OFFSET
1,2
COMMENTS
Table of differences re Table A182441.
This is a sequence of differences between rows k and k+1 of table A182441. That is if A182441(k+1,0)-A182441(k,0) = 1, a(n) = A182441(k+1,n+1) - A182441(k,n+1) for n = 0 to 3. The remainder of the sequence is a continuation using the recursive formula D(n) = 6D(n-1)- D(n-2) + 6.
It appears that for n > 0, a(n) is divisible by A213005(n).
It appears that if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p), and that if p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p).
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
FORMULA
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
From Colin Barker, Mar 05 2016: (Start)
a(n) = (-6+(5-3*sqrt(2))*(3+2*sqrt(2))^n + (3-2*sqrt(2))^n*(5+3*sqrt(2)))/4.
G.f.: x*(5+x) / ((1-x)*(1-6*x+x^2)).
(End)
MATHEMATICA
m = 12; n = 1; c = 0;
list3 = Reap[While[c < 22, t = 6 n - m + 6; Sow[t]; m = n; n = t; c++]][[2, 1]]
CoefficientList[ Series[x (5 + x)/(1 - 7x + 7x^2 - x^3), {x, 0, 20}], x] (* or *)
LinearRecurrence[{7, -7, 1}, {0, 5, 36}, 21] (* Robert G. Wilson v, Jun 24 2014 *)
PROG
(PARI) concat(0, Vec(x^2*(5+x)/((1-x)*(1-6*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 05 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Kenneth J Ramsey, May 14 2012
STATUS
approved