A160050 -id:A160050 - cp's OEIS frontend

A181318 a(n) = A060819(n)^2.

0, 1, 1, 9, 1, 25, 9, 49, 4, 81, 25, 121, 9, 169, 49, 225, 16, 289, 81, 361, 25, 441, 121, 529, 36, 625, 169, 729, 49, 841, 225, 961, 64, 1089, 289, 1225, 81, 1369, 361, 1521, 100, 1681, 441, 1849, 121, 2025, 529, 2209, 144, 2401, 625, 2601, 169, 2809, 729

Offset: 0

Paul Curtz, Jan 26 2011

The first sequence, p=0, of the family A060819(n)*A060819(n+p).
Hence array
p=0:  0, 1, 1,  9,  1, 25,  9,  49,    a(n)=A060819(n)^2,
p=1:  0, 1, 3,  3,  5, 15, 21,  14,     A064038(n),
p=2:  0, 3, 1, 15,  3, 35,  6,  63,     A198148(n),
p=3:  0, 1, 5,  9,  7, 10, 27,  35,     A160050(n),
p=4:  0, 5, 3, 21,  2, 45, 15,  77,     A061037(n),
p=5:  0, 3, 7,  6,  9, 25, 33,  21,     A178242(n),
p=6:  0, 7, 2, 27,  5, 55,  9,  91,     A217366(n),
p=7:  0, 2, 9, 15, 11, 15, 39,  49,     A217367(n),
p=8:  0, 9, 5, 33,  3, 65, 21, 105,     A180082(n).
Compare columns 2, 3 and 5, columns 4 and 7 and columns 6 and 8.
From Peter Bala, Feb 19 2019: (Start)
We make some general remarks about the sequence a(n) = numerator(n^2/(n^2 + k^2)) = (n/gcd(n,k))^2 for k a fixed positive integer (we suppress the dependence of a(n) on k). The present sequence corresponds to the case k = 4.
a(n) is a quasi-polynomial in n. In fact, a(n) = n^2/b(n) where b(n) = gcd(n^2,k^2) is a purely periodic sequence in n.
In addition to being multiplicative these sequences are also strong divisibility sequences, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, it follows that a(n) is a divisibility sequence: if n divides m then a(n) divides a(m).
By the multiplicativeness and strong divisibility property of the sequence a(n) it follows that if gcd(n,m) = 1 then a( a(n)*a(m) ) = a(a(n)) * a(a(m)), a( a(a(n))*a(a(m)) ) = a(a(a(n))) * a(a(a(m))) and so on.
The sequence a(n) has the rational generating function Sum_{d divides k} f(d)*F(x^d), where F(x) = x*(1 + x)/(1 - x)^3 = x + 4*x^2 + 9*x^3 + 16*x^4 + ... is the o.g.f. for the squares A000290, and where f(n) is the Dirichlet inverse of the Jordan totient function J_2(n) - see A007434. The function f(n) is multiplicative and is defined on prime powers p^k by f(p^k) = (1 - p^2). See A046970. Cf. A060819. (End)
a(n-4) is the constant needed to complete the n-polygonal numbers into squares (see A377851); a(-1) = 1, which completes the triangle numbers, is not shown in the data. - Jonathan Dushoff, Nov 12 2024

G. C. Greubel, Table of n, a(n) for n = 0..10000
Wikipedia, Completing the square.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A181829, A046970, A007434, A060819, A168077, A377851.

[n^2/GCD(n,4)^2: n in [0..100]]; // G. C. Greubel, Sep 19 2018

a:=n->n^2/gcd(n,4)^2: seq(a(n),n=0..60); # Muniru A Asiru, Feb 20 2019

Table[n^2/GCD[n,4]^2, {n,0,100}] (* G. C. Greubel, Sep 19 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,1,1,9,1,25,9,49,4,81,25,121},60] (* Harvey P. Dale, Jan 18 2025 *)

