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 A096024 Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 5 and (n+6) mod 8 <> 1. 5
 423, 1263, 2103, 2943, 3783, 4623, 5463, 6303, 7143, 7983, 8823, 9663, 10503, 11343, 12183, 13023, 13863, 14703, 15543, 16383, 17223, 18063, 18903, 19743, 20583, 21423, 22263, 23103, 23943, 24783, 25623, 26463, 27303, 28143, 28983, 29823 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Numbers n such that n mod 840 = 423. LINKS Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (2,-1). FORMULA a(n) = 2*a(n-1)-a(n-2). G.f.: 3*x*(139*x+141) / (x-1)^2. - Colin Barker, Apr 11 2013 a(n) = 840*n-417. [Bruno Berselli, Apr 11 2013] EXAMPLE 423 mod 2 = 424 mod 3 = 425 mod 4 = 426 mod 5 = 427 mod 6 = 428 mod 7 = 1 and 429 mod 8 = 5, hence 423 is in the sequence. PROG (PARI) {k=6; m=30000; for(n=1, m, j=0; b=1; while(b&&j

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