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A096026
Numbers k such that (k+j) mod (2+j) = 1 for j from 0 to 8 and (k+9) mod 11 <> 1.
6
2523, 5043, 7563, 10083, 12603, 15123, 17643, 20163, 22683, 25203, 30243, 32763, 35283, 37803, 40323, 42843, 45363, 47883, 50403, 52923, 57963, 60483, 63003, 65523, 68043, 70563, 73083, 75603, 78123, 80643, 85683, 88203, 90723, 93243
OFFSET
1,1
COMMENTS
Numbers k such that k mod 2520 = 3 and k mod 27720 <> 3.
FORMULA
G.f.: 3*x*(839*x^10 +840*x^9 +840*x^8 +840*x^7 +840*x^6 +840*x^5 +840*x^4 +840*x^3 +840*x^2 +840*x +841) / ((x -1)^2*(x +1)*(x^4 -x^3 +x^2 -x +1)*(x^4 +x^3 +x^2 +x +1)). - Colin Barker, Apr 11 2013
EXAMPLE
2523 mod 2 = 2524 mod 3 = 2525 mod 4 = 2526 mod 5 = 2527 mod 6 = 2528 mod 7 = 2529 mod 8 = 2530 mod 9 = 2531 mod 10 = 1 and 2532 mod 11 = 2, hence 2523 is in the sequence.
MATHEMATICA
Select[Range[94000], Union[Mod[#+Range[0, 8], Range[2, 10]]]=={1}&&Mod[ #+9, 11]!=1&] (* or *) LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {2523, 5043, 7563, 10083, 12603, 15123, 17643, 20163, 22683, 25203, 30243}, 40](* Harvey P. Dale, Sep 25 2019 *)
PROG
(PARI) {k=9; m=95000; 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..100000] | forall{j: j in [0..8] | IsOne((n+j) mod (2+j)) and (n+9) mod 11 ne 1}]; // Bruno Berselli, Apr 11 2013
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Jun 15 2004
STATUS
approved