A129194 -id:A129194 - cp's OEIS frontend

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

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

A140106 Number of noncongruent diagonals in a regular n-gon.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37

Offset: 1

Andrew McFarland, Jun 03 2008

Number of double-stars (diameter 3 trees) with n nodes. For n >= 3, number of partitions of n-2 into two parts. - Washington Bomfim, Feb 12 2011
Number of roots of the n-th Bernoulli polynomial in the left half-plane. - Michel Lagneau, Nov 08 2012
From Gus Wiseman, Oct 17 2020: (Start)
Also the number of 3-part non-strict integer partitions of n - 1. The Heinz numbers of these partitions are given by A285508. The version for partitions of any length is A047967, with Heinz numbers A013929. The a(4) = 1 through a(15) = 6 partitions are (A = 10, B = 11, C = 12):
  111  211  221  222  322  332  333  433  443  444  544  554
            311  411  331  422  441  442  533  552  553  644
                      511  611  522  622  551  633  661  662
                                711  811  722  822  733  833
                                          911  A11  922  A22
                                                    B11  C11
(End)

			The square (n=4) has two congruent diagonals; so a(4)=1. The regular pentagon also has congruent diagonals; so a(5)=1. Among all the diagonals in a regular hexagon, there are two noncongruent ones; hence a(6)=2, etc.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Washington Bomfim, Double-star corresponding to the partition [3,7]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Index entries for sequences related to trees

Cf. A000554, A007304, A007997, A013929, A022998.
Cf. A047967, A129194, A235451, A285508, A321773.
A001399(n-3) = A069905(n) = A211540(n+2) counts 3-part partitions.
Essentially the same as A004526.

A140106:= func< n | n eq 1 select 0 else Floor((n-2)/2) >;
[A140106(n): n in [1..80]]; // G. C. Greubel, Feb 10 2023

with(numtheory): for n from 1 to 80 do:it:=0:
y:=[fsolve(bernoulli(n,x) , x, complex)] : for m from 1 to nops(y) do : if Re(y[m])<0 then it:=it+1:else fi:od: printf(`%d, `,it):od:

a[1]=0; a[n_?OddQ] := (n-3)/2; a[n_] := n/2-1; Array[a, 100] (* Jean-François Alcover, Nov 17 2015 *)

a(n)=if(n>1,n\2-1,0) \\ Charles R Greathouse IV, Oct 16 2015

def A140106(n): return n-2>>1 if n>1 else 0 # Chai Wah Wu, Sep 18 2023

def A140106(n): return 0 if (n==1) else (n-2)//2
[A140106(n) for n in range(1,81)] # G. C. Greubel, Feb 10 2023

a(n) = floor((n-2)/2), for n > 1, otherwise 0. - Washington Bomfim, Feb 12 2011
G.f.: x^4/(1-x-x^2+x^3). - Colin Barker, Jan 31 2012
a(n) = floor(A129194(n-1)/A022998(n)), for n > 1. - Paul Curtz, Jul 23 2017
a(n) = A001399(n-3) - A001399(n-6). Compare to A007997(n) = A001399(n-3) + A001399(n-6). - Gus Wiseman, Oct 17 2020

More terms from Joseph Myers, Sep 05 2009

A084844 Denominators of the continued fraction n + 1/(n + 1/...) [n times].

Original entry on oeis.org

1, 2, 10, 72, 701, 8658, 129949, 2298912, 46866034, 1082120050, 27916772489, 795910114440, 24851643870041, 843458630403298, 30918112619119426, 1217359297034666112, 51240457936070359069, 2296067756927144738850, 109127748348241605689981

Offset: 1

Hollie L. Buchanan II, Jun 08 2003

The (n-1)-th term of the Lucas sequence U(n,-1). The numerator is the n-th term. Adjacent terms of the sequence U(n,-1) are relatively prime. - T. D. Noe, Aug 19 2004
From Flávio V. Fernandes, Mar 05 2021: (Start)
Also, the n-th term of the n-th metallic sequence (the diagonal through the array A073133, and its equivalents, which is rows formed by sequences beginning with A000045, A000129, A006190, A001076, A052918) as shown below (for n>=1):
   0    1    0    1     0      1  ... A000035
   0   [1]   1    2     3      5  ... A000045
   0    1   [2]   5    12     29  ... A000129
   0    1    3  [10]   33    109  ... A006190
   0    1    4   17   [72]   305  ... A001076
   0    1    5   26   135   [701] ... A052918. (End)

			a(4) = 72 since 4 + 1/(4 + 1/(4 + 1/4)) = 305/72.

