A000466 -id:A000466 - cp's OEIS frontend

A238087 A129779(n+2)/A006954(n).

-1, 1, -1, 1, -5, 63, -315, 99, -675675, 135135, -1640925, 83329155, -4583103525, 5791834125, -71152682225625, 14230536445125, -26797763435625, 24833960277501375, -73881031825566590625, 8546099690638125, -106610328924292590271875

Offset: 0

Paul Curtz, Feb 24 2014

sign

In A129779 or A097801 there is no mention of a link with the Bernoulli numbers.

It appears that a(n+6) is divisible by 4*(n+4)^2-1 and that the sum of the digits of a(n+5) is a multiple of 9.

			a(0) = -1/1, a(1) = 2/2, a(2) = -6/6, a(3) = 30/30, a(4) = -210/42 =-5.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000466.

a[0] = -1; a[1] = 1; a[n_] := (-1)^(n+1)*(2*n)!/(2^(n-1)*n!*Denominator[BernoulliB[2*n-2]]); Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 24 2014 *)

a(n+2) = -A004193(n+1)/A000367(n+1).

More terms from Jean-François Alcover, Feb 24 2014

A247829 a(3k) = k(k+1), a(3k+1) = (2k-1)(2k+1), a(3k+2) = (2k-1)(2k+3).

Original entry on oeis.org

0, -1, -3, 2, 3, 5, 6, 15, 21, 12, 35, 45, 20, 63, 77, 30, 99, 117, 42, 143, 165, 56, 195, 221, 72, 255, 285, 90, 323, 357, 110, 399, 437, 132, 483, 525, 156, 575, 621, 182, 675, 725, 210, 783, 837, 240, 899, 957, 272, 1023, 1085, 306, 1155, 1221, 342, 1295

Offset: 0

Paul Curtz, Dec 01 2014

A permutation of A061037(n) = -1, -3, 0, 5, 3, 21, 2, 45, 15, 77, 6, ... and of A214297(n) = -1, 0, -3, 2, 3, 6, 5, 12, ... .
Among consequences: b(n) = 4*a(n) + (sequence of period 3:repeat 1, 4, 16) = 1, 0, 4, 9, 16, 36, 25, 64, 100, ... , is a permutation of the squares of the nonnegative integers A000290(n).
And a(n)/b(n) = 0/1, -1/0, -3/4, 2/9, 3/16, 5/36, ... is a permutation of the Balmer series A061037(n)/A061038(n) = -1/0, -3/4, 0/1, 5/36, 3/16, ... .
a(5n) is divisible by 5.

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

Cf. A142717 (2 terms out of 3), A000290, A000466, A002378, A061037/A061038, A078371, A214297.

m:=50; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3*x^7-3*x^6-14*x^4-6*x^3-2*x^2+3*x+1)/((x-1)^3*(x^2 +x+1)^3))); // G. C. Greubel, Sep 20 2018

seq(op([k*(k+1),(2*k-1)*(2*k+1),(2*k-1)*(2*k+3)]), k=0..100); # Robert Israel, Dec 01 2014

Table[Sequence @@ {n*(n+1), (2*n-1)*(2*n+1), (2*n-1)*(2*n+3)}, {n, 0, 18}] (* Jean-François Alcover, Dec 16 2014 *)

concat(0, Vec(x*(3*x^7-3*x^6-14*x^4-6*x^3-2*x^2+3*x+1)/((x-1)^3*(x^2+x+1)^3) + O(x^100))) \\ Colin Barker, Dec 02 2014

a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9).
a(3*k) + a(3*k+1) + a(3*k+2) = 9*k^2 + 5*k - 4.
G.f.: x*(3*x^7 - 3*x^6 - 14*x^4 - 6*x^3 - 2*x^2 + 3*x + 1)/((x-1)^3*(x^2 +x+1)^3). - Robert Israel, Dec 01 2014
a(n) = -(n^2 + n + floor(n/3)*(27*floor(n/3)^3 - 18*(n+1)*floor(n/3)^2 + (3*n^2 + 21*n - 14)*floor(n/3) - (5*n^2 - n + 5)))/2. - Luce ETIENNE, Mar 13 2017
From Amiram Eldar, Oct 08 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/2.
Sum_{n>=1} (-1)^n/a(n) = Pi/4 + 1 - 2*log(2). (End)

A284359 Double triangle (2n+2 terms by row). Every row is 2n + 1 followed by 2n + 1 times 2n + 2.

