|
| |
|
|
A198410
|
|
((3^(n-1) + 1)^3 -1)/3^n.
|
|
16
|
|
|
|
7, 37, 271, 2269, 19927, 177877, 1596511, 14355469, 129159847, 1162320517, 10460530351, 94143710269, 847290203767, 7625602267957, 68630391713791, 617673439330669, 5559060695695687, 50031545486420197, 450283907053258831, 4052555156505760669, 36472996387631139607
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
2,1
|
|
|
COMMENTS
|
This sequence is generalizable :
Proposition : p^n divide (p^(n-1) + 1)^ p - 1
Proof : let two integers a, p such that p>=2, and k = gcd(a,p). Then ak divide (a + 1)^p - 1 because (a+1)^p - 1 = [a^p + binomial(p,1)*a^(p-1) + …+ binomial(p,p-2)*a^2] + pa ==binomial(p,1)*n^(n-1) == 0 (mod p^n) with a = p^(n-1) and k = p.
|
|
|
LINKS
|
Table of n, a(n) for n=2..22.
Index to sequences with linear recurrences with constant coefficients, signature (13,-39,27).
|
|
|
FORMULA
|
G.f. -x^2*(7-54*x+63*x^2) / ( (x-1)*(3*x-1)*(9*x-1) ). - R. J. Mathar, Oct 25 2011
|
|
|
EXAMPLE
|
a(2) = ((3 + 1)^3 - 1)/3^2 = 63/9 = 7.
|
|
|
MAPLE
|
A198410 := proc(n)
(3^(n-1)+1)^3 ;
(%-1)/3^n ;
end proc:
seq(A198410(n), n=2..10) ; # R. J. Mathar, Oct 25 2011
|
|
|
CROSSREFS
|
Cf. A060073.
Sequence in context: A159597 A217723 A100309 * A199192 A089677 A075996
Adjacent sequences: A198407 A198408 A198409 * A198411 A198412 A198413
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Michel Lagneau, Oct 24 2011
|
|
|
STATUS
|
approved
|
| |
|
|