Alois P. Heinz, Table of n, a(n) for n = 1..387
Eric Weisstein's World of Mathematics, Lucas Sequence

Cf. A084845 (numerators).

Cf. A000045, A097690, A097691, A117715, A290864 (primes in this sequence).

A084844 :=proc(n) combinat[fibonacci](n, n) end:
seq(A084844(n), n=1..30); # Zerinvary Lajos, Jan 03 2007

myList[n_] := Module[{ex = {n}}, Do[ex = {ex, n}, {n - 1}]; Flatten[ex]] Table[Denominator[FromContinuedFraction[myList[n]]], {n, 1, 20}]
Table[s=n; Do[s=n+1/s, {n-1}]; Denominator[s], {n, 20}] (* T. D. Noe, Aug 19 2004 *)
Table[Fibonacci[n, n], {n, 1, 20}] (* Vladimir Reshetnikov, May 07 2016 *)
Table[DifferenceRoot[Function[{y,m},{y[2+m]==n*y[1+m]+y[m],y[0]==0,y[1]==1}]][n],{n,1,20}] (* Benedict W. J. Irwin, Nov 03 2016 *)

from sympy import fibonacci
def a(n):
    return fibonacci(n, n)
print([a(n) for n in range(1, 31)]) # Indranil Ghosh, Aug 12 2017

a(n) = (s^n - (-s)^(-n))/(2*s - n), where s = (n + sqrt(n^2 + 4))/2. - Vladimir Reshetnikov, May 07 2016
a(n) = y(n,n), where y(m+2,n) = n*y(m+1,n) + y(m,n), with y(0,n)=0, y(1,n)=1 for all n. - Benedict W. J. Irwin, Nov 03 2016
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jun 03 2017
a(n) = A117715(n,n). - Bobby Jacobs, Aug 12 2017
a(n) = [x^n] x/(1 - n*x - x^2). - Ilya Gutkovskiy, Oct 10 2017
a(n) == 0 (mod n) for even n and 1 (mod n) for odd n. - Flávio V. Fernandes, Dec 08 2020
a(n) == 0 (mod n) for even n and 1 (mod n^2) for odd n; see A065599. - Flávio V. Fernandes, Dec 25 2020
a(n) == 0 (mod 2*(n/2)^2) for even n and 1 (mod n^2) for odd n; see A129194. - Flávio V. Fernandes, Feb 06 2021

A159600 E.g.f. C(x) satisfies: C(x) = (1 - 2*S(x)^2)^(1/4), where S'(x) = C(x)^3 and C'(x) = -S(x) with C(0)=1.

Original entry on oeis.org

1, -1, 3, -27, 441, -11529, 442827, -23444883, 1636819569, -145703137041, 16106380394643, -2164638920874507, 347592265948756521, -65724760945840254489, 14454276753061349098587

Offset: 0

Paul D. Hanna, May 07 2009

sign

See A104203 for the expansion of the sine lemniscate function sl(x).
E.g.f. C(x) is an even function; zero terms are omitted.
Radius of convergence is |x| <= r:
r = sqrt(2)*(Pi/2)^(3/2)/gamma(3/4)^2 with
C(r) = gamma(3/4)^2/(Pi/2)^(3/2) where:
r = L/sqrt(2) where L=Lemniscate constant;
r = 1.8540746773013719184338503471952600...
C(r) = 0.76275976350181318806232598096361579...