Original entry on oeis.org

1, 2, 3, 4, 4, 4, 5, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17

Offset: 0

Paul Curtz, Mar 25 2017

In essence the same as A167991. - R. J. Mathar, Mar 27 2017

			1,  2,
3,  4,  4,  4,
5,  6,  6,  6,  6,  6,
7,  8,  8,  8,  8,  8,  8,  8,
9, 10, 10, 10, 10, 10, 10, 10, 10, 10,
... .
The row sum is A000466(n+1).

Cf. A000466, A005408, A103517 (main diagonal), A167381.

Table[2 n + 2 - Boole[k == 1], {n, 0, 8}, {k, 2 n + 2}] // Flatten (* Michael De Vlieger, Mar 25 2017 *)

for(n=0, 10, for(k=1, 2*n + 2, print1(2*n + 2 - (k==1), ", ");); print();) \\ Indranil Ghosh, Mar 26 2017, translated from Mathematica code

for n in range(0, 11):
    print([2*n + 2 -(k==1) for k in range(1, 2*n + 3)])
# Indranil Ghosh, Mar 26 2017

a(n) = A167381(n+1) - A167381(n).

A320085 Triangle read by rows, 0 <= k <= n: T(n,k) is the numerator of the derivative of the k-th Bernstein basis polynomial of degree n evaluated at the interval midpoint t = 1/2; denominator is A320086.

Original entry on oeis.org

0, -1, 1, -1, 0, 1, -3, -3, 3, 3, -1, -1, 0, 1, 1, -5, -15, -5, 5, 15, 5, -3, -3, -15, 0, 15, 3, 3, -7, -35, -63, -35, 35, 63, 35, 7, -1, -3, -7, -7, 0, 7, 7, 3, 1, -9, -63, -45, -63, -63, 63, 63, 45, 63, 9, -5, -5, -135, -15, -105, 0, 105, 15, 135, 5, 5

Offset: 0

Franck Maminirina Ramaharo, Oct 05 2018

If n = 2*k, then T(n,k) = 0 since the k-th Bernstein basis polynomial of degree n has a single unique local maximum occurring at t = k/n, which coincides with the interval midpoint t = 1/2 (T(0,0) = 0 because the only 0 degree Bernstein basis polynomial is the constant 1).

			Triangle begins:
   0;
  -1,   1;
  -1,   0,    1;
  -3,  -3,    3,   3;
  -1,  -1,    0,   1,    1;
  -5, -15,   -5,   5,   15,  5;
  -3,  -3,  -15,   0,   15,  3,   3;
  -7, -35,  -63, -35,   35, 63,  35,  7;
  -1,  -3,   -7,  -7,    0,  7,   7,  3,   1;
  -9, -63,  -45, -63,  -63, 63,  63, 45,  63, 9;
  -5,  -5, -135, -15, -105,  0, 105, 15, 135, 5, 5;
  ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Rita T. Farouki, The Bernstein polynomial basis: A centennial retrospective, Computer Aided Geometric Design Vol. 29 (2012), 379-419.
Ron Goldman, Pyramid Algorithms. A Dynamic Programming Approach to Curves and Surfaces for Geometric Modeling, Morgan Kaufmann Publishers, 2002, Chap. 5.
Eric Weisstein's World of Mathematics, Bernstein Polynomial
Wikipedia, Bernstein polynomial

Inspired by A141692.

Cf. A007318, A128433, A128434, A319861, A319862, A320086.

