A259555 -id:A259555 - cp's OEIS frontend

A000466 a(n) = 4*n^2 - 1.

-1, 3, 15, 35, 63, 99, 143, 195, 255, 323, 399, 483, 575, 675, 783, 899, 1023, 1155, 1295, 1443, 1599, 1763, 1935, 2115, 2303, 2499, 2703, 2915, 3135, 3363, 3599, 3843, 4095, 4355, 4623, 4899, 5183, 5475, 5775, 6083, 6399, 6723, 7055, 7395

Offset: 0

Chan Siu Kee (skchan5(AT)hkein.ie.cuhk.hk)

Sum_{n>=1} (-1)^n*a(n)/n! = 1 - 1/e = A068996. - Gerald McGarvey, Nov 06 2007
Sequence arises from reading the line from -1, in the direction -1, 15, ... and the same line from 3, in the direction 3, 35, ..., in the square spiral whose nonnegative vertices are the squares A000290. - Omar E. Pol, May 24 2008
a(n) is the product of the consecutive odd integers 2n-1 and 2n+1 (cf. A005408). - Doug Bell, Mar 08 2009
For n>0: a(n) = A176271(2*n,n); cf. A016754, A053755. - Reinhard Zumkeller, Apr 13 2010
a(n+1) gives the curvature c(n) of the n-th circle touching the two equal semicircles of the symmetric arbelos (1/2, 1/2) and the (n-1)-st circle, with input c(0) = 3 =  A059100(1) (referring to the second circle of the Pappus chain), for n >= 0. - Wolfdieter Lang and Kival Ngaokrajang, Jul 03 2015
After 3, a(n) is pseudoprime to base 2n. For example: (2*2)^(a(2)-1) == 1 (mod a(2)), in fact 4^14 = 15*17895697+1. - Bruno Berselli, Sep 24 2015
Numbers m such that m+1 and (m+1)/4 are squares. - Bruno Berselli, Mar 03 2016
After -1, the least common multiple of 2*m+1 and 2*m-1. - Colin Barker, Feb 11 2017
This sequence contains all products of the twin prime pairs (see A037074). - Charles Kusniec, Oct 03 2019

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
L. B. W. Jolley, Summation of Series, Dover, 2nd ed., 1961.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
A. Languasco and A. Zaccagnini, Manuale di Crittografia, Ulrico Hoepli Editore (2015), p. 259.

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Kival Ngaokrajang, Illustration of the Pappus chain (downwards direction).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001539, A016286, A016742.
Factor of A160466. Superset of A037074.
Cf. A059100 (curvatures for a Pappus chain).

[4*n^2-1: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A000466:=n->4*n^2-1; seq(A000466(n), n=0..100); # Wesley Ivan Hurt, Nov 19 2013

4 Range[0,50]^2-1 (* Harvey P. Dale, Jan 23 2011 *)

makelist(4*n^2-1, n, 0, 50); /* Martin Ettl, Nov 12 2012 */

a(n)=4*n^2-1 \\ Charles R Greathouse IV, Oct 27 2011

[4*n^2-1 for n in (0..50)] # Bruno Berselli, Sep 24 2015

O.g.f.: ( 1-6*x-3*x^2 ) / (x-1)^3 . - R. J. Mathar, Mar 24 2011
E.g.f.: (-1 + 4*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 26 2016
Sum_{n>=1} 1/a(n) = 1/2 [Jolley eq. 233]. - Benoit Cloitre, Apr 05 2002
Sum_{n>=1} 2/a(n) = 1 = 2/3 + 2/15 + 2/35 + 2/63 + 2/99 + 2/143, ..., with partial sums: 2/3, 4/5, 6/7, 8/9, 10/11, 12/13, 14/15, ... - Gary W. Adamson, Jun 16 2003
1/3 + Sum_{n>=2} 4/a(n) = 1 = 1/3 + 4/15 + 4/35 + 4/63, ..., with partial sums: 1/3, 3/5, 5/7, 7/9, 9/11, ..., (2n+1)/(2n+3). - Gary W. Adamson, Jun 18 2003
Sum_{n>=0} 2/a(2*n+1) = Pi/4 = 2/3 + 2/35 + 2/99, ... = (1 - 1/3) + (1/5 - 2/7) + (1/9 - 1/11) + ... = Sum_{n>=0} (-1)^n/(2*n+1). - Gary W. Adamson, Jun 22 2003
Product(n>=1, (a(n)+1)/a(n)) = Pi/2 (Wallis formula). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
a(n)+2 = A053755(n). - Zak Seidov, Jan 16 2007
a(n)^2 + A008586(n)^2 = A053755(n)^2 (Pythagorean triple). - Zak Seidov, Jan 16 2007
a(n) = a(n-1) + 8*n - 4 for n > 0, a(0)=-1. - Vincenzo Librandi, Dec 17 2010
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 - 1/2 = (A019669-1)/2. [Jolley eq (366)]. - R. J. Mathar, Mar 24 2011
For n>0, a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^3*(cos(x))^(2*n-2)). - Francesco Daddi, Aug 02 2011
Nonlinear recurrence for c(n) = a(n+1) (see the arbelos comment above) from Descartes' three circle theorem (see the links under A259555): c(n) = 4 + c(n-1) + 4*sqrt(c(n-1) + 1), with input c(0) = 3  =  A059100(1), for n >= 0. The appropriate solution of this recurrence is c(n-1) + 1 = 4*n^2. - Wolfdieter Lang, Jul 03 2015
a(n) = 3*Pochhammer(5/2,n-1)/Pochhammer(1/2,n-1). Hence, the e.g.f. for a(n+1), i.e., dropping the first term, is 3* 1F1(5/2;1/2;x), with 1F1 being the confluent hypergeometric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/sqrt(2))/sqrt(2). - Amiram Eldar, Feb 04 2021

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.

