A142705 -id:A142705 - cp's OEIS frontend

A226023 A142705 (numerators of 1/4-1/(4n^2)) sorted to natural order.

0, 2, 3, 6, 12, 15, 20, 30, 35, 42, 56, 63, 72, 90, 99, 110, 132, 143, 156, 182, 195, 210, 240, 255, 272, 306, 323, 342, 380, 399, 420, 462, 483, 506, 552, 575, 600, 650, 675, 702, 756, 783, 812, 870, 899

Offset: 0

Paul Curtz, May 23 2013

A198442(n) without indices 4*n+2.
a(n)/A130823(n+1) = 0, 2,3,2, 4,5,4, 6,7,6, 8,9,8, ... (equal to A133310+1, after 0; see also A008611).
-1,   0,   2,   3, is divisible by  1 (for a(-1)=-1),
3,    6,  12,  15,                  3,
15,  20,  30,  35                   5,
35,  42,  56,  63                   7,
63,  72,  90,  99                   9,
99, 110, 132, 143,                 11, etc.
First column:  A000466(n),
second column: A002943(n),
third column:  A002939(n+1),
fourth column: A000466(n+1).
a(n) is also the numerator of 1/4-1/(4*n+2)^2: 0/1, 2/9, 3/16, 6/25, 12/49, 15/64, 20/81, 30/121, 35/144, 42/169, 56/225,...
The n-th denominator is equal to 4*a(n) + A146325(n+2).
Note that the differences of a(n-1): 1, 2, 1, 3, 6, 3, 5, 10, 5, 7, 14, 7, 9, 18, 9, 11, 22,... (from A043547 by pairs and 2*n+1) has the same recurrence.
(Of course every sequence which obeys a linear recurrence with constant coefficients has first differences that obey the same linear recurrence. - R. J. Mathar, Jun 14 2013)

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Trisections: A002939, A000466, A002943.

A226023 := proc(n)
    option remember;
    if n <=6 then
        op(n+1,[0,2,3,6,12,15,20]) ;
    else
        procname(n-1)+2*procname(n-3)-2*procname(n-4)-procname(n-6)+procname(n-7) ;
    end if;
end proc: # R. J. Mathar, Jun 28 2013

A226023[n_]:=Floor[(2n+1)/3]Floor[(2n+5)/3];
Array[A226023,100,0] (* Paolo Xausa, Dec 05 2023 *)

a(n) = floor( (2*n + 1)/3 ) * floor( (2*n + 5)/3 ) = A004396(n) * A004396(n+2).
Recurrences: a(n) = 3*a(n-3) -3*a(n-6) +a(n-9) = a(n-1) +2*a(n-3) -2*a(n-4) -a(n-6) +a(n-7).
a(n+15)  - a(n) = 10*A042968(n+8).
a(n+1) - a(n-2) = 2*A042968(n) with a(-2)=0, a(-1)=-1.
G.f.: x*(2+x+3*x^2+2*x^3+x^4-x^5)/((1-x)^3 * (1+x+x^2)^2). [Ralf Stephan, May 24 2013]

A142888 First differences of A142705.

Original entry on oeis.org

3, -1, 13, -9, 29, -23, 51, -43, 79, -69, 113, -101, 153, -139, 199, -183, 251, -233, 309, -289, 373, -351, 443, -419, 519, -493, 601, -573, 689, -659, 783, -751, 883, -849, 989, -953, 1101, -1063, 1219, -1179, 1343, -1301, 1473, -1429, 1609, -1563, 1751

Offset: 1

Paul Curtz, Sep 29 2008

Also obtained from A135370 if adjacent pairs are swapped and if the sequence is then multiplied by (-1)^(n+1).

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

Cf. A005843, A142705.

CoefficientList[Series[(3 + 2 x + 6 x^2 - x^4)/((1 + x)^3 (1 - x)^2), {x, 0, 60}], x] (* Vincenzo Librandi, May 25 2014 *)
LinearRecurrence[{-1,2,2,-1,-1},{3,-1,13,-9,29},50] (* Harvey P. Dale, Apr 02 2018 *)

Vec(x*(3+2*x+6*x^2-x^4)/((1+x)^3*(1-x)^2) + O(x^100)) \\ Colin Barker, Jan 26 2016

a(n) = A142705(n+1)-A142705(n).
a(n) = -a(n-1) +2*a(n-2) +2*a(n-3) -a(n-4) -a(n-5). - R. J. Mathar, Sep 12 2010
a(2n-1)+a(2n) = A005843(n).
G.f.: x(3+2x+6x^2-x^4)/((1+x)^3*(1-x)^2). - R. J. Mathar, Oct 24 2008, parenthesis added Sep 12 2010
From Colin Barker, Jan 26 2016: (Start)
a(n) = (5+3*(-1)^n+(10-6*(-1)^n)*n-6*(-1)^n*n^2)/8.
a(n) = (-3*n^2+2*n+4)/4 for n even.
a(n) = (3*n^2+8*n+1)/4 for n odd.
(End)

