A222465 -id:A222465 - cp's OEIS frontend

A114949 a(n) = n^2 + 6.

6, 7, 10, 15, 22, 31, 42, 55, 70, 87, 106, 127, 150, 175, 202, 231, 262, 295, 330, 367, 406, 447, 490, 535, 582, 631, 682, 735, 790, 847, 906, 967, 1030, 1095, 1162, 1231, 1302, 1375, 1450, 1527, 1606, 1687, 1770, 1855, 1942, 2031, 2122, 2215, 2310, 2407, 2506

Offset: 0

Cino Hilliard, Feb 21 2006

2/a(n) = R(n)/r, n >= 0, with R(n) the n-th radius of the counterclockwise Pappus chain of the arbelos with semicircle radii r, r1 = 2r/3, r2 = r - r1 = r/3. See the MathWorld link for such a Pappus chain. The clockwise chain companion has circle radii R'(n)/r = 2/A222465(n), n >= 0. - Wolfdieter Lang, Mar 01 2013

			The arbelos chain defined in a comment above has circle radii [1/3, 2/7, 1/5, 2/15, 1/11, 2/31, 1/21, 2/55, 1/35, 2/87, 1/53,...], for n >= 0. - _Wolfdieter Lang_, Mar 01 2013

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pappus chain.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A002522, A016742, A059100, A082044, A087475, A114964 (see comment), A117619, A117950, A117951, A222465.

A114949:=n->n^2+6: seq(A114949(n), n=0..100); # Wesley Ivan Hurt, Apr 28 2017

Range[0, 49]^2 + 6 (* Alonso del Arte, Jan 30 2013 *)

From R. J. Mathar, May 17 2009: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -(6 - 11*x + 7*x^2)/(x - 1)^3. (End)
a(n) = 2*n + a(n - 1) - 1, with n > 0, a(0)=6. - Vincenzo Librandi, Nov 13 2010
a(n) = A000290(n) + 6. - Omar E. Pol, Mar 02 2013
a(n) = ((n-2)^3 + (n-1)^3 + n^3 + (n+1)^3 + (n+2)^3)/(5*n) for n>=1. - Bruno Berselli, May 12 2014
For n >= 1, a(n) = (A016742(n) + A082044(n) - 1)/A000290(n). - Bruce J. Nicholson, Apr 19 2017
From Amiram Eldar, Nov 02 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(6)*Pi*coth(sqrt(6)*Pi))/12.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(6)*Pi*cosech(sqrt(6)*Pi))/12. (End)
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(5/6)*sinh(sqrt(5)*Pi)/sinh(sqrt(6)*Pi).
Product_{n>=0} (1 + 1/a(n)) = sqrt(7/6)*sinh(sqrt(7)*Pi)/sinh(sqrt(6)*Pi). (End)
E.g.f.: exp(x)*(6 + x + x^2). - Elmo R. Oliveira, Jan 17 2025

A164897 a(n) = 4n(n+1) + 3.

Original entry on oeis.org

3, 11, 27, 51, 83, 123, 171, 227, 291, 363, 443, 531, 627, 731, 843, 963, 1091, 1227, 1371, 1523, 1683, 1851, 2027, 2211, 2403, 2603, 2811, 3027, 3251, 3483, 3723, 3971, 4227, 4491, 4763, 5043, 5331, 5627, 5931, 6243, 6563, 6891, 7227, 7571, 7923, 8283, 8651, 9027, 9411

Offset: 0

Paul Curtz, Aug 30 2009

One-fourth the sum of the three terms produced by the division of complex numbers (2*n-3+(2*n-1)*i)/(2*n+1+(2*n+3)*i). For (b+c*i)/(d+e*i) the three terms in parentheses are ((b*d+c*e)+(c*d-b*e)*i)/(d^2+e^2). By substituting b=2*n-3, c=2*n-1, d=2*n+1, and e=2*n+3 one gets a(n). - J. M. Bergot, Sep 10 2015

The continued fraction expansion of sqrt(a(n)) is [2n+1; {2n+1, 4n+2}]. - Magus K. Chu, Sep 08 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..900
R. M. Green and Tianyuan Xu, 2-roots for simply laced Weyl groups, arXiv:2204.09765 [math.RT], 2022.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000124, A008590, A010731, A016743, A069894.

Odd-indexed terms of A059100.

