A261327 -id:A261327 - cp's OEIS frontend

A262397 a(n) = floor(A261327(n)/9).

0, 0, 0, 1, 0, 3, 1, 5, 1, 9, 2, 13, 4, 19, 5, 25, 7, 32, 9, 40, 11, 49, 13, 59, 16, 69, 18, 81, 21, 93, 25, 107, 28, 121, 32, 136, 36, 152, 40, 169, 44, 187, 49, 205, 53, 225, 58, 245, 64, 267, 69, 289, 75, 312, 81, 336, 87, 361, 93, 387, 100, 413, 106, 441

Offset: 0

Paul Curtz, Sep 21 2015

Hexasections:
  0,  1,  4,  9, 16,  25,  36, ... = A000290(n)
  0,  5, 19, 40, 69, 107, 152, ... = c(n)
  0,  1,  5, 11, 18,  28,  40, ... = d(n+1)
  1,  9, 25, 49, 81, 121, 169, ... = A016754(n)
  0,  2,  7, 13, 21,  32,  44, ... = A240438(n+1)
  3, 13, 32, 59, 93, 136, 187, ... = e(n+1).
The six sequences have the signature (2, -1, 1, -2, 1), that is, the signature of a(n) without the 0's.
It appears that d(n+1) and A240438(n+1) are connected via the following scheme.
Let x(n) be the sequence that concatenates terms of d(n+1) in reverse order with terms of A240438(n+1), both without their index_0 term:
   ..., 18, 11,  5,  1,  0,  0,  2,  7, 13, 21, 32, ...
And consider the first and second differences of this sequence:
   ..., -7, -6, -4, -1,  0,  2,  5,  6,  8, 11, 12, ...
   ...,  1,  2,  3,  1,  2,  3,  1,  2,  3,  1,  2, ...
In the first differences, we get A047234(n+1) and A047267(n+1). And in the second differences, we get A010882(n).
In the same way, c(n) and e(n+1) are connected via the first and second differences of this sequence, with both their index_0 term:
   ...,  69,  40,  19,   5,   0,   3,  13,  32,  59, ...
that are respectively:
   ..., -29, -21, -14,  -5,   3,  10,  19,  27,  34, ...
   ...,   8,   7,   9,   8,   7,   9,   8,   7,   9, ... .
Is it possible to find a direct definition for a(n)?

			a(0) = floor(1/9) = 0, a(1)= floor (5/9) = 0, a(2) = floor(2/9) = 0, a(3)= floor (13/9) = 1.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,-2,0,1).

Cf. A000290, A010882, A016754, A047234, A047267, A240438, A261327.

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 70]/9 // Floor (* Jean-François Alcover, Sep 26 2015, after Vincenzo Librandi in A261327 *)

a(n) = numerator((n^2+4)/4)\9; \\ Michel Marcus, Sep 22 2015

concat([0,0,0], Vec(-x^3*(x^4 +x^3 +x^2 +x +1)*(x^12 -x^11 +x^10 -x^8 +2*x^6 -x^4 +x^2 -x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)*(x^2 +x +1)*(x^6 -x^3 +1)*(x^6 +x^3 +1)) + O(x^100))) \\ Colin Barker, Sep 25 2015

a(n)=if(n%2,n^2+4,(n/2)^2+1)\9 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = (A261327(n) - A261327(n) mod 9)/9.
From Colin Barker, Sep 25 2015: (Start)
a(n) = floor((n^2+4)/36) for n even.
a(n) = floor((n^2+4)/9) for n odd.
G.f.: -x^3*(x^4 +x^3 +x^2 +x +1)*(x^12 -x^11 +x^10 -x^8 +2*x^6 -x^4 +x^2 -x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)*(x^2 +x +1)*(x^6 -x^3 +1)*(x^6 +x^3 +1)). (End)

New name suggested by Michel Marcus, Sep 22 2015

A280579 Square array read by antidiagonals downwards giving the first differences A261327(n+p) - A261327(n), with p >= 0.