a(n)=n^2/gcd(n,4)^2 \\ Charles R Greathouse IV, Dec 21 2011

[n^2/gcd(n, 4)^2 for n in (0..100)] # G. C. Greubel, Feb 20 2019

a(2*n) = A168077(n), a(2*n+1) = A016754(n).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
G.f.: x*(1 + x + 9*x^2 + x^3 + 22*x^4 + 6*x^5 + 22*x^6 + x^7 + 9*x^8 + x^9 + x^10)/(1-x^4)^3. - R. J. Mathar, Mar 10 2011
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 16)) = n^2/(gcd(n^2,16)) = (n/gcd(n,4))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 16, 1, 4, 1, 16, ...] is a purely periodic sequence of period 4.
a(n) is a quasi-polynomial in n: a(4*n) = n^2; a(4*n + 1) = (4*n + 1)^2; a(4*n + 2) = (2*n + 1)^2; a(4*n + 3) = (4*n + 3)^2.
O.g.f.: Sum_{d divides 4} A046970(d)*x^d*(1 + x^d)/(1 - x^d)^3 = x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3 - 3*x^4*(1 + x^4)/(1 - x^4)^3. (End)
Sum_{n>=1} 1/a(n) = 5*Pi^2/12. - Amiram Eldar, Aug 12 2022
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 4^max(0, e-2), and a(p^e) = p^(2*e) for p > 2.
Dirichlet g.f.: zeta(s-2)*(1 - 3/2^s - 3/4^s).
Sum_{k=1..n} a(k) ~ (37/192) * n^3. (End)
a(n) = (37 - 27*(-1)^n - 3*(-1)^(n*(n-1)/2) - 3*(-1)^(n*(n+1)/2)) * n^2/64. - Vaclav Kotesovec, Nov 14 2024

Edited by Jean-François Alcover, Oct 01 2012 and Jan 15 2013

More terms from Michel Marcus, Jun 09 2014

A125650 Numerator of n(n+3)/(4(n+1)(n+2)) = sum(k=1..n, 1/(k(k+1)(k+2)) ).

Original entry on oeis.org

0, 1, 5, 9, 7, 5, 27, 35, 11, 27, 65, 77, 45, 26, 119, 135, 38, 85, 189, 209, 115, 63, 275, 299, 81, 175, 377, 405, 217, 116, 495, 527, 140, 297, 629, 665, 351, 185, 779, 819, 215, 451, 945, 989, 517, 270, 1127, 1175, 306, 637, 1325, 1377, 715, 371, 1539, 1595, 413, 855

Offset: 0

Alexander Adamchuk, Nov 29 2006

3^2 divides a(3k). p divides a(p) for an odd prime p. p divides a(p-3) for prime p>3. p^k divides a(p^k) for an odd prime p. a(n) = m^2 is a perfect square for n = {1,3,24,147,864,5043,29400,171363,...} = A125651(n). Corresponding numbers m such that m^2 = a[ A125651(n) ] are listed in A125652(n) = {1,3,9,105,306,3567,10395,121173,...}.

			The rationals n(n+3)/(4(n+1)(n+2)) = a(n)/A230328(n) begin:
0, 1/6, 5/24, 9/40, 7/30, 5/21, 27/112, 35/144, 11/45, 27/110, 65/264, 77/312, 45/182, 26/105, 119/480, ... - _Wolfdieter Lang_, Mar 08 2018

L. B.W. Jolley, Summation of Series, Second revised ed., Dover, 1961, p.38, (201).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A125651, A125652. A160050, A230328 (denominators).

[Numerator(n*(n+3)/(4*(n+1)*(n+2))): n in [0..60]]; // Vincenzo Librandi, May 21 2012

Table[Numerator[n(n+3)/(4(n+1)(n+2))],{n,0,100}]