[4*n*(n+1)+3: n in [0..50]]; // Vincenzo Librandi, Apr 24 2011

A164897:=n->4*n*(n+1)+3: seq(A164897(n), n=0..100); # Wesley Ivan Hurt, Sep 10 2015

Table[4 n (n + 1) + 3, {n, 0, 50}] (* Harvey P. Dale, Jan 23 2011 *)

a(n)=4*n*(n+1)+3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A000124(2*n) + A000124(2*n+1) = A069894(n)+1.
a(n+1) - a(n) = 8n+8 = A008590(n+1) (first differences).
a(n+1) - 2*a(n) + a(n-1) = 8 = A010731(n) (second differences).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n>2.
G.f.: (3+2*x+3*x^2) / (1-x)^3.
Sum_{k=n+1..2*n+1} a(k) - Sum_{k=0..n} a(k) = (2*n+2)^3. - Bruno Berselli, Jan 24 2011
E.g.f.: (4x^2 + 8x + 1)*exp(x). - G. C. Greubel, Sep 22 2015
a(n)^2 = A222465(n)*A222465(n+1) - 12. - Ezhilarasu Velayutham, Mar 18 2020
Sum_{n>=0} 1/a(n) = tanh(Pi/sqrt(2))*Pi/(4*sqrt(2)). - Amiram Eldar, Aug 21 2022
a(n) = A059100(2*n+1). - Dimitri Papadopoulos, Nov 21 2023

Definition simplified by R. J. Mathar, Sep 16 2009

A242412 a(n) = (2*n-1)^2 + 14.

Original entry on oeis.org

15, 23, 39, 63, 95, 135, 183, 239, 303, 375, 455, 543, 639, 743, 855, 975, 1103, 1239, 1383, 1535, 1695, 1863, 2039, 2223, 2415, 2615, 2823, 3039, 3263, 3495, 3735, 3983, 4239, 4503, 4775, 5055, 5343, 5639, 5943, 6255, 6575, 6903, 7239, 7583, 7935, 8295, 8663, 9039, 9423, 9815

Offset: 1

Aaron David Fairbanks, May 13 2014

The previous definition was "a(n) = normalized inverse radius of the inscribed circle that is tangent to the left circle of the symmetric arbelos and the n-th and (n-1)-st circles in the Pappus chain".
See links section for image of these circles, via Wolfram MathWorld (there an asymmetric arbelos is shown).
The Rothman-Fukagawa article has another picture of the circles, based on a Japanese 1788 sangaku problem. - N. J. A. Sloane, Jan 02 2020

			For n = 1, the radius of the outermost circle divided by the radius of a circle drawn tangent to all three of the initial inner circle, the opposite inner circle (the 0th circle in the chain), and the 1st circle in the chain is 15.
For n = 2, the radius of the outermost circle divided by the radius of a circle drawn tangent to all three of the initial inner circle, the 1st circle in the chain, and the 2nd circle in the chain is 23.

N. J. A. Sloane, Table of n, a(n) for n = 1..1000
Brady Haran and Simon Pampena, Epic Circles, Numberphile video (2014).
Tony Rothman and Hidetoshi Fukagawa, Japanese temple geometry, Scientific American, Vol. 278, No. 5, May 1998, pp. 85-91.
Eric Weisstein's World of Mathematics, Image of inscribed circles (in red).
Eric Weisstein's World of Mathematics, Pappus Chain.
Wikipedia, Pappus chain.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000012, A060747, A059100, A114949, A222465, A259555.

[4*n^2 - 4*n + 15: n in [1..50]]; // Wesley Ivan Hurt, May 13 2014

A242412:=n->4*n^2 - 4*n + 15; seq(A242412(n), n=1..50); # Wesley Ivan Hurt, May 13 2014

Table[4 n^2 - 4 n + 15, {n, 50}] (* Wesley Ivan Hurt, May 13 2014 *)
LinearRecurrence[{3,-3,1},{15,23,39},50] (* Harvey P. Dale, Feb 22 2023 *)

a(n) = 4*n^2 - 4*n + 15 \\ Charles R Greathouse IV, May 14 2014

a(n) = 4*n^2 - 4*n + 15.
From Colin Barker, May 14 2014: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -x*(15*x^2 - 22*x + 15)/(x-1)^3. (End)
From Descartes three circle theorem:
a(n) = 2 + c(n) + c(n-1) + 2*sqrt(2*(c(n) + c(n-1)) + c(n)*c(n-1)), with c(n) = A059100(n) = n^2 + 2, n >= 1, which produces 4*n^2 - 4*n + 15. - Wolfdieter Lang, Jul 01 2015
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(4*x^2 + 15) - 15.
a(n) = A060747(n)^2 + 14. (End)

More terms from Wesley Ivan Hurt, May 13 2014
More terms and links from Robert G. Wilson v, May 13 2014
Edited: Name reformulated (with consent of the author). - Wolfdieter Lang, Jul 01 2015
Edited by N. J. A. Sloane, Jan 02 2020, simplifying the definition and adding a reference to the fact that this sequence arose in a sangaku problem from 1788 in a temple in Tokyo Prefecture.

cp's OEIS Frontend

A114949 a(n) = n^2 + 6.

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A164897 a(n) = 4*n*(n+1) + 3.

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A242412 a(n) = (2*n-1)^2 + 14.

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Magma

Maple

Mathematica

PARI

Formula

Extensions

A164897 a(n) = 4n(n+1) + 3.