Original entry on oeis.org

0, 0, 4, 0, -3, 1, 0, 11, 8, 12, 0, -8, 3, 0, 4, 0, 24, 16, 27, 24, 28, 0, -19, 5, -3, 8, 5, 9, 0, 43, 24, 48, 40, 51, 48, 52, 0, -36, 7, -12, 12, 4, 15, 12, 16, 0, 68, 32, 75, 56, 80, 72, 83, 80, 84, 0, -59, 9, -27

Offset: 0

Paul Curtz, Jan 05 2017

Successive rows:
p
0:  0,   0,   0,   0,   0,   0,   0, ...
1:  4,  -3,  11,  -8,  24, -19,  43, ...
2:  1,   8,   3,  16,   5,  24,   7, ...
3: 12,   0,  27,  -3,  48, -12,  75, ...
4:  4,  24,   8,  40,  12,  56,  16, ...
5: 28,   5,  51,   4,  80,  -3, 115, ...
6:  9,  48,  15,  72,  21,  96,  27, ...
... .
Main diagonal: alternate 3*n^2, -3.
From p>0, the rows are multiples of 1, 1, 3, 4, 1, 3, 1, 8, 3, 5, 1, 12, 1, 7, 3, 16, 1, ... . Sequences appearing after division: shifted A144433 or A195161, A064680. For p=3, we have (n+2)^2, -n^2.
First column: alternate n^2, 4*(n^2 + n + 1). Its first differences (4, -3, 11, -8, 24, ...) is the sequence of the square array for p=1.
Third column: 0, 3, 8, 15, ... is A005563(n).
Fifth column: 5, 21, 45, 77, ... is a bisection of A061037(n).
Seventh column: 7, 16, 40, 55, 91, 112, ... is a subsequence of A061039(n).
Etc. From the Rydberg spectra of the hydrogen atom (mentioned in A261327).
Starting for instance from p=-3,at the main antidiagonal,yields:
-3:              -12,   0,  -27,  3, ... see p=3
-2:          -1,  -8,  -3,  -16, -5, ...     p=2
-1:     -4,   3, -11,   8,  -24, 19, ...     p=1.

Cf. A000290, A002061, A005563, A061037, A061039, A064680, A112087, A144433, A195161, A261327.

A281098 a(n) is the GCD of the sequence d(n) = A261327(k+n) - A261327(k) for all k.

Original entry on oeis.org

0, 1, 1, 3, 4, 1, 3, 1, 8, 3, 5, 1, 12, 1, 7, 3, 16, 1, 9, 1, 20, 3, 11, 1, 24, 1, 13, 3, 28, 1, 15, 1, 32, 3, 17, 1, 36, 1, 19, 3, 40, 1, 21, 1, 44, 3, 23, 1, 48, 1, 25, 3, 52, 1, 27, 1, 56, 3, 29, 1, 60, 1, 31, 3, 64, 1, 33, 1, 68, 3, 35, 1, 72, 1, 37, 3, 76, 1, 39, 1

Offset: 0

Paul Curtz, Jan 14 2017

Successive sequences:
   0,   0,  0,    0, ...    = 0 * ( )
   4,  -3,  11,  -8, ...    = 1 * ( )
   1,   8,   3,  16, ...    = 1 * ( )                 A195161
  12,   0,  27,  -3, ...    = 3 * (4, 0, 9, -1, ...)
   4,  24,   8,  40, ...    = 4 * (1, 6, 2, 10, ...)  A064680
5;   28,   5,  51,   4, ...    = 1 * ( )
   9,  48,  15,  72, ...    = 3 * (3, 16, 5, 24, ...) A195161
  52,  12,  83,  13, ...    = 1 * ( )
  16,  80,  24, 112, ...    = 8 * (2, 10, 3, 14, ...) A064080
  84   21, 123,  24, ...    = 3 * (28, 7, 41, 8, ...)
 25, 120,  35, 160, ...    = 5 * (5, 24, 7, 32, ...) A195161

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for linear recurrences with constant coefficients, signature (0,-1,0,1,0,2,0,1,0,-1,0,-1).

