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A164900 revision #40

A164900
a(2n) = 4*n*(n+1) + 3; a(2n+1) = 2*n*(n+2) + 3.
3
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
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
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
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Aug 30 2009
STATUS
proposed