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A213472 Period 20, repeat [1, 4, 0, 9, 1, 6, 4, 5, 9, 6, 6, 9, 5, 4, 6, 1, 9, 0, 4, 1]. 0
1, 4, 0, 9, 1, 6, 4, 5, 9, 6, 6, 9, 5, 4, 6, 1, 9, 0, 4, 1, 1, 4, 0, 9, 1, 6, 4, 5, 9, 6, 6, 9, 5, 4, 6, 1, 9, 0, 4, 1, 1, 4, 0, 9, 1, 6, 4, 5, 9, 6, 6, 9, 5, 4, 6, 1, 9, 0, 4, 1, 1, 4, 0, 9, 1, 6, 4, 5, 9, 6, 6, 9, 5, 4, 6, 1, 9, 0, 4, 1, 1, 4, 0, 9, 1, 6 (list; graph; refs; listen; history; text; internal format)
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) = A010879(A005448(n)).
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
Sequence in context: A306954 A187606 A138478 * A305742 A199000 A339530
KEYWORD
nonn,easy
AUTHOR
Ant King, Jun 12 2012
STATUS
approved

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Last modified April 20 00:00 EDT 2024. Contains 371798 sequences. (Running on oeis4.)