Cf. A064680, A144433 or A195161.

Cf. A000012, A005408, A008586, A010701, A109007 (bisection), A016825, A165988 (via A022998), A166138, A166304, A280579.

CoefficientList[Series[(-x (-1 - x - 4 x^2 - 5 x^3 - 3 x^4 - 6 x^5 + 3 x^6 - 5 x^7 + 4 x^8 - x^9 + x^10))/((x^2 - x + 1) (1 + x + x^2) (x - 1)^2*(1 + x)^2*(1 + x^2)^2), {x, 0, 79}], x] (* Michael De Vlieger, Feb 02 2017 *)

f(n) = numerator((4 + n^2)/4);
a(n) = gcd(vector(1000, k, f(k+n) - f(k))); \\ Michel Marcus, Jan 15 2017

A281098(n) = if(n%2, gcd((n\2)-1,3), n>>(bitand(n,2)/2)); \\ Antti Karttunen, Feb 15 2023

G.f.: -x*( -1 - x - 4*x^2 - 5*x^3 - 3*x^4 - 6*x^5 + 3*x^6 - 5*x^7 + 4*x^8 - x^9 + x^10 )/( (x^2 - x + 1)*(1 + x + x^2)*(x - 1)^2*(1 + x)^2*(1 + x^2)^2 ). - R. J. Mathar, Jan 31 2017
a(2*k)   = A022998(k).
a(2*k+1) = A109007(k-1).
a(3*k)   = interleave 3*k*(3 +(-1)^k)/2, 3.
a(3*k+1) = interleave 1, A166304(k).
a(3*k+2) = interleave A166138(k), 1.
a(4*k)   = 4*k.
a(4*k+1) = period 3: repeat [1, 1, 3].
a(4*k+2) = 1 + 2*k.
a(4*k+3) = period 3: repeat [3, 1, 1].
a(n+12) - a(n) = 6*A131743(n+3).
a(n) = (18*n + 40 - 16*cos(n*Pi/3) + 9*n*cos(n*Pi/2) + 32*cos(2*n*Pi/3) + (18*n - 40)*cos(n*Pi) + 3*n*cos(3*n*Pi/2) - 16*cos(5*n*Pi/3))/48. - Wesley Ivan Hurt, Oct 04 2018

Corrected and extended by Michel Marcus, Jan 15 2017

A306454 a(n) = A261327(n)/A013946(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 169, 1, 1, 1, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 289, 1, 1, 1, 1, 1, 841, 1, 1, 1, 25, 1, 1, 25, 1, 1, 1

Offset: 1

Paul Curtz, Feb 16 2019

nonn

Are all terms odd squares?
b(n) = A013946(n)*A261327(n) = 25, 4, 169, 25, 841, 100, 2809, 289, 7225, 676, 625, ... . Are all terms squares?
a(n) = A008833(n^2+4) if n is odd and A008833((n^2+4)/4) if n is even, so a(n) is always an odd square. - Jianing Song, Feb 27 2019
Are the square roots only primes?
The sequence of period 4: repeat [25, 1, 1, 25] appears apparently every 25 terms.
From Robert Israel, Mar 20 2019: (Start)
The first term whose square root is not 1 or a prime is a(261) = 25^2.
a(11+25*k) is divisible by 25. The first term where a(11+25*k) > 25 is a(261)=a(11+25*10)=625.
The first term where a(12+25*k) > 1 is a(1212)=a(12+25*48)=169.
The first term where a(13+25*k) > 1 is a(213)=a(13+25*8)=289.
a(14+25*k) is divisible by 25. The first term where a(14+25*k) > 25 is a(364)=a(14+25*14)=625.
All prime factors of members of the sequence are in A002144. For any p in A002144, there is k with 1 <= k < p^2/2 such that p^2 | a(n) if and only if n == k or -k (mod p^2). (End)

			A261327(n) = 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, ... .
A013946(n) = 5, 2, 13, 5, 29, 10, 53, 17, 85, 26,   5, 37, 173,  2, ... .

