

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
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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 YinYang 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(n1)/A026741(n), n>0.
G.f.: ( 33*x2*x^23*x^4x^5 ) / ( (x1)^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