a(n)=n*(n+3)/2^min(3,valuation(n*(n+3),2)); \\ Max Alekseyev, Jan 11 2007

a(n) = Sum_{k=1..n} 1/(k(k+1)(k+2)).
a(n) = n*(n+3)/2^min(3,valuation(n*(n+3),2)). a(n)=n*(n+3)/4 for n=1 or 4 (mod 8); a(n)=n*(n+3)/8 for n=0 or 5 (mod 8); a(n) = n*(n+3)/2 for n=2, 3, 6, or 7 (mod 8). - Max Alekseyev, Jan 11 2007
a(n) = A106609(n)*A106609(n+3). - Paul Curtz, Jan 13 2011
G.f.: x*(x^19 -2*x^18 +3*x^17 -5*x^16 +3*x^15 -6*x^14 +7*x^13 -11*x^12 -12*x^11 +24*x^10 -36*x^9 +24*x^8 -38*x^7 +28*x^6 -18*x^5 -3*x^4 -2*x -1) / ((x-1)^3*(x^2+1)^3*(x^4+1)^3). - Colin Barker, Feb 21 2013
G.f. for rationals r(n) = a(n)/A230328(n): (1/4)*(1 - hypergeometric([1, 2], [3], -x/(1-x)))/(1-x) = (- 2*x + 3*x^2 + 2*(2*x - (1 + x^2))*log(1-x))/(4*(1-x)*x^2). For the r(n) formula see Jolley's general remark (201) on p.38. Thanks to Gary Detlefs for pointing to this remark. - Wolfdieter Lang, Mar 08 2018

A178242 Numerator of n(5+n)/((n+1)(n+4)).

Original entry on oeis.org

0, 3, 7, 6, 9, 25, 33, 21, 26, 63, 75, 44, 51, 117, 133, 75, 84, 187, 207, 114, 125, 273, 297, 161, 174, 375, 403, 216, 231, 493, 525, 279, 296, 627, 663, 350, 369, 777, 817, 429, 450, 943, 987, 516, 539, 1125, 1173, 611, 636, 1323, 1375, 714, 741, 1537, 1593

Offset: 0

Paul Curtz, Dec 20 2010

Sequence of differences denominator(n) - numerator(n) = 1,2,2,1... = A014695(n).

Denominator: A160050(n+2).

			The reduced fractions are 0, 3/5, 7/9, 6/7, 9/10, 25/27, 33/35, 21/22, 26/27, 63/65, 75/77, 44/45, ..

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Wikipedia, Quasi-polynomial.
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A001793, A014695, A028557, A060819, A160050, A176027.

[Floor(n*(n+5)*((-1)^((2*n-(-1)^n-3)/4)+3)/8) : n in [0..50]]; // Vincenzo Librandi, Oct 08 2011

A178242 := proc(n) n*(5+n)/(n+1)/(n+4) ;  numer(%) ;end proc:
seq(A178242(n),n=0..80) ; # R. J. Mathar, Dec 20 2010

f[n_] := n/GCD[n, 4]; Array[f[#] f[# + 5] &, 50, 0]
Table[Numerator[n*(5+n)/((n+1)*(n+4))], {n,0,50}] (* G. C. Greubel, Sep 21 2018 *)

vector(50, n, n--; numerator(n*(5+n)/((n+1)*(n+4)))) \\ G. C. Greubel, Sep 21 2018

a(n) = numerator(A176027(n)/A001793(n+1)).
a(n) = A060819(n)*A060819(n+5).
a(n) = +3*a(n-1) -6*a(n-2) +10*a(n-3) -12*a(n-4) +12*a(n-5) -10*a(n-6) +6*a(n-7) -3*a(n-8) +a(n-9).
a(n) = 3*a(n-4) -3*a(n-8) +a(n-12).
G.f.:  x*(-3+2*x-3*x^2-3*x^3+x^7) / ( (x-1)^3*(x^2+1)^3 ).
a(n) = n*(n+5)*((-1)^((2*n-(-1)^n-3)/4)+3)/8 = n*(n+5)*(3-i^(n*(n+1)))/8, where i=sqrt(-1); also a(n) = a(n-4)*A028557(n)/A028557(n-4) for n>4. - Bruno Berselli, Dec 30 2010
From Peter Bala, Aug 07 2022: (Start)
a(n) = numerator of n*(n+5)/4.
a(n) is quasi-polynomial in n: a(4*n) = n*(4*n+5) = A343560(n+1); a(4*n+1) = (2*n+3)*(4*n+1); a(4*n+2) = (2*n+1)*(4*n+7); a(4*n+3) = (n+2)*(4*n+3) = A180863(n+2). (End)
Sum_{n>=1} 1/a(n) = 112/75 - Pi/10. - Amiram Eldar, Aug 16 2022

A181890 a(n) = 8n^2 + 14n + 5.