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A013946, A261327.

Cf. A002144, A008833.

core:= proc(n) local t; mul(t[1],t=select(s -> s[2]::odd, ifactors(n)[2])) end proc:
map(n -> numer((4+n^2)/4)/core(n^2+4), [$1..100]); # Robert Israel, Mar 20 2019

core[n_] := Times @@ Select[FactorInteger[n], OddQ[#[[2]]]&][[All, 1]];
a[n_] := Numerator[(n^2+4)/4]/core[n^2+4];
Array[a, 100] (* Jean-François Alcover, Jan 04 2022 *)

A013946(n) = core(n^2+4); \\ From A013946
A261327(n) = if(n%2, n^2+4, (n/2)^2+1); \\ From A261327
A306454(n) = (A261327(n)/A013946(n)); \\ Antti Karttunen, Feb 28 2019

A306454(n) = { my(k=((n^2)+4)/if(n%2,1,4)); k/core(k); }; \\ Antti Karttunen, Feb 28 2019, after Jianing Song's formula

A349118 Row sums of a triangle based on A261327.

Original entry on oeis.org

1, 5, 3, 18, 8, 47, 18, 100, 35, 185, 61, 310, 98, 483, 148, 712, 213, 1005, 295, 1370, 396, 1815, 518, 2348, 663, 2977, 833, 3710, 1030, 4555, 1256, 5520, 1513, 6613, 1803, 7842, 2128, 9215, 2490, 10740, 2891, 12425, 3333, 14278, 3818, 16307, 4348, 18520, 4925

Offset: 2

Paul Curtz, Nov 08 2021

The following triangle has A261327 as its diagonals:
  1
      5
  1       2
      5       13
  1       2        5
      5       13       29
  1       2        5        10
      5       13       29        53
  1       2        5        10        17
      5       13       29        53        85
  ...
a(0) = a(1) = 0.
a(n)'s final digit: neither 4 nor 9.
First full bisection difference table:
  0,  1,  3,  8,  18,  35,  61,  98, ...    = 0, A081489 = b(n)
  1,  2,  5, 10,  17,  26,  37,  50, ...    = A002522
  1,  3,  5,  7,   9,  11,  13,  15, ...    = A005408
  2,  2,  2,  2,   2,   2,   2,   2, ...    = A007395
  0,  0,  0,  0,   0,   0,   0,   0, ...    = A000004
Second full bisection difference table:
  0,   5,  18,  47, 100, 185, 310, 483, ...    = c(n)
  5,  13,  29,  53,  85, 125, 173, 229, ...    = A078370
  8,  16,  24,  32,  40,  48,  56,  64, ...    = A008590(n+1)
  8,   8,   8,   8,   8,   8,   8,   8, ...    = A010731
  0,   0,   0,   0,   0,   0,   0,   0, ...    = A000004
Both bisections are cubic polynomials.
c(-n) = -c(n).

Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A002522, A005408, A007395, A078370, A081489 (first bisection).

Cf. also A008590, A010731, A261327.

LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {1, 5, 3, 18, 8, 47, 18, 100}, 50] (* Amiram Eldar, Nov 08 2021 *)

G.f.: (5*x^5+2*x^4-2*x^3-x^2+5*x+1)/((x-1)^4*(x+1)^4).

A069894 Centered square numbers: a(n) = 4n^2 + 4n + 2.

