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Auxiliary r(n) sequence used to prove some properties about Rowland's sequence: r(1) = 1, and r(n) = 1/2*(c(n)+1), where c(n) is A190894, for n>1.
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%I #27 Aug 11 2023 10:02:45

%S 1,5,6,11,12,23,24,47,48,50,51,101,102,105,110,111,117,233,234,467,

%T 468,470,471,941,942,945,1889,1890,3779,3780,7559,7560,7566,15131,

%U 15132,15158,15159,15162,30323,30324,60647,60648,60650,60651,60701,60702,121403,121404,242807,242808,242810

%N Auxiliary r(n) sequence used to prove some properties about Rowland's sequence: r(1) = 1, and r(n) = 1/2*(c(n)+1), where c(n) is A190894, for n>1.

%C This sequence is matched with another auxiliary sequence called c(n) (A190894). Rowland's sequence (A106108) can be easily described in terms of c(n) and r(n). Also, they can be used to prove easily that the difference between two consecutive terms is always 1 or a prime.

%C This sequence is related to Rowland's sequence (A106108) with initial condition a(1)=7.

%C Sequence r(n) satisfies 2r(n) - 1 = c(n), for any n>1.

%C For further information, see the references.

%H F. Chamizo, D. Raboso, and S. Ruiz-Cabello, <a href="https://doi.org/10.37236/2006">On Rowland's sequence</a>, Electronic J. Combin., Vol. 18(2), 2011, #P10.

%H E. S. Rowland, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL11/Rowland/rowland21.html">A natural prime-generating recurrence</a>, J. Integer Seq., 11(2): Article 08.2.8, 13, 2008.

%e For n = 2, r(2) = 1/2 * (c(2) + 1) = 1/2 * (9 + 1) = 5.

%e For n = 3, r(3) = 1/2 * (c(3) + 1) = 1/2 * (11 + 1) = 6.

%Y Cf. A106108, A190894.

%K nonn

%O 1,2

%A _SerafĂ­n Ruiz-Cabello_, May 23 2011