Original entry on oeis.org

5, 27, 65, 119, 189, 275, 377, 495, 629, 779, 945, 1127, 1325, 1539, 1769, 2015, 2277, 2555, 2849, 3159, 3485, 3827, 4185, 4559, 4949, 5355, 5777, 6215, 6669, 7139, 7625, 8127, 8645, 9179, 9729, 10295, 10877, 11475, 12089, 12719, 13365, 14027, 14705, 15399, 16109, 16835, 17577

Offset: 0

Paul Curtz, Feb 01 2011

A160050(4*n+1) = A033954(n); A160050(4*n+2) = A001107(n); the third quadrisection is a(n).
First 16 terms of clockwise spiral for odd numbers are as follows:
.
  13--15--17--19
   |           |
  11   1---3  21
   |       |   |
   9---7---5  23
               |
  31--29--27--25
.
a(n) comes from the third vertical.
Sequence found by reading the line from 5, in the direction 5, 27, in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Dec 25 2011

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

Cf. A000217, A001107, A014106, A014635, A033954, A064038, A160050.

[8*n^2+14*n+5: n in [0..700]]; // Vincenzo Librandi, Feb 01 2011

Table[(2n+1)(4n+5),{n,0,40}]  (* Harvey P. Dale, Feb 06 2011 *)
CoefficientList[Series[(5 + 12 x - x^2)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, May 23 2014 *)

a(n)=8*n^2+14*n+5 \\ Charles R Greathouse IV, Dec 21 2011

a(n) = A160050(4*n+3).
a(n) = (2*n+1)*(4*n+5).
a(n) = a(n-1) + 16*n + 6.
a(n) = 2*a(n-1) - a(n-2) + 16.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (5 + 12*x - x^2)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 25 2011
a(n) = A014635(n+1) - 1. - Omar E. Pol, Dec 25 2011
From Vaclav Kotesovec, Aug 18 2018: (Start)
Sum_{n>=0} 1/a(n) = 2/3 - Pi/12 - log(2)/6 = 0.289342748774193011891907697817...
Sum_{n>=0} (-1)^n / a(n) = (1 + sqrt(2))*Pi/12 - 2/3 - sqrt(2)*log(tan(Pi/8))/6 = 0.173114712692423461587883724528539... (End)
a(n) = A014106(2*n+1). - Rick L. Shepherd, Aug 06 2019
E.g.f.: (5 + 22*x + 8*x^2)*exp(x). - Elmo R. Oliveira, Oct 19 2024

A198148 a(n) = n(n+2)(9 - 7*(-1)^n)/16.

Original entry on oeis.org

0, 3, 1, 15, 3, 35, 6, 63, 10, 99, 15, 143, 21, 195, 28, 255, 36, 323, 45, 399, 55, 483, 66, 575, 78, 675, 91, 783, 105, 899, 120, 1023, 136, 1155, 153, 1295, 171, 1443, 190, 1599, 210, 1763, 231, 1935, 253, 2115, 276, 2303, 300, 2499, 325

Offset: 0

Paul Curtz, Oct 21 2011

See, in A181318(n), A060819(n)*A060819(n+p): A060819(n)^2, A064038(n), a(n), A160050(n), A061037(n), A178242(n). The second differences a(n+2)-2*a(n+1)+a(n) = -5, 16, -26, 44, -61, 86, -110, 142, -173, 212, -250, 296, -341, 394, -446, 506, taken modulo 9 are periodic with the palindromic period 4, 7, 1, 8, 2, 5, 7, 7, 7, 5, 2, 8, 1, 7, 4.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. quadrisections A014105, A001539, A000384, A141759.

Cf. A060819, A000217, A000466, A142705, A000034, A005563, A010689, A028552, A050534.

