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A008954 Final digit of triangular number n*(n+1)/2. 10
0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

LINKS

Table of n, a(n) for n=0..99.

Index entries for sequences related to final digits of numbers

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

a(n) = a(n-5) - a(n-10) + a(n-15). G.f.: -(x^12+3*x^11+6*x^10+5*x^8+5*x^6+5*x^4+6*x^2+3*x+1)*x / ( (x-1)*(x^2+1)*(x^4+x^3+x^2+x+1)*(x^8-x^6+x^4-x^2+1) ). - R. J. Mathar, Apr 15 2010

MATHEMATICA

Table[ Mod[ n*(n + 1)/2, 10 ], {n, 0, 80} ]

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

CROSSREFS

Cf. A000217, A061501, A008953.

First differences of A111072.

Sequence in context: A182196 A278488 A038023 * A169890 A181907 A141703

Adjacent sequences:  A008951 A008952 A008953 * A008955 A008956 A008957

KEYWORD

nonn,base,easy

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified October 23 01:24 EDT 2018. Contains 316518 sequences. (Running on oeis4.)