OFFSET
0,2
COMMENTS
Units digits of the centered triangular numbers A005448(n).
The cyclic part of this sequence is palindromic.
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,0,0,0,-1,0,0,0,0,1).
FORMULA
a(n) = a(n-5)-a(n-10)+a(n-15).
a(n) = a(n-20).
a(n) = 45-a(n-1)-a(n-2)-a(n-3)-a(n-4)-a(n-10)-a(n-11)-a(n-12)-a(n-13)-a(n-14).
a(n) = 90 - Sum_{i=1..19} a(n-i), with n > 19.
a(n) = (3n^2/2+3n/2+1) mod 10.
G.f.: (1+x+x^2)*(1+3*x-4*x^2+10*x^3-5*x^4+5*x^6-5*x^8+10*x^9-4*x^10+3*x^11+x^12) / ((1-x)*(1+x^2)*(1+x+x^2+x^3+x^4)*(1-x^2+x^4-x^6+x^8)). - Bruno Berselli, Jun 13 2012
EXAMPLE
As the seventh centered triangular number is A005448(7)=64, which has units’ digit 4, then a(7)=4
MATHEMATICA
Mod[1/2(3#^2-3#+2), 10] &/@Range[86]
PROG
(PARI) a(n)=(3*n^2+3*n+2)/2%10 \\ Charles R Greathouse IV, Jul 21 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ant King, Jun 12 2012
STATUS
approved