T:=proc(n,k) n*(binomial(n-1,k-1)-binomial(n-1,k))/gcd(n*(binomial(n-1,k-1)-binomial(n-1,k)),2^(n-1)); end proc: seq(seq(T(n,k),k=0..n),n=0..11); # Muniru A Asiru, Oct 06 2018

Table[Numerator[n*(Binomial[n-1, k-1] - Binomial[n-1, k])/2^(n-1)], {n, 0, 12}, {k, 0, n}]//Flatten

T(n, k) := n*(binomial(n - 1, k - 1) - binomial(n - 1, k))/gcd(n*(binomial(n - 1, k - 1) - binomial(n - 1, k)), 2^(n - 1))$
tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))$

def A320085(n,k): return numerator(n*(binomial(n-1, k-1) - binomial(n-1, k))/2^(n-1))
flatten([[A320085(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 19 2021

T(n, k) = numerator of 2*A141692(n,k)/A000079(n).
T(n, k) = n*(binomial(n-1, k-1) - binomial(n-1, k))/gcd(n*(binomial(n-1, k-1) - binomial(n-1, k)), 2^(n-1)).
T(n, n-k) = -T(n,k).
T(n, 0) = -n.
T(2*n+1, 1) = -A000466(n).
T(2*n, 1) = -A069834(n-1), n > 1.
T(n, k)/A320086(n,k) = 4*n*(k/n - 1/2)*A319861(n,k)/A319861(n,k).
Sum_{k=0..n} k*T(n,k)/A320086(n,k) = n.
Sum_{k=0..n} k^2*T(n,k)/A320086(n,k) = n^2.
Sum_{k=0..n} k*(k-1)*T(n,k)/A320086(n,k) = n*(n - 1).

A340728 a(n) is the number of divisors d of n such that n/d - d is prime.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 1, 0, 0, 3, 0, 1, 0, 0, 0, 3, 0, 1, 0, 1, 0, 3, 0, 1, 0, 0, 1, 1, 0, 2, 0, 1, 0, 3, 0, 2, 0, 0, 0, 3, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 4, 0, 2, 1, 0, 0, 3, 0, 2, 0, 1, 0, 1, 0, 1, 0, 0, 0, 3, 0, 3, 0, 0, 0, 3, 0, 1, 0, 1, 0, 3, 0, 1, 0, 0, 0, 1, 0, 3, 1

Offset: 1

J. M. Bergot and Robert Israel, Jan 17 2021

nonn

If n is odd, then a(n) = 0 unless n is in A000466, in which case a(n) = 1.

			a(8) = 2; among the divisors {1,2,4,8} of 8, there are two cases where 8/d-d is prime: 8/1-1 = 7 and 8/2-2 = 2.

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

Cf. A000466, A010051, A179993, A340729.

f:= proc(n) local D,i,m;
  D:= sort(convert(numtheory:-divisors(n),list));
  m:= nops(D);
  nops(select(i -> isprime(D[m+1-i]-D[i]), [$1..(m+1)/2]));
end proc:
map(f, [$1..100]);

a(n) = sumdiv(n, d, isprime(n/d-d)); \\ Michel Marcus, Jan 18 2021

a(n) = Sum_{d|n} c(n/d-d), where c is the prime characteristic (A010051). - Wesley Ivan Hurt, Jan 18 2021

A341780 Starts of runs of 3 consecutive anti-tau numbers (A046642).

Original entry on oeis.org

3, 15, 195, 255, 483, 783, 1023, 1155, 1295, 1443, 1599, 2703, 3363, 4623, 4899, 5183, 6399, 6723, 7395, 7743, 8463, 8835, 10815, 11235, 11663, 12099, 12543, 15375, 16383, 16899, 17955, 18495, 20163, 24963, 25599, 26895, 27555, 31683, 33855, 35343, 36099, 37635

Offset: 1

Amiram Eldar, Feb 19 2021

nonn

Since the even anti-tau numbers (A268066) are square numbers, all the terms are of the form 4*k^2 - 1, and there cannot be a run of more than 3 consecutive anti-tau numbers.