			E.g.f.: C(x) = 1 - x^2/2! + 3*x^4/4! - 27*x^6/6! + 441*x^8/8! -+ ...
C(x)^2 = 1 - 2*x^2/2! + 12*x^4/4! - 144*x^6/6! + 3024*x^8/8! -+ ...
C(x)^3 = 1 - 3*x^2/2! + 27*x^4/4! - 441*x^6/6! + 11529*x^8/8! -+ ...
C(x)^4 = 1 - 4*x^2/2! + 48*x^4/4! - 1008*x^6/6! + 32256*x^8/8! -+ ...
C(x)^4 + 2*S(x)^2 = 1 where:
S(x) = x - 3*x^3/3! + 27*x^5/5! - 441*x^7/7! + 11529*x^9/9! + ...
S(x)^2 = 2*x^2/2! - 24*x^4/4! + 504*x^6/6! - 16128*x^8/8! +-...
From _Paul D. Hanna_, Jul 29 2011: (Start)
O.g.f.: 1 - x + 3*x^2 - 27*x^3 + 441*x^4 - 11529*x^5 + 442827*x^6 -+ ... + a(n)*x^n + ...
O.g.f.: 1/(1 + x/(1 + 2*x/(1 + 9*x/(1 + 8*x/(1 + 25*x/(1 + 18*x/(1 + 49*x/(1 + 32*x/(1-...))))))))) (continued fraction). (End)

Cf. A159601 (S(x)); A193541, A193544: All of these have the same |a(n)|. - M. F. Hasler, Aug 31 2012

Cf. A129194.

a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ JacobiCN[ x, 1/2], {x, 0, m}]]]; (* Michael Somos, Apr 25 2011 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, With[ {s = InverseSeries[ Integrate[ Series[(1 - x^4 / 4) ^ (-1/2), {x, 0, m + 1}], x]]}, m! SeriesCoefficient[ Sqrt[ (2 - s^2) / (2 + s^2)], {x, 0, m}]]]]; (* Michael Somos, Apr 25 2011 *)

{a(n)=local(S=x,C);for(i=0,2*n,S=intformal((1-2*S^2+O(x^(2*n+2)))^(3/4))); C=(1-2*S^2)^(1/4) ;(2*n)!*polcoeff(C,2*n)}

{a(n) = my(A, m); if( n<0, 0, m = 2*n; A = serreverse( intformal( (1 - x^4 / 4 + x * O(x^m)) ^ (-1/2))); m! * polcoeff( sqrt( (2 - A^2) / (2 + A^2)), m))}; /* Michael Somos, Apr 25 2011 */

{a(n) = local(C=1+x); for(i=1,n, C = exp( intformal( C * intformal(-1/C^3 + x*O(x^n)) ) ) ); n!*polcoeff(C,n)}
for(n=0,20,print1(a(2*n),", "))

{a(n) = local(C=1+x); for(i=1,n, C = exp( intformal( -1/C * intformal(C^3 + x*O(x^n)) ) ) ); n!*polcoeff(C,n)}
for(n=0,20,print1(a(2*n),", "))

E.g.f. C(x) satisfies: C(x)^4 + 2*S(x)^2 = 1 where S(x) = Integral [1 - 2*S(x)^2]^(3/4) dx with S(0)=0; Left-shift of the Laplace transform of e.g.f. C(x) equals the Laplace transform of S(x).
E.g.f.: Sum_{k>=0} 2^k * a(k) * x^(2*k) / (2*k)! = cos lemn(x) where cos lemn(x) is the cosine lemniscate function of Gauss. - Michael Somos, Apr 25 2011
O.g.f.: 1/(1 + 1^2*x/(1 + 2^2/2*x/(1 + 3^2*x/(1 + 4^2/2*x/(1 + 5^2*x/(1 + 6^2/2*x/(1 + 7^2*x/(1 + 8^2/2*x/(1+...))))))))) (continued fraction). - Paul D. Hanna, Jul 29 2011
a(n) = (-1)^floor(n/2) * A193544(n) = (-1)^ceiling(n/2) * A193544(n) = -A159601(n). - M. F. Hasler, Aug 31 2012
G.f.: 1/Q(0) where p=1/2, Q(k) = 1 + x*(2*k+1)^2/( 1 + p*x*(2*k+2)^2/Q(k+1) ); (continued fraction due to Stieltjes T.J.). - Sergei N. Gladkovskii, Mar 22 2013
From Paul D. Hanna, Jun 02 2015: (Start)
E.g.f. C(x) satisfies:
(1) C(x) = exp( Integral C(x) * Integral -1/C(x)^3 dx dx ).
(2) C(x) = exp( Integral -1/C(x) * Integral C(x)^3 dx dx ).
(End)
G.f.: 1 / (1 + b(1)*x / (1 + b(2)*x / (1 + b(3)*x / ... ))) where b = A129194. - Michael Somos, Jan 03 2013

A193832 Irregular triangle read by rows in which row n lists 2n-1 copies of 2n-1 and n copies of 2n, for n >= 1.