List([0..60], n -> n*(n+2)*(9-7*(-1)^n)/16); # G. C. Greubel, Feb 21 2019

[n*(n+2)*(9-7*(-1)^n)/16: n in [0..60]]; // Vincenzo Librandi, Nov 25 2011

A198148:=n->n*(n+2)*(9-7*(-1)^n)/16; seq(A198148(k), k=0..100); # Wesley Ivan Hurt, Oct 16 2013

LinearRecurrence[{0,3,0,-3,0,1},{0,3,1,15,3,35},60] (* Vincenzo Librandi, Nov 25 2011 *)

a(n)=n*(n+2)*(9-7*(-1)^n)/16 \\ Charles R Greathouse IV, Oct 16 2015

[n*(n+2)*(9-7*(-1)^n)/16 for n in (0..60)] # G. C. Greubel, Feb 21 2019

a(n) = A060819(n)*A060819(n+2).
a(2n) = n*(n+1)/2 = A000217(n).
a(2n+1) = (2*n+1)*(2*n+3) = A000466(n+1).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6), n>5.
a(n+1) - a(n) = (7*(-1)^n *(2*n^2+6*n+3) +18*n +27)/16.
a(n) = A142705(n) / A000034(n+1).
a(n) = A005563(n) / A010689(n+1). - Franklin T. Adams-Watters, Oct 21 2011
G.f. x*(3 +x +6*x^2 -x^4)/(1-x^2)^3. - R. J. Mathar, Oct 25 2011
a(n)*a(n+1) = a(A028552(n)) = A050534(n+2). - Bruno Berselli, Oct 26 2011
a(n) = numerator( binomial((n+2)/2,2) ). - Wesley Ivan Hurt, Oct 16 2013
E.g.f.: x*((24+x)*cosh(x) + (3+8*x)*sinh(x))/8. - G. C. Greubel, Sep 20 2018
Sum_{n>=1} 1/a(n) = 5/2. - Amiram Eldar, Aug 12 2022

A181407 a(n) = (n-4)(n-3)2^(n-2).

Original entry on oeis.org

3, 3, 2, 0, 0, 16, 96, 384, 1280, 3840, 10752, 28672, 73728, 184320, 450560, 1081344, 2555904, 5963776, 13762560, 31457280, 71303168, 160432128, 358612992, 796917760, 1761607680, 3875536896, 8489271296, 18522046464, 40265318400, 87241523200, 188441690112

Offset: 0

Paul Curtz, Jan 28 2011

Binomial transform of (3, 0, -1, followed by A005563).
The sequence and its successive differences are:
   3,  3,  2,   0,   0,   16,   96,  384,      a(n),
   0, -1, -2,   0,  16,   80,  288,  896,      A178987,
  -1, -1,  2,  16,  64,  208,  608, 2688,     -A127276,
   0,  3, 14,  48, 144,  400, 1056, 2688,      A176027,
   3, 11, 34,  96, 256,  656, 1632, 3968,      A084266(n+1)
   8, 23, 62, 160, 400,  976, 2336, 5504,
  15, 39, 98, 240, 576, 1360, 3168, 7296.
Division of the k-th column by abs(A174882(k)) gives
   3,  3,  1,  0,  0,  1,  3,  3,  5,  15,  21, 14,   A064038(n-3),
   0, -1, -1,  0,  1,  5,  9,  7, 10,  27,  35, 22,   A160050(n-3),
  -1, -1,  1,  2,  4, 13, 19, 13, 17,  43,  53, 32,   A176126,
   0,  3,  7,  6,  9, 25, 33, 21, 26,  63,  75, 44,   A178242,
   3, 11, 17, 12, 16, 41, 51, 31, 37,  87, 101, 58,
   8  23, 31, 20, 25, 61, 73, 43, 50, 115, 131, 74,
  15, 39, 49, 30, 36, 85, 99, 57, 65, 147, 165, 92.
Columns are (or from) A005563, A142463, A056220, A002378, A000290, A001844, A058331, A002061, A002522, A097080, A093328, A014206.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A176027, A181318.

List([0..40], n-> (n-4)*(n-3)*2^(n-2)); # G. C. Greubel, Feb 21 2019

[(n-4)*(n-3)*2^(n-2): n in [0..40] ]; // Vincenzo Librandi, Feb 01 2011

Table[(n-4)*(n-3)*2^(n-2), {n,0,40}] (* G. C. Greubel, Feb 21 2019 *)

vector(40, n, n--; (n-4)*(n-3)*2^(n-2)) \\ G. C. Greubel, Feb 21 2019

[(n-4)*(n-3)*2^(n-2) for n in (0..40)] # G. C. Greubel, Feb 21 2019

a(n) = 16*A001788(n-4).
a(n+1) - a(n) = A178987(n).
G.f.: (3 - 15*x + 20*x^2) / (1-2*x)^3. - R. J. Mathar, Jan 30 2011
E.g.f.: (x^2 - 3*x + 3)*exp(2*x). - G. C. Greubel, Feb 21 2019

A182868 a(n) = -1 + n + 4*n^2.