Original entry on oeis.org

2, 10, 26, 50, 82, 122, 170, 226, 290, 362, 442, 530, 626, 730, 842, 962, 1090, 1226, 1370, 1522, 1682, 1850, 2026, 2210, 2402, 2602, 2810, 3026, 3250, 3482, 3722, 3970, 4226, 4490, 4762, 5042, 5330, 5626, 5930, 6242, 6562, 6890, 7226, 7570, 7922, 8282

Offset: 0

Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 10 2002

Any number may be substituted for y to yield similar sequences. The number set used determines values given (i.e., integer yields integer). All centered square integers in the set of integers may be found by this formula.
1/2 + 1/10 + 1/26 + ... = (Pi/4)*tanh(Pi/2) [Jolley]. - Gary W. Adamson, Dec 21 2006
For n > 0, a(n-1) is the number of triples (w, x, y) having all terms in {0, ..., n} and min(|w - x|, |x - y|) = 1. - Clark Kimberling, Jun 12 2012
Consider the primitive Pythagorean triples (x(n), y(n), z(n) = y(n) + 1) with n >= 0, and x(n) = 2*n + 1, y(n) = 2*n*(n + 1), z(n) = 2*n*(n + 1) + 1. The sequence, a(n), is 2*z(n). - George F. Johnson, Oct 22 2012
Ulam's spiral (SE corner). See the Wikipedia link. - Kival Ngaokrajang, Jul 25 2014
Conference matrix orders (A000952) of the form n-1 is a perfect square are all in this sequence. All values less than 1000 are conference matrices except for 226 which is still an open question (Balonin & Seberry 2014). - Colin Hall, Nov 21 2018
For n > 0, a(n-1) is the number of maximum number of regions into which the plane can be divided using n convex quadrilaterals. Related: A077588 A077591. - Keyang Li, Jun 17 2022

			If y = 3, then 81 + 144 = 225; if y = 4, then 12^2 + 16^2 = 20^2; 7^2 + 24^2 = 25^2 = 15^2 + 20^2.

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
N. A. Balonin and Jennifer Seberry, A Review and New Symmetric Conference Matrices, Research Online, Faculty of Engineering and Information Sciences, University of Wollongong, 2014.
Keyang Li, Figure for n=1,2,3,4,5
Tintarn, n convex quadrilaterals in the plane
Wikipedia, Ulam Spiral Construction.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001105, A001844, A002522, A016754, A016825, A033996, A261377.

[4*n^2+4*n+2 : n in [0..50]]; // Wesley Ivan Hurt, Jul 26 2014

A069894:=n->4*n^2+4*n+2: seq(A069894(n), n=0..50); # Wesley Ivan Hurt, Jul 26 2014

Mathematica
```
Table[4n(n + 1) + 2, {n, 0, 45}]
```

vector(100, n, (2*n-1)^2+1); \\ Derek Orr, Jul 27 2014

[(2*n+1)^2 + 1 for n in range(50)] # G. C. Greubel, Nov 21 2018

(y*(2*x + 1))^2 + (y*(2*x^2 + 2*x))^2 = (y*(2*x^2 + 2*x + 1))^2, where y = 2. If a^2 + b^2 = c^2, then c^2 = y^2*(4*x^4 + 8*x^3 + 8*x^2 + 4*x + 1). Also 2*A001844.
a(n) = (2*n + 1)^2 + 1. - Vladimir Joseph Stephan Orlovsky, Nov 10 2008 [Corrected by R. J. Mathar, Sep 16 2009]
a(n) = 8*n + a(n-1) for n > 0, a(0)=2. - Vincenzo Librandi, Aug 08 2010
From George F. Johnson, Oct 22 2012: (Start)
G.f.: 2*(1 + x)^2/(1 - x)^3, a(0) = 2, a(1) = 10.
a(n+1) = a(n) + 4 + 4*sqrt(a(n) - 1).
a(n-1) * a(n+1) = (a(n)-4)^2 + 16.
a(n) - 1 = (2*n+1)^2 = A016754(n) for n > 0.
(a(n+1) - a(n-1))/8 = sqrt(a(n) - 1).
a(n+1) = 2*a(n) - a(n-1) + 8 for n > 2, a(0)=2, a(1)=10, a(2)=26.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2) for n > 3; a(0)=2, a(1)=10, a(2)=26, a(3)=50.
a(n) = A033996(n) + 2 = A002522(2n + 1).
a(n)^2 = A033996(n)^2 + A016825(n)^2. (End)
a(n) = A001105(n) + A001105(n+1). - Bruno Berselli, Jul 03 2017
E.g.f.: 2*(1 + 4*x + 2*x^2)*exp(x). - G. C. Greubel, Nov 21 2018
a(n) = A261327(4*n+2). - Paul Curtz, Dec 23 2021
a(n) = 2*A001844(n) = 4*A000217(n) + 2*A002061(n+1). - Klaus Purath, Aug 13 2025