Original entry on oeis.org

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

Offset: 1

Omar E. Pol, Aug 22 2011

Sequence of successive positive integers k in which if k is odd then k appears k times, otherwise if k is even then k appears k/2 times.
Note that an arrangement of the blocks of this sequence shows the growth of the generalized pentagonal numbers A001318 (see example).
The sums of each block give the positive integers of A129194: 1, 2, 9, 8, 25, 18, 49,...
Partial sums of A080995. - Paolo P. Lava, Aug 23 2011.
Concatenations of rows of triangles A001650 and A111650; also, seen as a flat list, the row lengths of triangle A260672 and the first differences of its row sums (cf. A260706). - Reinhard Zumkeller, Nov 17 2015
Also a(n) = number of squares in the arithmetic progression {24k + 1: 0 <= k <= n-1} [Granville]. - N. J. A. Sloane, Dec 13 2017

			a) If written as a triangle the initial rows are
  1, 2,
  3, 3, 3, 4, 4,
  5, 5, 5, 5, 5, 6, 6, 6,
  7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8,
  9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10,
  ...
Row sums give A126587.
b) An application using the blocks of this sequence: the illustration of the growth of an arrangement which represents the generalized pentagonal numbers A001318. For example; the first 9 positive initial terms: 1, 2, 5, 7, 12, 15, 22, 26, 35.
.
.         9
.       8 9
.     8 7 9
.   8 6 7 9
. 8 6 5 7 9
. 6 4 5 7 9
. 4 3 5 7 9
. 2 3 5 7 9
. 1 3 5 7 9
...

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
Andrew Granville, Squares in arithmetic progressions and infinitely many primes, arXiv:1708.06951 [math.NT], 2017.
Andrew Granville, Squares in arithmetic progressions and infinitely many primes, The American Mathematical Monthly, 124.10 (2017): 951-954. See p. 952.

Cf. A001318, A080995, A126587, A129194, A221671, A221672.

Cf. A001650, A111650, A260672, A260706.

a193832 n k = a193832_tabf !! (n-1) !! (k-1)
a193832_row n = a193832_tabf !! (n-1)
a193832_tabf = zipWith (++) a001650_tabf a111650_tabl
a193832' n = a193832_list !! (n - 1)
a193832_list = concat a193832_tabf
-- Reinhard Zumkeller, Nov 15 2015

Array[Join @@ MapIndexed[ConstantArray[#, #/(1 + Boole[First@ #2 == 2])] &, {2 # - 1, 2 #}] &, 7] // Flatten (* or *)
Table[If[k <= 2 n - 1, 2 n - 1, 2 n], {n, 7}, {k, 3 n - 1}] // Flatten (* Michael De Vlieger, Dec 14 2017 *)

a(n) = sqrt(8n/3) plus or minus 1 [Granville] - N. J. A. Sloane, Dec 13 2017

If 8 <= n <= 52, then a(n-1) < a(n) if and only if n is in A221672. - Jonathan Sondow, Dec 14 2017

Edited by N. J. A. Sloane, Dec 13 2017

A212760 Number of (w,x,y,z) with all terms in {0,...,n}, w even, and x = y + z.

Original entry on oeis.org

1, 3, 12, 20, 45, 63, 112, 144, 225, 275, 396, 468, 637, 735, 960, 1088, 1377, 1539, 1900, 2100, 2541, 2783, 3312, 3600, 4225, 4563, 5292, 5684, 6525, 6975, 7936, 8448, 9537, 10115, 11340, 11988, 13357, 14079, 15600, 16400, 18081, 18963, 20812, 21780, 23805

Offset: 0

Clark Kimberling, May 29 2012

A signed version is A122576.
For a guide to related sequences, see A211795.
Partial sums of the positive elements of A129194. - Omar E. Pol, Dec 28 2013

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

Cf. A211795.

