OFFSET
0,1
COMMENTS
Tribonacci sequence beginning 5, 7, 9.
In general, the ordinary generating function for the recurrence relation b(n) = b(n-1) + b(n-2) + b(n-3), with n>2 and b(0)=k, b(1)=m, b(2)=q, is (k + (m-k)*x + (q-m-k)*x^2)/(1 - x - x^2 - x^3).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Tribonacci Number
Index entries for linear recurrences with constant coefficients, signature (1,1,1)
FORMULA
G.f.: (5 + 2*x - 3*x^2)/(1 - x - x^2 - x^3).
a(n) = 3*K(n) - 4*T(n+1) + 8*T(n), where K(n) = A001644(n) and T(n) =A000073(n+1). - G. C. Greubel, Apr 23 2019
MATHEMATICA
LinearRecurrence[{1, 1, 1}, {5, 7, 9}, 40]
RecurrenceTable[{a[0]==5, a[1]==7, a[2]==9, a[n]==a[n-1]+a[n-2]+a[n-3]}, a, {n, 40}]
PROG
(Magma) I:=[5, 7, 9]; [n le 3 select I[n] else Self(n-1)+Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 04 2016
(PARI) my(x='x+O('x^40)); Vec((5+2*x-3*x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 23 2019
(Sage) ((5+2*x-3*x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
(GAP) a:=[5, 7, 9];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
CROSSREFS
Cf. similar sequences with initial values (p,q,r): A000073 (0,0,1), A081172 (1,1,0), A001590 (0,1,0; also 1,2,3), A214899 (2,1,2), A001644 (3,1,3), A145027 (2,3,4), A000213 (1,1,1), A141036 (2,1,1), A141523 (3,1,1), A214727 (1,2,2), A214825 (1,3,3), A214826 (1,4,4), A214827 (1,5,5), A214828 (1,6,6), A214829 (1,7,7), A214830 (1,8,8), A214831 (1,9,9).
KEYWORD
nonn,easy,less
AUTHOR
Ilya Gutkovskiy, Feb 04 2016
STATUS
approved