Edited by Robert G. Wilson v, Apr 11 2002

Offset corrected by Charles R Greathouse IV, Jul 25 2010

A262927 a(n+9) = a(n) + 10*(n+4) + 9. a(0)=0, a(1)=1, a(2)=3, a(3)=6, a(4)=10, a(5)=15, a(6)=23, a(7)=30, a(8)=39.

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 23, 30, 39, 49, 60, 72, 85, 99, 114, 132, 149, 168, 188, 209, 231, 254, 278, 303, 331, 358, 387, 417, 448, 480, 513, 547, 582, 620, 657, 696, 736, 777, 819, 862, 906, 951, 999, 1046, 1095, 1145, 1196, 1248, 1301, 1355, 1410, 1468, 1525

Offset: 0

Paul Curtz, Oct 04 2015

The main (or principal) sequence for the 11 steps recurrence is 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 30, 33, 36, ..., the partial sums of A054898.

a(n) mod 9 is a sequence of period 90.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,1,-2,1).

Cf. A054898, A261327, A262397.

I:=[0,1,3,6,10,15,23,30,39]; [n le 9 select I[n] else (Self(n-9)+10*(n-6)+9): n in [1..60]]; // Vincenzo Librandi, Oct 06 2015

LinearRecurrence[{2, -1, 0, 0, 0, 0, 0, 0, 1, -2, 1}, {0, 1, 3, 6, 10, 15, 23, 30, 39, 49, 60}, 60] (* Vincenzo Librandi, Oct 06 2015 *)
RecurrenceTable[{a[n+9] == a[n] + 10*(n+4) + 9, a[0]=0, a[1]=1, a[2]=3, a[3]=6, a[4]=10, a[5]=15, a[6]=23, a[7]=30, a[8]=39},a,{n,0,1000}] (* G. C. Greubel, Oct 16 2015 *)

a(n) = numerator(((2*n)^2+4)/4)\9 + numerator(((2*n+1)^2+4)/4)\9;
vector(100, n, a(n-1)) \\ Altug Alkan, Oct 04 2015

concat(0, Vec(-x*(x^8+2*x^7-x^6+3*x^5+x^4+x^3+x^2+x+1)/((x-1)^3*(x^2+x+1)*(x^6+x^3+1)) + O(x^100))) \\ Colin Barker, Oct 04 2015

a(n)=((2*n+1)^2+4)\9+(n^2+1)\9 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = A262397(2n) + A262397(2n+1).
a(n) = 2*a(n-1) - a(n-2) + a(n-9) - 2*a(n-10) + a(n-11), n>10.
G.f.: -x*(x^8+2*x^7-x^6+3*x^5+x^4+x^3+x^2+x+1) / ((x-1)^3*(x^2+x+1)*(x^6+x^3+1)). - Colin Barker, Oct 04 2015
a(n) = (5n^2 + 4n)/9 + O(1), or more precisely (5n^2 + 4n + 3)/9 <= a(n) <= (5n^2 + 4n - 10)/9. - Charles R Greathouse IV, Oct 16 2015

A329583 Numerators of 1 + n^2/4 + period 3: repeat [-1, 1, 1].