Original entry on oeis.org

-1, 4, 17, 38, 67, 104, 149, 202, 263, 332, 409, 494, 587, 688, 797, 914, 1039, 1172, 1313, 1462, 1619, 1784, 1957, 2138, 2327, 2524, 2729, 2942, 3163, 3392, 3629, 3874, 4127, 4388, 4657, 4934, 5219, 5512, 5813, 6122, 6439, 6764, 7097, 7438, 7787, 8144, 8509, 8882, 9263, 9652

Offset: 0

Paul Curtz, Feb 01 2011

First quadrisection of A176126(n). Take clockwise (square) spiral from A023443(n)=n-1: a(n) is on the negative x-axis. Fourth quadrisection (-1-n+4*n^2) is on the negative y-axis.
Conjecture: the 4 quadrisections of (the family) A064038, A160050, A176126, A178242 (see A181407) come from square spiral.
a(n) mod 9 has period 9: 8,4,8,2,4,5,5,4,2. a(n) mod 10 has period 10: 9,4,7,8,7,4,9,2,3,2. Each polynomial modulo some constant c has a period of length c (and perhaps shorter ones). - Paul Curtz and Bruno Berselli, Feb 05 2011

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

[-1+n+4*n^2: n in [0..700] ] // Vincenzo Librandi, Feb 01 2011

f[n_]:=-1+n+4*n^2;f[Range[0,100]] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2011 *)

a(n)=-1+n+4*n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = A176126(4*n).
a(n) = 4*n^2 + n - 1.
a(n) = a(n-1) - 3 + 8*n.
a(n) = 2*a(n) - a(n-2) + 8.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -(1 - 7*x - 2*x^2)/(1-x)^3. - Bruno Berselli, Feb 05 2011

A226089 Denominators of the series b(n+1) = (b(n)+k)/(1+b(n)*k); where k = 1/(n+1), b(1) = 0.

Original entry on oeis.org

1, 2, 7, 11, 8, 11, 29, 37, 23, 28, 67, 79, 46, 53, 121, 137, 77, 86, 191, 211, 116, 127, 277, 301, 163, 176, 379, 407, 218, 233, 497, 529, 281, 298, 631, 667, 352, 371, 781, 821, 431, 452, 947, 991, 518, 541, 1129, 1177, 613, 638, 1327, 1379, 716, 743, 1541

Offset: 1

Rick G. Rosner, Christian N. K. Anderson, and Kevin L. Schwartz, May 25 2013

The sequence shares numerators with the Harary numbers, A160050.

This is the sequence 0 + 1/2 + 1/3 + ... + 1/n using relativistic velocity addition, where the addition of velocities a and b = (a + b)/(1 + a*b/c^2). That is, for objects traveling at 0 + c/2 + c/3 + ... + c/n relative to each other, the n-th object has velocity (A160050(n)/a(n))*c relative to a stationary observer.

			a(11) = a(10) + 1/11 using relativistic velocity addition. Since a(10) = 27/28, the sum is (27/28 + 1/11) / (1 + 27/28 * (1/11)) = (325 / 308) / (335/308) = 65/67.

Christian N. K. Anderson, Table of n, a(n) for n = 1..10000
Christian N. K. Anderson, Ulam Spiral of the first 4000 terms.
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A160050, A001844.

LinearRecurrence[{3, -6, 10, -12, 12, -10, 6, -3, 1}, {1, 2, 7, 11, 8, 11, 29, 37, 23}, 100] (* or *)
Module[{m = 1}, Denominator[NestList[(++m*# + 1)/(m + #) &, 0, 99]]] (* Paolo Xausa, Nov 06 2024 *)

Vec(-x*(x^8-2*x^7+4*x^6-6*x^5+7*x^4-3*x^3+2*x^2+x+2) / ((x-1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jul 18 2015

library(gmp); reladd<-function(x,y) (x+y)/(1+x*y)y=as.bigq(rep(1,100)); y[1]=y[1]/2; for(i in 2:100) y[i]=reladd(y[i-1],y[i]/(i+1)); denominator(y)

