OFFSET
0,3
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
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(1) = 1, a(n+1) = (a(n) + n + 1) mod 10.
Periodic with period 20: repeat [0,1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3, 1,0]. - Franklin T. Adams-Watters, Mar 13 2006
It follows that all triangular numbers end with a digit of 0, 1, 3, 5, 6, or 8, and thus none end with a digit of 2, 4, 7, or 9. - Harvey P. Dale, Dec 31 2014
a(n) = n*(n+1)/2 mod 10. - Ant King, Apr 26 2009
From R. J. Mathar, Apr 15 2010: (Start)
a(n) = a(n-5) - a(n-10) + a(n-15).
G.f.: x*(1 +3*x +6*x^2 +5*x^4 +5*x^6 +5*x^8 +6*x^10 +3*x^11 +x^12)/(1 -x^5 +x^10 -x^15). (End)
MAPLE
seq(mod(binomial(n+1, 2), 10), n = 0 .. 100); # G. C. Greubel, Sep 14 2019
MATHEMATICA
Table[Mod[n*(n+1)/2, 10], {n, 0, 100}]
LinearRecurrence[{0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1}, {0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5}, 110] (* Harvey P. Dale, Dec 31 2014 *)
PROG
(PARI) a(n)=n*(n+1)/2%10 \\ Charles R Greathouse IV, Mar 05 2014
(Magma) [Binomial(n+1, 2) mod 10: n in [0..100]]; // G. C. Greubel, Sep 14 2019
(Sage) [Mod(binomial(n+1, 2), 10) for n in (0..100)] # G. C. Greubel, Sep 14 2019
(GAP) List([0..100], n-> (Binomial(n+1, 2) mod 10) ); # G. C. Greubel, Sep 14 2019
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved