A133884 a(n) = binomial(n+4,n) mod 4.

1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3, 3, 1, 1, 0, 0, 0, 0, 1, 1, 3, 3, 2, 2, 2, 2, 3

Offset: 0

Author

Hieronymus Fischer, Oct 10 2007

Keywords

• nonn
• easy

Comments

Periodic with length 4^2=16.

Examples

			For n=2, binomial(6,2) = 6*5/2 = 15, which is 3 (mod 4) so a(2) = 3. - _Michael B. Porter_, Jul 19 2016
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, -1, 1).

Crossrefs

Cf. A000040, A133630, A038509.

Cf. A133620, A133621, A133622, A133623, A133624, A133625

Cf. A133634, A133635, A133636.

Cf. A133874, A133880, A133890, A133900, A133910.

Programs

Magma

[Binomial(n+4,n) mod 4: n in [0..100]]; // Vincenzo Librandi, Jul 15 2016

Mathematica

Table[Mod[Binomial[n + 4, 4], 4], {n, 0, 100}] (* Vincenzo Librandi, Jul 15 2016 *)

Formula

a(n) = binomial(n+4,4) mod 4.

G.f.: (1 + x + 3*x^2 + 3*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 3*x^8 + 3*x^9 + x^10 + x^11)/(1 - x^16) = (1 + 2*x^2 + 2*x^6 + x^8)/((1 - x)*(1 + x^4)*(1 + x^8)).

a(n) = A000505(n+5) mod 4. - John M. Campbell, Jul 14 2016

a(n) = A000506(n+6) mod 4. - John M. Campbell, Jul 15 2016

Extensions

G.f. corrected by Bruno Berselli, Jul 19 2016