OFFSET
1,1
COMMENTS
Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 2 and (n+3) mod 5 <> 1.
This is one of a family of sequences which are defined (or could be defined) according to the same scheme: Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to k-1 and (n+k) mod (2+k) <> 1. We have A007310 for k = 1, A017629 for k = 2, this one (A096022) for k = 3, A096023 for k = 5, A096024 for k = 6, A096025 for k = 7, A096026 for k = 9, A096027 for k = 11. Remarkably these sequences are empty for k = 4, 8, 10, ... (i.e., if k+1 is a term of A080765).
Numbers n such that n mod 12 = 3 and n mod 60 <> 3.
Subsequence of A017557: 12n+3.
LINKS
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).
FORMULA
G.f.: 3*x*(5+4*x+4*x^2+4*x^3+3*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 04 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = 3*(10*n-3-i^(2*n)-(1-i)*i^(-n)-(1+i)*i^n)/2 where i=sqrt(-1). (End)
E.g.f.: 3*(3 + sin(x) - cos(x) + (5*x - 1)*sinh(x) - (2 - 5*x)*cosh(x)). - Ilya Gutkovskiy, Jun 05 2016
EXAMPLE
51 mod 2 = 52 mod 3 = 53 mod 4 = 1 and 54 mod 5 = 4, hence 51 is in the sequence; 3 mod 2 = 4 mod 3 = 5 mod 4 = 6 mod 5 = 1, hence 3 is not in the sequence.
MAPLE
A096022:=n->3*(10*n-3-I^(2*n)-(1-I)*I^(-n)-(1+I)*I^n)/2: seq(A096022(n), n=1..80); # Wesley Ivan Hurt, Jun 04 2016
MATHEMATICA
Table[3*(10n-3-I^(2n)-(1-I)*I^(-n)-(1+I)*I^n)/2, {n, 80}] (* Wesley Ivan Hurt, Jun 04 2016 *)
PROG
(PARI) {k=3; m=800; for(n=1, m, j=0; b=1; while(b&&j<k, if((n+j)%(2+j)==1, j++, b=0)); if(b&&(n+k)%(2+k)!=1, print1(n, ", ")))}
(Magma) [ n : n in [1..1500] | n mod 60 in [15, 27, 39, 51] ] // Vincenzo Librandi, Mar 24 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Jun 15 2004
EXTENSIONS
New definition from Ralf Stephan, Dec 01 2004
STATUS
approved