A259054 a(n) = 4n^2 - 4n + 19, n >= 1.

Original entry on oeis.org

19, 27, 43, 67, 99, 139, 187, 243, 307, 379, 459, 547, 643, 747, 859, 979, 1107, 1243, 1387, 1539, 1699, 1867, 2043, 2227, 2419, 2619, 2827, 3043, 3267, 3499, 3739, 3987, 4243, 4507, 4779, 5059, 5347, 5643, 5947, 6259, 6579, 6907, 7243, 7587, 7939, 8299, 8667, 9043, 9427, 9819

Offset: 1

Wolfdieter Lang and Kival Ngaokrajang, Jul 02 2015

a(n) gives twice the inverse radius of the circles touching the large Arbelos (2/3,1/3) circle (radius 1) and the n-th and (n-1)-th circles of the counterclockwise Pappus chain.
For twice the curvatures (inverse radii) of the counterclockwise Pappus chain of the (2/3,1/3) arbelos see A114949, also for the MathWorld link to Pappus chain.
For the small curvatures touching the left circle of the (2/3,1/3) arbelos and the n-th and (n-1)-st circles of the counterclockwise Pappus chain see A259555.
The curvatures of the circles can be computed with Descartes' three (actually 5) circle theorem. See A259555 for links to Descartes' theorem.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A114949, A259555.

Array[4 #^2 - 4 # + 19 &, 40] (* Michael De Vlieger, Jul 02 2015 *)
LinearRecurrence[{3,-3,1},{19,27,43},40] (* Harvey P. Dale, May 17 2016 *)

vector(60, n, 4*n^2 - 4*n + 19) \\ Michel Marcus, Jul 03 2015

a(n) = 4*n^2 - 4*n + 19, n >= 1.
O.g.f.: x*(19 - 30*x + 19*x^2)/(1 - x)^3.
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(4*x^2 + 19) - 19.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

Terms a(37) and beyond from Andrew Howroyd, May 01 2020

A259055 a(n) = 9n^2 + 18n + 7.

Original entry on oeis.org

7, 34, 79, 142, 223, 322, 439, 574, 727, 898, 1087, 1294, 1519, 1762, 2023, 2302, 2599, 2914, 3247, 3598, 3967, 4354, 4759, 5182, 5623, 6082, 6559, 7054, 7567, 8098, 8647, 9214, 9799, 10402, 11023, 11662, 12319, 12994, 13687, 14398, 15127

Offset: 0

Wolfdieter Lang and Kival Ngaokrajang, Jul 03 2015

a(n) gives twice the curvature of the n-th circle touching the two semicircles of the (2/3,1/3) arbelos and the (n-1)-th circle, with input circle of twice the curvature a(0) = A114949(1) = 7 (referring to the second circle of the counterclockwise Pappus chain).

Kival Ngaokrajang, Illustration of one half of the initial terms.
Eric Weisstein's World of Mathematics, Descartes Circle theorem.
Eric Weisstein's World of Mathematics, Pappus chain.
Wikipedia, Descartes' Theorem.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A114949, A259054, A259555.

A259055:=n->9*n^2+18*n+7: seq(A259055(n), n=0..100); # Wesley Ivan Hurt, Feb 04 2017

Table[9 n^2 + 18 n + 7, {n, 0, 40}] (* Michael De Vlieger, Jul 03 2015 *)
LinearRecurrence[{3,-3,1},{7,34,79},50] (* Harvey P. Dale, Sep 05 2018 *)

a(n)=9*n^2+18*n+7 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 9*(n+1)^2 - 2, n >= 0.
O.g.f.: (-2*x^2+13*x+7)/(1-x)^3.
Recurrence: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 3, with a(0)=7, a(1)=34, and a(2)=79.
Descartes' three (actually five) circle theorem (see links) leads to a nonlinear recurrence for twice the curvatures: a(n) = 2*(3 + 3/2) + a(n-1) + 4*sqrt((3 + 3/2)*a(n-1)/2 + 9/2) = 9 + a(n-1) + 6*sqrt(a(n-1) + 2), with input a(0) = 7 = 2*A114949(1). This leads to a quadratic equation with the relevant solution a(n) = 9*n^2 + 18*n + 7.
E.g.f.: exp(x)*(9*x*(x + 3) + 7). - Elmo R. Oliveira, Oct 20 2024

cp's OEIS Frontend

A000466 a(n) = 4*n^2 - 1.

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

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A259054 a(n) = 4*n^2 - 4*n + 19, n >= 1.

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A259055 a(n) = 9*n^2 + 18*n + 7.

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A259054 a(n) = 4n^2 - 4n + 19, n >= 1.

A259055 a(n) = 9n^2 + 18n + 7.