G.f.: -x*(x^8-2*x^7+4*x^6-6*x^5+7*x^4-3*x^3+2*x^2+x+2) / ((x-1)^3*(x^2+1)^3). - Colin Barker, Jul 18 2015
A160050(n)/a(n) = tanh(Sum_{k=2..n} arctanh(1/k)), a(n) = A160050(n) + (1,1,2,2). - Thomas Ordowski, Oct 23 2024
a(4k) = 4k^2 + 3k + 1, a(4k+1) = 4k^2 + 5k + 2, a(4k+2) = 8k^2 + 14k + 7, a(4k+3) = 8k^2 + 18k + 11. - David Radcliffe, Oct 25 2024

Edited by Thomas Ordowski, Oct 23 2024

Edited by Paolo Xausa, Nov 06 2024

A215189 Array t(n,k) of the family ((n+k)/gcd(n+k,4))*(n/gcd(n,4)), read by antidiagonals.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 9, 3, 3, 0, 1, 3, 1, 1, 0, 25, 5, 15, 5, 5, 0, 9, 15, 3, 9, 3, 3, 0, 49, 21, 35, 7, 21, 7, 7, 0, 4, 14, 6, 10, 2, 6, 2, 2, 0, 81, 18, 63, 27, 45, 9, 27, 9, 9, 0, 25, 45, 10, 35, 15, 25, 5, 15, 5, 5, 0, 121, 55, 99, 22, 77, 33, 55, 11, 33, 11, 11, 0, 9, 33, 15, 27, 6, 21, 9, 15, 3, 9, 3, 3, 0

Offset: 0

Jean-François Alcover, Jun 12 2013

Identification of rows and columns:
Row 2, n=1: A060819,
row 3, n=2: A060819 (shifted),
row 4, n=3: A068219,
row 5, n=4: A060819 (shifted),
row 6, n=5: A060819 (shifted and multiplied by 5),
row 7, n=6: A068219 (shifted),
row 8, n=7: A060819 (shifted and multiplied by 7);
column 1, k=0: A181318,
column 2, k=1: A064038,
column 3, k=2: A198148,
column 4, k=3: A160050,
column 5, k=4: A061037,
column 6, k=5: A178242,
column 7, k=6: A217366,
column 8, k=7: A217367.
This array is the transposition of the array given by Paul Curtz in the comments in A181318.

			Array begins:
   0,  0,  0,  0,  0,  0,  0, ...
   1,  1,  3,  1,  5,  3,  7, ...
   1,  3,  1,  5,  3,  7,  2, ...
   9,  3, 15,  9, 21,  6, 27, ...
   1,  5,  3,  7,  2,  9,  5, ...
  25, 15, 35, 10, 45, 25, 55, ...
   9, 21,  6, 27, 15, 33,  9, ...
  49, 14, 63, 35, 77, 21, 91, ...
  ...
Triangle begins:
    0;
    1,  0;
    1,  1,  0;
    9,  3,  3,  0;
    1,  3,  1,  1,  0;
   25,  5, 15,  5,  5,  0;
    9, 15,  3,  9,  3,  3,  0;
   49, 21, 35,  7, 21,  7,  7,  0;
    4, 14,  6, 10,  2,  6,  2,  2,  0;
   81, 18, 63, 27, 45,  9, 27,  9,  9,  0;
   25, 45, 10, 35, 15, 25,  5, 15,  5,  5,  0;
  121, 55, 99, 22, 77, 33, 55, 11, 33, 11, 11,  0;
    9, 33, 15, 27,  6, 21,  9, 15,  3,  9,  3,  3,  0;
  ...

G. C. Greubel, Antidiagonals n=0..100 of triangle, flattened

Cf. A060819, A181318, A064038, A198148, A160050, A061037, A178242, A217366, A217367.

/* As triangle: */ [[(n-k)/GCD(n-k, 4)*n/GCD(n, 4): k in [0..n]]: n in [0..12]]; // Bruno Berselli, Jun 13 2013

t[n_, k_] := (n+k)/GCD[n+k, 4]*n/GCD[n, 4];  Table[t[n-k, k], {n, 0, 12}, {k, 0, n}] // Flatten

t(n,k) = ((n+k)/gcd(n+k,4))*(n/gcd(n,4)).

