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A096027 Numbers k such that (k+j) mod (2+j) = 1 for j from 0 to 10 and (k+11) mod 13 <> 1.

%I A096027 #14 Apr 18 2024 18:20:21
%S A096027 27723,55443,83163,110883,138603,166323,194043,221763,249483,277203,
%T A096027 304923,332643,388083,415803,443523,471243,498963,526683,554403,
%U A096027 582123,609843,637563,665283,693003,748443,776163,803883,831603,859323,887043
%N A096027 Numbers k such that (k+j) mod (2+j) = 1 for j from 0 to 10 and (k+11) mod 13 <> 1.
%C A096027 Numbers k such that k mod 27720 = 3 and k mod 360360 <> 3.
%H A096027 <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,0,0,1,-1).
%F A096027 G.f.: 3*x*(9239*x^12 +9240*x^11 +9240*x^10 +9240*x^9 +9240*x^8 +9240*x^7 +9240*x^6 +9240*x^5 +9240*x^4 +9240*x^3 +9240*x^2 +9240*x +9241) / ((x -1)^2*(x +1)*(x^2 -x +1)*(x^2 +1)*(x^2 +x +1)*(x^4 -x^2 +1)). - _Colin Barker_, Apr 11 2013
%e A096027 27723 mod 2 = 27724 mod 3 = 27725 mod 4 = 27726 mod 5 = 27727 mod 6 = 27728 mod 7 = 27729 mod 8 = 27730 mod 9 = 27731 mod 10 = 27731 mod 11 = 27731 mod 12 = 1 and 27732 mod 13 = 3, hence 27723 is in the sequence.
%o A096027 (PARI) {k=11;m=900000;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,",")))}
%o A096027 (Magma) [n: n in [1..900000] | forall{j: j in [0..10] | IsOne((n+j) mod (2+j)) and (n+11) mod 13 ne 1}]; // _Bruno Berselli_, Apr 11 2013
%Y A096027 Cf. A007310, A017629, A096022, A096023, A096024, A096025, A096026.
%K A096027 nonn,easy
%O A096027 1,1
%A A096027 _Klaus Brockhaus_, Jun 15 2004