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 A164900 a(2n) = 4*n*(n+1) + 3; a(2n+1) = 2*n*(n+2) + 3. 2
 3, 3, 11, 9, 27, 19, 51, 33, 83, 51, 123, 73, 171, 99, 227, 129, 291, 163, 363, 201, 443, 243, 531, 289, 627, 339, 731, 393, 843, 451, 963, 513, 1091, 579, 1227, 649, 1371, 723, 1523, 801, 1683, 883, 1851, 969, 2027, 1059, 2211, 1153 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS a(n) = largest odd divisor of A059100(n+1). Proof: Observe that a(2n) = A059100(2n+1) and a(2n+1) = (A059100(2n+2))/2 and note that (A059100(m))/2 is odd for even m. - Jeremy Gardiner, Aug 25 2013 a(n) is also the denominator of the (n+1)-st largest circle in a special case of the Pappus chain inspired by the Yin-Yang symbol. See illustration in the links. - Kival Ngaokrajang, Jun 20 2015 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..10000 Kival Ngaokrajang, Illustration of initial terms. Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1) FORMULA a(2n) = A164897(n); a(2n+1) = A058331(n+1). a(n) = A164845(n-1)/A026741(n), n>0. G.f.: ( -3-3*x-2*x^2-3*x^4-x^5 ) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Jan 21 2011 a(n) = ((-1)^n+3)*(n^2+2*n+3)/4. - Bruno Berselli, Jan 21 2011 From Amiram Eldar, Aug 09 2022: (Start) a(n) = numerator(((n+1)^2 + 2)/2). Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(2))*Pi/sqrt(2) + tanh(Pi/sqrt(2))*Pi/(2*sqrt(2)) - 1)/2. (End) MATHEMATICA LinearRecurrence[{0, 3, 0, -3, 0, 1}, {3, 3, 11, 9, 27, 19}, 50] (* Amiram Eldar, Aug 09 2022 *) PROG (Magma) [((-1)^n+3)*(n^2+2*n+3)/4: n in [0..50]]; // Vincenzo Librandi, Aug 07 2011 (PARI) vector(100, n, n--; (1/4)*((-1)^n+3)*(n^2+2*n+3)) \\ Derek Orr, Jun 27 2015 CROSSREFS Cf. A026741, A058331, A059100, A164845, A164897. Sequence in context: A309692 A107229 A302510 * A304082 A122167 A095019 Adjacent sequences: A164897 A164898 A164899 * A164901 A164902 A164903 KEYWORD nonn,easy AUTHOR Paul Curtz, Aug 30 2009 STATUS approved

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Last modified December 6 10:03 EST 2022. Contains 358630 sequences. (Running on oeis4.)