A268410 a(n) = a(n - 1) + a(n - 2) + a(n - 3) for n>2, a(0)=5, a(1)=7, a(2)=9.

5, 7, 9, 21, 37, 67, 125, 229, 421, 775, 1425, 2621, 4821, 8867, 16309, 29997, 55173, 101479, 186649, 343301, 631429, 1161379, 2136109, 3928917, 7226405, 13291431, 24446753, 44964589, 82702773, 152114115, 279781477, 514598365, 946493957

Offset: 0

Author

Ilya Gutkovskiy, Feb 04 2016

Keywords

• nonn
• easy
• less

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)

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).

Programs

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

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

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}]

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

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