A227168 a(n) = gcd(2n, n(n+1)/2)^2.

Original entry on oeis.org

1, 1, 36, 4, 25, 9, 196, 16, 81, 25, 484, 36, 169, 49, 900, 64, 289, 81, 1444, 100, 441, 121, 2116, 144, 625, 169, 2916, 196, 841, 225, 3844, 256, 1089, 289, 4900, 324, 1369, 361, 6084, 400

Offset: 1

Paul Curtz, Jul 03 2013

a(n) is defined as A062828(n)^2 for n >= 1. If we extend the sequence to n=0 and negative n by use of the recurrence that relates a(n) to a(n+12), a(n+8) and a(n+4), we obtain a(0)=0, a(-1)=4 and a(-n) = A176743(n-2)^2 for n >= 2.
Define c(n) = a(n+2) - a(n-2) for c >= 0. Because a(n) is a shuffle of three interleaved 2nd-order polynomials, c(n) is a shuffle of three interleaved 1st-order polynomials: c(n) = 4* A062828(n)*(periodically repeated 1, 8, 1, 1).
The sequence a(n) is case p=0 of the family A062828(n)*A062828(n+p):
0, 1, 1, 36,  4, 25,  9, 196, ... = a(n).
0, 1, 6, 12, 10, 15, 42,  56, ... = A130658(n)*A000217(n) = A177002(n-1)*A064038(n+1).
0, 6, 2, 30,  6, 70, 12, 126, ... = 2*A198148(n)
0, 2, 5, 18, 28, 20, 27,  70, ... = A177002(n+2)*A160050(n+1) = A014695(n+2)*A000096(n).

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A062828, A016814, A017138, A005565, A006752, A061037, A061038.

[GCD(2*n, n*(n+1)/2)^2: n in [1..50]]; // G. C. Greubel, Sep 20 2018

A227168 := proc(n)
    A062828(n)^2 ;
end proc: # R. J. Mathar, Jul 25 2013

a[n_] := GCD[2*n, n*(n + 1)/2]^2; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Jul 03 2013 *)

a(n)=if(n%2, n*if(n%4>2, 2, 1), n/2)^2 \\ Charles R Greathouse IV, Jul 07 2013

a(n) = A062828(n)^2.
a(4n) = (4*n+1)^2; a(2n+1) = (n+1)^2; a(4n+2) = 4*(4*n+3)^2.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) * (period 4: repeat 4, 1, 1, 4) = A061038(n).
A005565(n-3) = a(n+1) * A061037(n). - Corrected by R. J. Mathar, Jul 25 2013
a(n) = A130658(n-1)^2 * A181318(n). - Corrected by R. J. Mathar, Aug 01 2013
G.f.: -x*(1 + x + 36*x^2 + 4*x^3 + 22*x^4 + 6*x^5 + 88*x^6 + 4*x^7 + 9*x^8 + x^9 + 4*x^10) / ( (x-1)^3*(1+x)^3*(x^2+1)^3 ). - R. J. Mathar, Jul 20 2013
Sum_{n>=1} 1/a(n) = 47*Pi^2/192 + 3*G/8, where G is Catalan's constant (A006752). - Amiram Eldar, Aug 21 2022

cp's OEIS Frontend

A181318 a(n) = A060819(n)^2.

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A125650 Numerator of n(n+3)/(4(n+1)(n+2)) = sum(k=1..n, 1/(k(k+1)(k+2)) ).

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A178242 Numerator of n*(5+n)/((n+1)*(n+4)).

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A181890 a(n) = 8*n^2 + 14*n + 5.

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A198148 a(n) = n*(n+2)*(9 - 7*(-1)^n)/16.

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A181407 a(n) = (n-4)*(n-3)*2^(n-2).

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A178242 Numerator of n(5+n)/((n+1)(n+4)).

A181890 a(n) = 8n^2 + 14n + 5.

A198148 a(n) = n(n+2)(9 - 7*(-1)^n)/16.

A181407 a(n) = (n-4)(n-3)2^(n-2).

A227168 a(n) = gcd(2n, n(n+1)/2)^2.