A178242 -id:A178242 - cp's OEIS frontend

A178370 The trisection A178242(3n+2).

7, 25, 26, 44, 133, 187, 125, 161, 403, 493, 296, 350, 817, 943, 539, 611, 1375, 1537, 854, 944, 2077, 2275, 1241, 1349, 2923, 3157, 1700, 1826, 3913, 4183, 2231, 2375, 5047, 5353, 2834, 2996, 6325, 6667, 3509, 3689, 7747, 8125, 4256, 4454, 9313, 9727, 5075, 5291

Offset: 0

Paul Curtz, Dec 21 2010

         For n =  0,  1,  2,  3,   4,   5,   6,   7, ...,
        a(n-1) = -1,  7, 25, 26,  44, 133, 187, 125, ...
  + A177049(n) =  1,  5, 14, 55,  91,  68,  95, 253, ...
         gives    0, 12, 39, 81, 135, 201, 282, 378, ...
which are increasing multiples of 3.
a(n) mod 9 = period 4: repeat 7,7,8,8.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A060819, A176672, A177049.

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (7+4*x-7*x^2+46*x^3-9*x^4+8*x^5+4*x^6+2*x^7-x^8)/((1-x)^3*(1 + x^2)^3 ) )); // G. C. Greubel, Feb 26 2020

m:=50; S:=series((7+4*x-7*x^2+46*x^3-9*x^4+8*x^5+4*x^6+2*x^7-x^8)/((1-x)^3*(1 + x^2)^3 ), x, m+1): seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Feb 26 2020

LinearRecurrence[{3,-6,10,-12,12,-10,6,-3,1}, {7,25,26,44,133,187,125,161,403}, 50] (* Harvey P. Dale, May 21 2015 *)

Vec( (7+4*x-7*x^2+46*x^3-9*x^4+8*x^5+4*x^6+2*x^7-x^8)/((1-x)^3*(1 + x^2)^3 ) +O('x^50) ) \\ G. C. Greubel, Feb 26 2020

def A178370_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (7+4*x-7*x^2+46*x^3-9*x^4+8*x^5+4*x^6+2*x^7-x^8)/((1-x)^3*(1 + x^2)^3 ) ).list()
A178370_list(50) # G. C. Greubel, Feb 26 2020

a(n) = A060819(2+3*n)*(A060819(7+3*n) + A176672(n+2))/2. - corrected by G. C. Greubel, Feb 26 2020
G.f.: (7 +4*x -7*x^2 +46*x^3 -9*x^4 +8*x^5 +4*x^6 +2*x^7 -x^8)/((1-x)^3 * (1 + x^2)^3 ). - R. J. Mathar, Jan 16 2011
From G. C. Greubel, Feb 26 2020: (Start)
a(n) = (6 + i^n*(1 - i + (-1)^n*(1 + i)))*(9*n^2 + 27*n + 14)/16.
E.g.f.: ( 3*(14+36*x+9*x^2)*exp(x) + (14+36*x-9*x^2)*cos(x) + (14-36*x-9*x^2)*sin(x) )/8. (End)
Sum_{n>=0} 1/a(n) = 1 - (3 + 4*sqrt(3))*Pi/45. - Amiram Eldar, Aug 12 2022

More terms from Jinyuan Wang, Feb 26 2020

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

Original entry on oeis.org

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

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

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)).

cp's OEIS Frontend

A178370 The trisection A178242(3n+2).

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

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

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

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

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