A171216 a(n) = (4^(5*n+1) + 7)/11.

373, 381301, 390451573, 399822410101, 409418147942773, 419244183493398901, 429306043897240473973, 439609388950774245347701, 450160014285592827236045173, 460963854628447055089710256501, 472026987139529784411863302656373

Offset: 1

Author

A.K. Devaraj, Dec 05 2009

Keywords

• nonn
• easy

Comments

In A165806, A165808 & A165809 a congruence property of polynomial functions was demonstrated. In the present sequence a congruence property of exponential functions is demonstrated. Let the function be f(n) = 2^n + 7. Then f(n + k*phi(f(n))) is congruent to 0 mod(f(n)). This is a sequence of quotients generated by (f(n + k*phi f(n)))/f(n) when n = 2.

References

• A. K. Devaraj, "Euler's generalisation of Fermat's theorem - a further generalisation" - Hawaii International conference on Mathematics & Statistics (2004). [ISSN 15503747]

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..300
• Index entries for linear recurrences with constant coefficients, signature (1025,-1024).

Programs

Magma

[(4^(5*n+1) + 7)/11 : n in [1..15]]; // Vincenzo Librandi, Oct 06 2011

Mathematica

(4^(5*Range[15]+1)+7)/11 (* Paolo Xausa, Mar 20 2024 *)

PARI

a(n)=(4^(5*n+1) + 7)/11 \\ Charles R Greathouse IV, Oct 05 2011

Formula

G.f. -x*(-373+1024*x) / ( (1024*x-1)*(x-1) ). - R. J. Mathar, Oct 08 2011