A386394 a(n) = (5^(2*n) + 3*2^(5*n-2))/7.

7, 199, 5743, 168151, 4990207, 149920999, 4553331343, 139603087351, 4314710904607, 134256051681799, 4200826222176943, 132042253318646551, 4165747461421299007, 131813802646096802599, 4180788781690478542543, 132853918439479834845751, 4228042325697967752173407

Offset

1,1

Comments

a(n) is integer for n > 0.

Proof: It is sufficient to prove that 5^(2*n) + 3*2^(5*n-2) is divisible by 7. Since from 5 == -2 (mod 7) follows that 5^(2*n) == 2^(2*n) (mod 7) and from 2^5 == 2^2 (mod 7) follows 2^(5*n-2) == 2^(2*(n-1)) (mod 7), one gets that 5^(2*n) + 3*2^(5*n-2) == 2^(2*n) + 3*2^(2*(n-1)) (mod 7) and 2^(2*n) + 3*2^(2*(n-1)) = 2^(2*(n-1))*(4 + 3) == 0 (mod 7). QED

References

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 5.1.24 on page 158.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Index entries for linear recurrences with constant coefficients, signature (57,-800).

Formula

a(n) = 57*a(n-1) - 800*a(n-2) for n > 2.

G.f.: x*(7 - 200*x)/((1 - 25*x)*(1 - 32*x)).

E.g.f.: (3*exp(32*x) + 4*exp(25*x) - 7)/28.

a(n) = (A009969(n) + 3*A013824(n-1))/7.

Mathematica

a[n_]:=(5^(2n)+3*2^(5n-2))/7; Array[a, 17]

Prog

(Magma) [(5^(2*n)+3*2^(5*n-2))/7 : n in [1..20]]; // Vincenzo Librandi, Jul 21 2025

Crossrefs

Cf. A009969, A013824.

Sequence in context: A068233 A154936 A206473 * A376461 A300616 A370084

Adjacent sequences: A386391 A386392 A386393 * A386395 A386396 A386397

Keyword

nonn,easy

Author

Stefano Spezia, Jul 20 2025

Status

approved