OFFSET
1,2
COMMENTS
Or, numbers k such that k^2 == 1 (mod 9).
Or, numbers k such that the iterative cycle j -> sum of digits of j^2 when started at k contains a 1. E.g., 8 -> 6+4 = 10 -> 1+0+0 = 1 and 17 -> 2+8+9 = 19 -> 3+6+1 = 10 -> 1+0+0 = 1. - Asher Auel, May 17 2001
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
FORMULA
a(1) = 1; a(n) = 9(n-1) - a(n-1). - Rolf Pleisch, Jan 31 2008 [Offset corrected by Jon E. Schoenfield, Dec 22 2008]
From R. J. Mathar, Feb 10 2008: (Start)
O.g.f.: 1 + 5/(4(x+1)) + 27/(4(-1+x)) + 9/(2(-1+x)^2).
a(n+1) - a(n) = A010697(n). (End)
a(n) = (9*A132355(n) + 1)^(1/2). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = a(n-2) + 9, for n > 2.
a(n) = 9*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), n > 1. (End)
E.g.f.: 1 + ((18*x - 9)*exp(x) + 5*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/9) (A332437).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/9)*cosec(Pi/9). (End)
MATHEMATICA
Select[ Range[ 300 ], PowerMod[ #, 2, 3^2 ]==1& ]
PROG
(PARI) a(n)=9*(n>>1)+if(n%2, 1, -1) \\ Charles R Greathouse IV, Jun 29 2011
(PARI) for(n=1, 40, print1(9*n-8, ", ", 9*n-1, ", ")) \\ Charles R Greathouse IV, Jun 29 2011
(Haskell)
a056020 n = a056020_list !! (n-1)
a05602_list = 1 : 8 : map (+ 9) a056020_list
-- Reinhard Zumkeller, Jan 07 2012
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Robert G. Wilson v, Jun 08 2000
STATUS
approved