A386397 - OEIS

A386397

a(n) = (3^(n+2) + 4^(2*n+1))/13.

1, 7, 85, 1279, 20221, 322807, 5162725, 82597039, 1321532941, 21144468007, 338311310965, 5412980443999, 86607685509661, 1385722963371607, 22171567399596805, 354745078350502159, 5675921253478894381, 90814740055274889607, 1453035840883235972245, 23248573454128288771519

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OFFSET

0,2

COMMENTS

a(n) is integer for n >= 0.

Proof: It is sufficient to prove that 3^(n+2) + 4^(2*n+1) is divisible by 13. Since from 3^2 == -4 (mod 13) follows that 3^(n+2) == -4*3^n (mod 13) and from 4^2 == 3 (mod 13) follows 4^(2*n+1) == 4*3^n (mod 13), one gets from their sum that 3^(n+2) + 4^(2*n+1) == 0 (mod 13). QED

REFERENCES

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 5.1.25 on page 158.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Index entries for linear recurrences with constant coefficients, signature (19,-48).

FORMULA

a(n) = 19*a(n-1) - 48*a(n-2) for n > 1.

G.f.: (1 - 12*x)/((1 - 3*x)*(1 - 16*x)).

E.g.f.: exp(3*x)*(9 + 4*exp(13*x))/13.

a(n) = (A000244(n+2) + A013709(n))/13.

MATHEMATICA

a[n_]:=(3^(n+2)+4^(2n+1))/13; Array[a, 20, 0]