A096025 Numbers n such that (n+j) mod (2+j) = 1 for j from 0 to 6 and (n+7) mod 9 <> 1.

843, 1683, 3363, 4203, 5883, 6723, 8403, 9243, 10923, 11763, 13443, 14283, 15963, 16803, 18483, 19323, 21003, 21843, 23523, 24363, 26043, 26883, 28563, 29403, 31083, 31923, 33603, 34443, 36123, 36963, 38643, 39483, 41163, 42003, 43683

Offset

1,1

Comments

Numbers n such that n mod 840 = 3 and n mod 2520 <> 3.

Links

Table of n, a(n) for n=1..35.

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

Formula

a(n) = -3*(209+70*(-1)^n-420*n). a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: 3*x*(279*x^2+280*x+281) / ((x-1)^2*(x+1)). - Colin Barker, Apr 11 2013

Example

843 mod 2 = 844 mod 3 = 845 mod 4 = 846 mod 5 = 847 mod 6 = 848 mod 7 = 849 mod 8 = 1 and 850 mod 9 = 4, hence 843 is in the sequence.

Mathematica

LinearRecurrence[{1, 1, -1}, {843, 1683, 3363}, 40] (* Harvey P. Dale, Nov 22 2015 *)

Prog

(PARI) {k=7; m=44000; 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..44000] | forall{j: j in [0..6] | IsOne((n+j) mod (2+j)) and (n+7) mod 9 ne 1}]; // Bruno Berselli, Apr 11 2013

Crossrefs

Cf. A007310, A017629, A096022, A096023, A096024, A096026, A096027.

Sequence in context: A252568 A045243 A368778 * A004969 A004949 A252061

Adjacent sequences: A096022 A096023 A096024 * A096026 A096027 A096028

Keyword

nonn,easy

Author

Klaus Brockhaus, Jun 15 2004

Status

approved