A133894 Numbers m such that binomial(m+4,m) mod 4 = 0.

12, 13, 14, 15, 28, 29, 30, 31, 44, 45, 46, 47, 60, 61, 62, 63, 76, 77, 78, 79, 92, 93, 94, 95, 108, 109, 110, 111, 124, 125, 126, 127, 140, 141, 142, 143, 156, 157, 158, 159, 172, 173, 174, 175, 188, 189, 190, 191, 204, 205, 206, 207, 220, 221, 222, 223, 236, 237

Offset

1,1

Comments

Also numbers m such that floor(1+(m/4)) mod 4 = 0.

Partial sums of the sequence 12,1,1,1,13,1,1,1,13, ... which has period 4.

Numbers congruent to {12,13,14,15} mod 16. Numbers n such that n xor 12 = n - 12. [Brad Clardy, May 06 2013]

Links

Table of n, a(n) for n=1..58.

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

Formula

a(n) = 4*n + 12 - 3*(n mod 4).

G.f.: 12/(1-x)+x(1+x+x^2+13x^3)/((1-x^4)(1-x)) = (12+x+x^2+x^3+x^4)/((1-x^4)(1-x)) = (12-11x-x^5)/((1-x^4)(1-x)^2).

a(n) = 4*n+3*((1-i)*i^n+(1+i)*(-i)^n+(-1)^n+5)/2, where i=sqrt(-1). - Bruno Berselli, Apr 08 2011

Prog

(PARI) a(n)=(n+3)\4*16-[1, 4, 3, 2][n%4+1] \\ Charles R Greathouse IV, May 06 2013

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635, A133874, A133884, A133890, A133900, A133910.

Sequence in context: A364733 A270041 A272270 * A209725 A045879 A257073

Adjacent sequences: A133891 A133892 A133893 * A133895 A133896 A133897

Keyword

nonn,easy

Author

Hieronymus Fischer, Oct 20 2007

Status

approved