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 A109718 Periodic sequence with period {0,1,0,3}, or n^3 mod 4. 2
 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Since n^(2k+1) mod 4 = n^3 mod 4 for k>1 this sequence also represents n^5 mod 4; and n^7 mod 4; etc LINKS Index entries for linear recurrences with constant coefficients, signature (0,0,0,1). FORMULA a(n) = n^3 mod 4. G.f. = (x+3x^3)/(1-x^4). a(n) = (1/12)*{11*(n mod 4)-7*[(n+1) mod 4]+5*[(n+2) mod 4]-[(n+3) mod 4]} - Paolo P. Lava, Nov 21 2006 a(n) = (n mod 2)*(n mod 4) = (1+(-1)^(n+1))*(2+i^(n+1))/2 with i=sqrt(-1).  - Bruno Berselli, Mar 14 2011 PROG (Sage) [power_mod(n, 3, 4 )for n in xrange(0, 105)]# - Zerinvary Lajos, Oct 29 2009 (MAGMA) &cat[[0, 1, 0, 3]: k in [0..26]];  // Bruno Berselli, Mar 14 2011 (PARI) a(n)=n^3%4 \\ Charles R Greathouse IV, Apr 06 2016 CROSSREFS n mod 4 = A010873; n^2 mod 4 = A000035. Cf. A110270; A131743.  - Bruno Berselli, Mar 14 2011 Sequence in context: A155522 A007524 A204689 * A053385 A213543 A247254 Adjacent sequences:  A109715 A109716 A109717 * A109719 A109720 A109721 KEYWORD easy,nonn AUTHOR Bruce Corrigan (scentman(AT)myfamily.com), Aug 09 2005 STATUS approved

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Last modified October 17 18:51 EDT 2019. Contains 328127 sequences. (Running on oeis4.)