Original entry on oeis.org

0, 6, 3, 12, 6, 30, 9, 54, 18, 84, 27, 126, 36, 174, 51, 228, 66, 294, 81, 366, 102, 444, 123, 534, 144, 630, 171, 732, 198, 846, 225, 966, 258, 1092, 291, 1230, 324, 1374, 363, 1524, 402, 1686, 441, 1854, 486, 2028, 531, 2214, 576, 2406, 627

Offset: 0

Paul Curtz, Nov 17 2019

First bisection is 3*A008810.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,2,3,0,-3,-2,1,1).

Cf. A008810, A042965, A047395, A130823, A131561, A261327.

MapIndexed[#1 - 2 Boole[Mod[First@ #2, 3] == 1] + 1 &, CoefficientList[Series[(1 + 5 x - x^2 - 2 x^3 + 2 x^4 + 5 x^5)/(1 - x^2)^3, {x, 0, 44}], x]] (* Michael De Vlieger, Nov 18 2019 *)

concat(0, Vec(3*x*(2 + 3*x + x^2 - 2*x^3 + x^4 + 3*x^5 + 2*x^6) / ((1 - x)^3*(1 + x)^3*(1 + x + x^2)) + O(x^40))) \\ Colin Barker, Nov 24 2019

a(n) = A261327(n) + A131561(n+2) = (n^2 + 4)*(5 - 3*(-1)^n)/8 + (-1)^((n+1) mod 3).
From Colin Barker, Nov 24 2019: (Start)
G.f.: 3*x*(2 + 3*x + x^2 - 2*x^3 + x^4 + 3*x^5 + 2*x^6) / ((1 - x)^3*(1 + x)^3*(1 + x + x^2)).
a(n) = -a(n-1) + 2*a(n-2) + 3*a(n-3) - 3*a(n-5) - 2*a(n-6) + a(n-7) + a(n-8) for n>8. (End)

Incorrect 129 replaced with 123 by Colin Barker, Nov 24 2019

A267942 Interleave (n-1)^2 + 2 and (n+1)^2 + 2.

Original entry on oeis.org

3, 3, 2, 6, 3, 11, 6, 18, 11, 27, 18, 38, 27, 51, 38, 66, 51, 83, 66, 102, 83, 123, 102, 146, 123, 171, 146, 198, 171, 227, 198, 258, 227, 291, 258, 326, 291, 363, 326, 402, 363, 443, 402, 486, 443, 531, 486, 578, 531, 627, 578, 678, 627, 731, 678, 786, 731

Offset: 0

Paul Curtz, Jan 22 2016

Trisections:
3,  6,  6,  27, 27, 66,  66, ... = 3*(1, 2, 2, 9, 9, 22, 22, ... ). See A056105.
3,  3,  18, 18, 51, 51, 102, ... = 3*(1, 1, 6, 6, 17, 17, ... ). See A056109.
2, 11,  11, 38, 38, 83,  83, ...   (== 2 (mod 9)).
The trisections also have the signature (1,2,-2,-1,1). The corresponding main sequence is 0, 0, 0, 0, 1, 1, 3, 3, ... = A161680(n) with each term duplicated.

			a(0) = (2+13)/5, a(1) = (13+2)/5, a(2) = (5+5)/5, a(3) = (29+1)/5, ... (using first formula).

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

Cf. A000290, A007395, A010701, A022998, A056105, A056109, A059100, A161680, A174474, A193356, A255844, A261327.