			3 is a term since 3, 4 and 5 are all anti-tau numbers: gcd(3, tau(3)) = gcd(3, 2) = 1, gcd(4, tau(4)) = gcd(4, 3) = 1 and gcd(5, tau(5)) = gcd(5, 2) = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A000466, A046642 and A341779.

Cf. A000010, A009191, A268066.

antiTauQ[n_] := CoprimeQ[n, DivisorSigma[0, n]]; Select[4*Range[100]^2 - 1, AllTrue[# + {0, 1, 2}, antiTauQ] &]

A069076 a(n) = (4*n^2 - 1)^3.

Original entry on oeis.org

27, 3375, 42875, 250047, 970299, 2924207, 7414875, 16581375, 33698267, 63521199, 112678587, 190109375, 307546875, 480048687, 726572699, 1070599167, 1540798875, 2171747375, 3004685307, 4088324799, 5479701947, 7245075375

Offset: 1

Benoit Cloitre, Apr 05 2002

Konrad Knopp, Theory and application of infinite series, Dover, p. 269.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000466, A069075.

(4Range[30]^2-1)^3 (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{27,3375,42875,250047,970299,2924207,7414875},30] (* Harvey P. Dale, Jan 20 2012 *)

Sum_{n>=1} 1/a(n) = (32 - 3*Pi^3)/64.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7); a(1)=27, a(2)=3375, a(3)=42875, a(4)=250047, a(5)=970299, a(6)=2924207, a(7)=7414875. - Harvey P. Dale, Jan 20 2012
G.f: x*(x^6 - 34*x^5 - 3165*x^4 - 19852*x^3 - 19817*x^2 - 3186*x - 27)/(x-1)^7. - Harvey P. Dale, Jan 20 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^3/128 + 3*Pi/32 - 1/2. - Amiram Eldar, Feb 25 2022

A123754 Positive numbers of the form 4*n^2 - 1 which are not semiprimes.

Original entry on oeis.org

3, 63, 99, 195, 255, 399, 483, 575, 675, 783, 1023, 1155, 1295, 1443, 1599, 1935, 2115, 2303, 2499, 2703, 2915, 3135, 3363, 3843, 4095, 4355, 4623, 4899, 5475, 5775, 6083, 6399, 6723, 7055, 7395, 7743, 8099, 8463, 8835, 9215, 9603, 9999, 10815

Offset: 1

Roger L. Bagula, Nov 16 2006

nonn

Positive numbers of the form 4*n^2 - 1 which are semiprimes can be found in A037074.

Or, all positive products of the form A014076(i)*[A014076(i)+-2], duplicates removed. - R. J. Mathar, Aug 08 2007

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A000466, A037074.

Select[4*(Range[54])^2-1, Not[PrimeQ[Sqrt[(#+ 1)]-1] && PrimeQ[Sqrt[(#+1)]+1]]&]
Select[4*Range[100]^2-1,PrimeOmega[#]!=2&] (* Harvey P. Dale, Jul 24 2016 *)

Equals ( A000466 - {-1} ) - A001358. - R. J. Mathar, Aug 08 2007

Edited by N. J. A. Sloane, Aug 03 2007

A185669 a(n) = 4n^2 + 3n + 2.

Original entry on oeis.org

2, 9, 24, 47, 78, 117, 164, 219, 282, 353, 432, 519, 614, 717, 828, 947, 1074, 1209, 1352, 1503, 1662, 1829, 2004, 2187, 2378, 2577, 2784, 2999, 3222, 3453, 3692, 3939, 4194, 4457, 4728, 5007, 5294, 5589, 5892, 6203, 6522, 6849, 7184, 7527, 7878, 8237, 8604, 8979, 9362, 9753, 10152, 10559, 10974, 11397, 11828