Edited by R. J. Mathar, Oct 24 2008

More terms from Vincenzo Librandi, May 25 2014

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

Original entry on oeis.org

-1, -15, -3, -7, 0, 9, 5, 33, 3, 65, 21, 105, 2, 153, 45, 209, 15, 273, 77, 345, 6, 425, 117, 513, 35, 609, 165, 713, 12, 825, 221, 945, 63, 1073, 285, 1209, 20, 1353, 357, 1505, 99, 1665, 437, 1833, 30, 2009, 525, 2193, 143

Offset: 0

Paul Curtz, May 21 2013

Denominators are in A226008.
Fractions in lowest terms for n>0: -15/4, -3/4, -7/36, 0/1, 9/100, 5/36, 33/196, 3/16, 65/324, 21/100, 105/484, 2/9, 153/676, 45/196, 209/900, 15/64,...
If t(n) is the sequence with period 8: 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, ... (see A226044), then A226008(n) = 4*a(n) + t(n).

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

Cf. A000466, A028566, A061037, A061041, A078371, A106609, A142705, A145923, A226008.

[-1] cat [Numerator(1/4-4/n^2): n in [1..50]]; // Bruno Berselli, May 22 2013

Join[{-1}, Table[Numerator[1/4 - 4/n^2], {n, 50}]] (* Bruno Berselli, May 24 2013 *)

concat([-1], vector(100, n, numerator(1/4 - 4/n^2))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-8) -3*a(n-16) +a(n-24).
a(2n) = A061037(n), a(2n+1) = A145923(n-2) for A145923(-2)=-15, A145923(-1)=-7.
a(4n)   = A142705(n) for A142705(0)=-1, a(8n) = A000466(n);
a(4n+1) = A028566(4n-3) for A028566(-3)=-15;
a(4n+2) = A078371(n-1) for A078371(-1)=-3;
a(4n+3) = A028566(4n-1) for A028566(-1)=-7.
a(n+4)  = A106609(n) * A106609(n+8).
G.f.: -(1 +15*x +3*x^2 +7*x^3 -9*x^5 -5*x^6 -33*x^7 -6*x^8 -110*x^9 -30*x^10 -126*x^11 -2*x^12 -126*x^13 -30*x^14 -110*x^15 -3*x^16 -33*x^17 -5*x^18 -9*x^19 +7*x^21 +3*x^22 +15*x^23)/(1-x^8)^3. - Bruno Berselli, May 22 2013
a(n) = (n^2-16)*(6*cos(Pi*n/4)-54*cos(Pi*n/2)+6*cos(3*Pi*n/4)-219*(-1)^n+293)/512. - Bruno Berselli, May 22 2013
a(n+10) = a(n+2)*(n+14)/(n-2) for n=0,1 and n>2. - Bruno Berselli, May 22 2013

Edited by Bruno Berselli, May 22 2013

A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 4, 9, 10, 11, 6, 13, 14, 15, 8, 17, 18, 19, 10, 21, 22, 23, 12, 25, 26, 27, 14, 29, 30, 31, 16, 33, 34, 35, 18, 37, 38, 39, 20, 41, 42, 43, 22, 45, 46, 47, 24, 49, 50, 51, 26, 53, 54, 55, 28, 57, 58, 59, 30, 61, 62, 63, 32, 65, 66, 67, 34, 69, 70, 71, 36, 73, 74, 75, 38, 77, 78, 79, 40, 81, 82, 83, 42, 85, 86, 87, 44, 89, 90, 91, 46, 93, 94, 95, 48, 97, 98, 99

Offset: 1

R. J. Mathar, Feb 25 2011

a(n) is the length of the period of the sequence k^2 mod n, k=1,2,3,4,..., i.e., the length of the period of A000035 (n=2), A011655 (n=3), A000035 (n=4), A070430 (n=5), A070431 (n=6), A053879 (n=7), A070432 (n=8), A070433 (n=9), A008959 (n=10), A070434 (n=11), A070435 (n=12) etc.
From Franklin T. Adams-Watters, Feb 24 2011: (Start)
Clearly if gcd(n,m) = 1, a(nm) = lcm(a(n),a(m)), so it suffices to establish this for prime powers.
If p is a prime, the period must divide p, but k^2 mod p is not constant, so a(p) = p.
a(p^e), e > 1, must be divisible by a(p^(e-1)), and must divide p^e. If p != 2, (p^(e-1)+1)^2 =  p^(2e-2)+2p^(e-1)+1 == 2p^(e-1)+1 (mod p^2), so a(p^e) != p^(e-1); it must then be e.
By inspection, a(4) = 2 and a(8) = 4.
This leaves a(2^e), e > 3. But then (2^(e-2)+1)^2 = 2^(2e-4)+2^(e-1)+1 == 2^(e-1)+1 (mod 2^e), so a(n) > 2^(e-2). On the other hand, (2^(e-1)+c)^2 = 2^(2e-2)+c2^e+c^2 == c^2 (mod 2^e). Hence the period is 2^(e-1). (End)

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

Cf. A000224 (size of the set of moduli of k^2 mod n), A019554, A060819, A061037, A090129, A142705, A164115, A283971.

