

A332060


a(n) = 3*a(n1) + a(n2) after initial values a(0..5) = (0, 1, 2, 3, 5, 13).


1



0, 1, 2, 3, 5, 13, 44, 145, 479, 1582, 5225, 17257, 56996, 188245, 621731, 2053438, 6782045, 22399573, 73980764, 244341865, 807006359, 2665360942, 8803089185, 29074628497, 96026974676, 317155552525, 1047493632251, 3459636449278, 11426402980085
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OFFSET

0,3


COMMENTS

These numbers arise as borders of intervals [a(n), a(n+1)] = [b(k)=k, b(m)=m] with m := b(k) + b(k1) after the "holes" between the borders have been filled according to b(k+1) = b(k) + b(m) and b(m1) = b(k+1) + b(n) for any such interval of length m  k > 2, i.e., starting from k = 5, m = 13.
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(k1) = 5 + 8 = 13. See Example for more.


LINKS

Table of n, a(n) for n=0..28.
Index entries for linear recurrent sequences with constant coefficients, signature (3,1).


FORMULA

G.f.: x*(1  x  4*x^2  6*x^3  5*x^4)/(1  3*x  x^2).


EXAMPLE

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(k1), either.)
We start with [a(1), a(2)] = [1, 2].
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).
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).
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.
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.


PROG

(PARI)
for(n=1+#a=[0, 1, 2, 3, 5, 13], #a=Vec(a, 30), a[n]=a[n1]*3+a[n2]); a \\ Remove initial 0 to get a[1] = 1 etc.
apply( {A332060(n)=if(n>3, [5, 13]*([0, 1; 1, 3]^(n4))[, 1], n)}, [0..20])


CROSSREFS

Cf. A000045 (Fibonacci numbers F(n+1) = F(n) + F(n1)); A006190, A052924, A006497 (a(n+1) = 3*a(n) + a(n1)); A000129 (Pell numbers a(n+1) = 2*a(n) + a(n1)).
Sequence in context: A192241 A093999 A042445 * A048634 A012899 A215102
Adjacent sequences: A332057 A332058 A332059 * A332061 A332062 A332063


KEYWORD

nonn


AUTHOR

M. F. Hasler, Mar 04 2020


STATUS

approved