Offset: 0

Paul Curtz, Feb 09 2011

Natural numbers A000027 written clockwise as a square spiral:
.
  43--44--45--46--47--48--49
   |
  42  21--22--23--24--25--26
   |   |                   |
  41  20   7---8---9--10  27
   |   |   |           |   |
  40  19   6   1---2  11  28
   |   |   |       |   |   |
  39  18   5---4---3  12  29
   |   |               |   |
  38  17--16--15--14--13  30
   |                       |
  37--36--35--34--33--32--31
.
Walking in straight lines away from the center:
1,  2, 11, ... = A054552(n)     = 1 -3*n+4*n^2,
1,  8, 23, ... = A033951(n)     = 1 +3*n+4*n^2,
1,  3, 13, ... = A054554(n+1)   = 1 -2*n-4*n^2,
1,  7, 21, ... = A054559(n+1)   = 1 +2*n+4*n^2,
1,  4, 15, ... = A054556(n+1)   = 1   -n+4*n^2,
1,  6, 19, ... = A054567(n+1)   = 1   +n+4*n^2,
1,  5, 17, ... = A053755(n)     = 1     +4*n^2,
1,  9, 25, ... = A016754(n)     = 1 +4*n+4*n^2 = (1+2*n)^2,
2,  8, 22, ... = 2*A084849(n)   = 2 +2*n+4*n^2,
2, 12, 30, ... = A002939(n+1)   = 2 +6*n+4*n^2,
2,  9, 24, ... = a(n)           = 2 +3*n+4*n^2,
2, 10, 26, ... = A069894(n)     = 2 +4*n+4*n^2,
3, 11, 27, ... = A164897(n)     = 3 +4*n+4*n^2,
3, 12, 29, ... = A054552(n+1)+1 = 3 +5*n+4*n^2,
3, 14, 33, ... = A033991(n+1)   = 3 +7*n+4*n^2,
3, 15, 35, ... = A000466(n+1)   = 3 +8*n+4*n^2,
4, 14, 32, ... = 2*A130883(n+1) = 4 +6*n+4*n^2,
4, 16, 36, ... = A016742(n+1)   = 4 +8*n+4*n^2 = (2+2*n)^2,
5, 18, 39, ... = A007742(n+1)   = 5 +9*n+4*n^2,
5, 19, 41, ... = A125202(n+2)   = 5+10*n+4*n^2.

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

[2+3*n+4*n^2: n in [0..80]];  // Vincenzo Librandi, Feb 09 2011

Table[4n^2 + 3n + 2, {n,0,50}] (* G. C. Greubel, Jul 09 2017 *)
LinearRecurrence[{3,-3,1},{2,9,24},60] (* Harvey P. Dale, Aug 11 2021 *)

a(n)=4*n^2+3*n+2 \\ Charles R Greathouse IV, Apr 14 2014

a(n) = a(n-1) + 8*n - 1.
a(n) = 2*a(n-1) - a(n-2) + 8.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (2 +3*x +3*x^2)/(1-x)^3 . - R. J. Mathar, Feb 11 2011
a(n) = A033954(n) + 2. - Bruno Berselli, Apr 10 2011
E.g.f.: (4*x^2 + 7*x + 2)*exp(x). - G. C. Greubel, Jul 09 2017

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

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

Cf. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

cp's OEIS Frontend

A238087 A129779(n+2)/A006954(n).

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A247829 a(3*k) = k*(k+1), a(3*k+1) = (2*k-1)*(2*k+1), a(3*k+2) = (2*k-1)*(2*k+3).

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A284359 Double triangle (2*n+2 terms by row). Every row is 2*n + 1 followed by 2*n + 1 times 2*n + 2.

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A320085 Triangle read by rows, 0 <= k <= n: T(n,k) is the numerator of the derivative of the k-th Bernstein basis polynomial of degree n evaluated at the interval midpoint t = 1/2; denominator is A320086.

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A340728 a(n) is the number of divisors d of n such that n/d - d is prime.

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A341780 Starts of runs of 3 consecutive anti-tau numbers (A046642).

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A069076 a(n) = (4*n^2 - 1)^3.

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A247829 a(3k) = k(k+1), a(3k+1) = (2k-1)(2k+1), a(3k+2) = (2k-1)(2k+3).

A284359 Double triangle (2n+2 terms by row). Every row is 2n + 1 followed by 2n + 1 times 2n + 2.

A185669 a(n) = 4n^2 + 3n + 2.