A186646 := proc(n) if n mod 4 = 0 then n/2 ; else n ; end if; end proc ;

Flatten[Table[{n,n+1,n+2,(n+3)/2},{n,1,101,4}]] (* or *) LinearRecurrence[ {0,0,0,2,0,0,0,-1},{1,2,3,2,5,6,7,4},100] (* Harvey P. Dale, May 30 2014 *)
Table[n (7 - (-1)^n - 2 Cos[n Pi/2])/8, {n, 100}] (* Federico Provvedi , Jan 02 2018 *)

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

def A186646(n): return n if n&3 else n>>1 # Chai Wah Wu, Jan 10 2023

a(n) = 2*a(n-4) - a(n-8).
a(4n) = 2n; a(4n+1) = 4n+1; a(4n+2) = 4n+2; a(4n+3) = 4n+3.
a(n) = n/A164115(n).
G.f.: x*(1 + 2*x + 3*x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6) / ( (x-1)^2*(1+x)^2*(x^2+1)^2 ).
Dirichlet g.f.: (1-2/4^s)*zeta(s-1).
A019554(n) | a(n). - Charles R Greathouse IV, Feb 24 2011
a(n) = n*(7 - (-1)^n - (-i)^n - i^n)/8, with i=sqrt(-1). - Bruno Berselli, Feb 25 2011
Multiplicative with a(p^e)=2^e if p=2 and e<=1; a(p^e)=2^(e-1) if p=2 and e>=2; a(p^e)=p^e otherwise. - David W. Wilson, Feb 26 2011
a(n) * A060819(n+2) = A142705(n+1) = A061037(2n+2). - Paul Curtz, Mar 02 2011
a(n) = n - (n/2)*floor(((n-1) mod 4)/3). - Gary Detlefs, Apr 14 2013
a(2^n) = A090129(n+1). - R. J. Mathar, Oct 09 2014
a(n) = n*(7 - (-1)^n - 2*cos(n*Pi/2))/8. - Federico Provvedi, Jan 02 2018
E.g.f.: (1/4)*x*(4*cosh(x) + sin(x) + 3*sinh(x)). - Stefano Spezia, Jan 26 2020
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

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

A226044 Period of length 8: 1, 64, 16, 64, 4, 64, 16, 64.

Original entry on oeis.org

1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, 64

Offset: 0

Paul Curtz, May 24 2013

A002378(n)/A016754(n) gives 0/1, 2/9, 6/25, 12/49, 20/81, 30/121, 42/169, 56/225,..., where A016754(n) = 4*A002378(n) + 1;
A142705(n)/A154615(n+1) gives 0/1, 3/16,  2/9,  15/64,  6/25,  35/144, 12/49,  63/256,..., where A142705(n) = 4*A154615(n+1) + A010685(n);
A061037(n)/A061038(n) gives 0/1, 5/36,  3/16, 21/100, 2/9,   45/196, 15/64,  77/324,..., where A061038(n) = 4*A061037(n) + A177499(n);
A225948(n)/A226008(n) gives 0/1, 9/100, 5/36, 33/196, 3/16,  65/324, 21/100, 105/484,..., where A226008(n) = 4*A225948(n) + a(n).
See also the triangle in Example lines.

			Triangle in which the terms of each line are repeated:
A000012: 1,   ...
A010685: 1,   4,  ...
A177499: 1,  16,  4,  16,  ...
A226044: 1,  64, 16,  64,  4,  64, 16,  64, ...
         1, 256, 64, 256, 16, 256, 64, 256, 4, 256, 64, 256, 16, 256, 64, 256, ...

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

Cf. A010685, A177499, A205383, A225948.

Table[{1, 64, 16, 64, 4, 64, 16, 64}, {7}] // Flatten (* Jean-François Alcover, May 24 2013 *)

a(n) = A205383(n+7)^2.

G.f.: (1+64*x+16*x^2+64*x^3+4*x^4+64*x^5+16*x^6+64*x^7)/((1-x)*(1+x)*(1+x^2)*(1+x^4)). [Bruno Berselli, May 25 2013]

cp's OEIS Frontend

A226023 A142705 (numerators of 1/4-1/(4n^2)) sorted to natural order.

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A142888 First differences of A142705.

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A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

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A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

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A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

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

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