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A008954
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Final digit of triangular number n*(n+1)/2.
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12
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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
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OFFSET
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0,3
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,0,0,0,-1,0,0,0,0,1).
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FORMULA
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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*(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)
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MAPLE
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seq(mod(binomial(n+1, 2), 10), n = 0 .. 100); # G. C. Greubel, Sep 14 2019
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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