A157474 -id:A157474 - cp's OEIS frontend

A173511 a(n) = 4*n^2 + floor(n/2).

0, 4, 17, 37, 66, 102, 147, 199, 260, 328, 405, 489, 582, 682, 791, 907, 1032, 1164, 1305, 1453, 1610, 1774, 1947, 2127, 2316, 2512, 2717, 2929, 3150, 3378, 3615, 3859, 4112, 4372, 4641, 4917, 5202, 5494, 5795, 6103, 6420, 6744, 7077, 7417, 7766, 8122

Offset: 0

Reinhard Zumkeller, Feb 20 2010

			a(6) = 147; 4(6)^2 + floor(6/3) = 144 + 3 = 147.

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A002943, A007494, A007742, A110654, A157474, A173512.

A173511:=n->4*n^2 + floor(n/2); seq(A173511(k), k=0..100); # Wesley Ivan Hurt, Nov 01 2013

Table[4n^2 + Floor[n/2], {n,0,100}] (* Wesley Ivan Hurt, Nov 01 2013 *)
LinearRecurrence[{2,0,-2,1},{0,4,17,37},50] (* Harvey P. Dale, Nov 23 2019 *)

a(n) = 4*n^2 + n\2 \\ Charles R Greathouse IV, Jun 11 2015

def A173511(n): return (n**2<<2)+(n>>1) # Chai Wah Wu, Jan 18 2023

a(n) = floor((2*n + 1/8)^2).
a(n+1) - a(n) = A173512(n).
a(n) = A002943(n) - A007494(n) = A007742(n) - A110654(n).
a(2*n) = A157474(n) for n>0.
From - R. J. Mathar, Feb 21 2010: (Start)
a(n)= 2*a(n-1) -2*a(n-3) +a(n-4).
G.f.: -x*(4+9*x+3*x^2)/((1+x)*(x-1)^3). (End)
E.g.f.: (x*(8*x + 9)*cosh(x) + (8*x^2 + 9*x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2024

A157475 a(n) = 512n + 16.

Original entry on oeis.org

528, 1040, 1552, 2064, 2576, 3088, 3600, 4112, 4624, 5136, 5648, 6160, 6672, 7184, 7696, 8208, 8720, 9232, 9744, 10256, 10768, 11280, 11792, 12304, 12816, 13328, 13840, 14352, 14864, 15376, 15888, 16400, 16912, 17424, 17936, 18448, 18960

Offset: 1

Vincenzo Librandi, Mar 01 2009

The identity (2048*n^2+128*n+1)^2-(16*n^2+n)*(512*n+16)^2=1 can be written as A157476(n)^2-A157474(n)*a(n)^2=1 (see also second comment in A157476). [rewritten by Bruno Berselli, Aug 22 2011]

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A157474, A157476.

512*Range[40]+16 (* or *) LinearRecurrence[{2,-1},{528,1040},40] (* Harvey P. Dale, Dec 07 2011 *)

G.f.: 16*x*(33-x)/(1-x)^2. - Bruno Berselli, Aug 22 2011

a(1)=528, a(2)=1040, a(n) = 2*a(n-1)-a(n-2). - Harvey P. Dale, Dec 07 2011

A157476 a(n) = 2048n^2 + 128n + 1.

Original entry on oeis.org

2177, 8449, 18817, 33281, 51841, 74497, 101249, 132097, 167041, 206081, 249217, 296449, 347777, 403201, 462721, 526337, 594049, 665857, 741761, 821761, 905857, 994049, 1086337, 1182721, 1283201, 1387777, 1496449, 1609217, 1726081, 1847041

Offset: 1

Vincenzo Librandi, Mar 01 2009

The identity (2048*n^2+128*n+1)^2-(16*n^2+n)*(512*n+16)^2=1 can be written as a(n)^2-A157474(n)*A157475(n)^2=1. [rewritten by Bruno Berselli, Aug 22 2011]

This is the case s=4 of the identity (8*n^2*s^4+8*n*s^2+1)^2 - (n^2*s^2+n)*(8*n*s^3+4*s)^2 = 1. - Bruno Berselli, Jan 25 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A157474, A157475.

Table[2048n^2+128n+1,{n,30}] (* or *) LinearRecurrence[{3,-3,1},{2177,8449,18817},30] (* Harvey P. Dale, Aug 15 2011 *)

a(n)=2048*n^2+128*n+1 \\ Charles R Greathouse IV, Jun 17 2017

From Harvey P. Dale, Aug 15 2011: (Start)
G.f.: x*(-x^2-1918*x-2177)/(x-1)^3.
a(1)=2177, a(2)=8449, a(3)=18817, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). (End)

A157716 One-eighth of triangular numbers (integers only).

Original entry on oeis.org

0, 15, 17, 62, 66, 141, 147, 252, 260, 395, 405, 570, 582, 777, 791, 1016, 1032, 1287, 1305, 1590, 1610, 1925, 1947, 2292, 2316, 2691, 2717, 3122, 3150, 3585, 3615, 4080, 4112, 4607, 4641, 5166, 5202, 5757, 5795, 6380, 6420, 7035, 7077, 7722, 7766, 8441

Offset: 1

Vladimir Joseph Stephan Orlovsky, Mar 04 2009

From Lamine Ngom, Oct 27 2020: (Start)
Numbers of the form (4*k)^2-k (A157446) or (4*k)^2+k (A157474).
Also numbers k such that 1+64*k is a square. (End)
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(32*n))*(1 - q^(32*n-15))*(1 - q^(32*n-17)) = 1 - q^15 - q^17 + q^62 + q^66 - q^141 - q^147 + + - - .... - Peter Bala, Dec 24 2024

			The first three members of A000217 that are divisible by 8 are A000217(0), A000217(15) and A000217(16), so a(1) = A000217(0)/8 = 0, a(2) = A000217(15)/8 = 15, a(3) = A000217(16)/8 = 17.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A101877, A219257.

seq((2*n-1 + 7/8*(-1)^n)^2 - 1/64, n = 1 .. 1000); # Robert Israel, Apr 20 2014

Array[(2 # - 1 + 7/8*(-1)^#)^2 - 1/64 &, 46] (* or *)
Rest@ CoefficientList[Series[x^2*(15 + 2 x + 15 x^2)/((1 + x)^2*(1 - x)^3), {x, 0, 46}], x] (* Michael De Vlieger, Nov 05 2020 *)

G.f.: x^2*(15+2*x+15*x^2)/((1+x)^2*(1-x)^3 ). [Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009; checked and corrected by R. J. Mathar, Sep 16 2009]
a(n) =  (2*n-1 + 7/8*(-1)^n)^2 -1/64. - Robert Israel, Apr 20 2014
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Nov 10 2020
Sum_{n>=2} 1/a(n) = 16 - (sqrt(2*(2+sqrt(2))) + sqrt(2) + 1)*Pi. - Amiram Eldar, Mar 17 2022

Definition edited by N. J. A. Sloane, Mar 08 2009

cp's OEIS Frontend

A173511 a(n) = 4*n^2 + floor(n/2).

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A157475 a(n) = 512n + 16.

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A157476 a(n) = 2048n^2 + 128n + 1.

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A157716 One-eighth of triangular numbers (integers only).

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