&cat [[(n-1)^2+2, (n+1)^2+2]: n in [0..50]]; // Vincenzo Librandi, Jan 23 2016

Flatten[Table[{n^2 - 2 n + 3, n^2 + 2 n + 3}, {n, 0, 30}]] (* Vincenzo Librandi, Jan 23 2016 *)
CoefficientList[Series[(3 - 7 x^2 + 4 x^3 + 2 x^4)/((1 - x)^3 (1 + x)^2), {x, 0, 56}], x] (* Michael De Vlieger, Jan 24 2016 *)

Vec((3-7*x^2+4*x^3+2*x^4)/((1-x)^3*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 22 2016

a(n) = (A261327(n+2) + A261327(n-3))/5.
a(n+1) = a(n) + (-1)^n * A022998(n), a(0)=3.
a(n+3) = a(n) + 3*A193356(n), a(0)=a(1)=3, a(2)=2.
a(n) = 3 + A174474(n).
a(2n) + a(2n+1) = A255844(n).
From Colin Barker, Jan 22 2016: (Start)
a(n) = (2*n^2 - 6*(-1)^n*n - 2*n + 3*(-1)^n + 21)/8.
a(n) = (n^2 - 4*n + 12)/4 for n even.
a(n) = (n^2 + 2*n + 9)/4 for n odd.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4.
G.f.: (3 - 7*x^2 + 4*x^3 + 2*x^4) / ((1-x)^3*(1+x)^2).
(End)

More terms from Colin Barker, Jan 22 2016

A323178 a(n) = 1 + 100*n^2 for n >= 0.

Original entry on oeis.org

1, 101, 401, 901, 1601, 2501, 3601, 4901, 6401, 8101, 10001, 12101, 14401, 16901, 19601, 22501, 25601, 28901, 32401, 36101, 40001, 44101, 48401, 52901, 57601, 62501, 67601, 72901, 78401, 84101, 90001, 96101, 102401, 108901

Offset: 0

Paul Curtz, Jan 06 2019

nonn

Terms of A261327 ending in 1 (01 for n > 0.)
a(n) mod 9 = period 9: repeat [1, 2, 5, 1, 8, 8, 1, 5, 2] = A275704(n+3).
(Analogous sequence: b(n) = 29 + 100*n*(n+1) = A261327(A017329) = 29, 229, 629, ... .)

Cf. A000290, A008602 (20*n), A261327, A275704.

Subsequence of A017281.

a[n_] := 1 + 100*n^2 ; Array[a, 50, 0] (* or *)
CoefficientList[Series[(-1 - 98 x - 101 x^2)/(-1 + x)^3, {x, 0, 50}], x] (* or *)
CoefficientList[Series[E^x (1 + 100 x + 100 x^2), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* Stefano Spezia, Jan 06 2019 *)

a(n) = A261327(A008602(n)).
Recurrence: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2 with initial values a(0) = 1, a(1) = 101 and a(2) = 401.
From Stefano Spezia, Jan 06 2019: (Start)
O.g.f.: (-1 - 98*x - 101*x^2)/(-1 + x)^3.
E.g.f.: exp(x)*(1 + 100*x + 100*x^2).
(End)

Corrected and extended (recurrence formula) by Werner Schulte, Feb 18 2019

cp's OEIS Frontend

A262397 a(n) = floor(A261327(n)/9).

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A280579 Square array read by antidiagonals downwards giving the first differences A261327(n+p) - A261327(n), with p >= 0.

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A281098 a(n) is the GCD of the sequence d(n) = A261327(k+n) - A261327(k) for all k.

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A306454 a(n) = A261327(n)/A013946(n).

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A349118 Row sums of a triangle based on A261327.

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A069894 Centered square numbers: a(n) = 4*n^2 + 4*n + 2.

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A262927 a(n+9) = a(n) + 10*(n+4) + 9. a(0)=0, a(1)=1, a(2)=3, a(3)=6, a(4)=10, a(5)=15, a(6)=23, a(7)=30, a(8)=39.

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A069894 Centered square numbers: a(n) = 4n^2 + 4n + 2.