] has upper bound a(6) = a(5) + 31 = 44.
%o A332060 (PARI)
%o A332060 for(n=1+#a=[0,1,2,3,5,13],#a=Vec(a,30),a[n]=a[n-1]*3+a[n-2]);a \\ Remove initial 0 to get a[1] = 1 etc.
%o A332060 apply( {A332060(n)=if(n>3,[5,13]*([0,1;1,3]^(n-4))[,1],n)}, [0..20])
%Y A332060 Cf. A000045 (Fibonacci numbers F(n+1) = F(n) + F(n-1)); A006190, A052924, A006497 (a(n+1) = 3*a(n) + a(n-1)); A000129 (Pell numbers a(n+1) = 2*a(n) + a(n-1)).
%K A332060 nonn
%O A332060 0,3
%A A332060 _M. F. Hasler_, Mar 04 2020

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