Cf. A001318, A260706.

a212760 = a260706 . fromInteger . a001318 . (+ 1)
-- Reinhard Zumkeller, Nov 17 2015

[(n+1)*(n+2)*(2*n+3+(-1)^n)/8 : n in [0..50]]; // Wesley Ivan Hurt, Jul 22 2014

A212760:=n->(n+1)*(n+2)*(2*n+3+(-1)^n)/8: seq(A212760(n), n=0..50); # Wesley Ivan Hurt, Jul 22 2014

t = Compile[{{n, _Integer}}, Module[{s = 0}, (Do[If[(Mod[w, 2] == 0) && x == y + z, s++],
{w, 0, n}, {x, 0, n}, {y, 0, n}, {z, 0, n}]; s)]];
Map[t[#] &, Range[0, 50]]  (* A212760 *)
Table[(n + 1) (n + 2) (2 n + 3 + (-1)^n)/8, {n, 0, 50}] (* Wesley Ivan Hurt, Jul 22 2014 *)
CoefficientList[Series[(1 + 2 x + 6 x^2 + 2 x^3 + x^4)/((1 + x)^3 (1 - x)^4), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 23 2014 *)

a(n) = a(n-1)+3*a(n-2)-3*a(n-3)-3*a(n-4)+3*a(n-5)+a(n-6)-a(n-7).
G.f.: ( 1+2*x+6*x^2+2*x^3+x^4 ) / ( (1+x)^3*(1-x)^4 ).
a(n) = (n+1)*(n+2)*(2*n+3+(-1)^n)/8. - Wesley Ivan Hurt, Jul 22 2014
a(n) = A260706(A001318(n+1)). - Reinhard Zumkeller, Nov 17 2015
a(n) = Sum_{i=1..n+1} numerator(i^2/2). - Wesley Ivan Hurt, Feb 26 2017

A309337 a(n) = n^3 if n odd, 3*n^3/4 if n even.

Original entry on oeis.org

0, 1, 6, 27, 48, 125, 162, 343, 384, 729, 750, 1331, 1296, 2197, 2058, 3375, 3072, 4913, 4374, 6859, 6000, 9261, 7986, 12167, 10368, 15625, 13182, 19683, 16464, 24389, 20250, 29791, 24576, 35937, 29478, 42875, 34992, 50653, 41154, 59319, 48000, 68921, 55566, 79507, 63888, 91125

Offset: 0

Ilya Gutkovskiy, Jul 24 2019

Moebius transform of A078307.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A000578, A016755, A059376, A078307, A129194, A193356, A309338.

a[n_] := If[OddQ[n], n^3, 3 n^3/4]; Table[a[n], {n, 0, 45}]
nmax = 45; CoefficientList[Series[x (1 + 6 x + 23 x^2 + 24 x^3 + 23 x^4 + 6 x^5 + x^6)/(1 - x^2)^4, {x, 0, nmax}], x]
LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {0, 1, 6, 27, 48, 125, 162, 343}, 46]
Table[n^3 (7 - (-1)^n)/8, {n, 0, 45}]

G.f.: x * (1 + 6*x + 23*x^2 + 24*x^3 + 23*x^4 + 6*x^5 + x^6)/(1 - x^2)^4.
G.f.: Sum_{k>=1} J_3(k) * x^k/(1 + x^k), where J_3() is the Jordan function (A059376).
Dirichlet g.f.: zeta(s-3) * (1 - 2^(1-s)).
a(n) = n^3 * (7 - (-1)^n)/8.
a(n) = Sum_{d|n} (-1)^(n/d + 1) * J_3(d).
Sum_{n>=1} 1/a(n) = 25*zeta(3)/24 = 1.252142607457910713958...
Multiplicative with a(2^e) = 3*2^(3*e-2), and a(p^e) = p^(3*e) for odd primes p. - Amiram Eldar, Oct 26 2020
a(n) = Sum_{1 <= i, j, k <= n} (-1)^(1 + gcd(i,j,k,n)) = Sum_{d | n} (-1)^(d+1) * J_3(n/d). Cf. A129194. - Peter Bala, Jan 16 2024

A122576 Expansion of g.f.: -x(1 - 2x + 6x^2 - 2x^3 + x^4)/((1-x)^3*(1+x)^4).

Original entry on oeis.org

-1, 3, -12, 20, -45, 63, -112, 144, -225, 275, -396, 468, -637, 735, -960, 1088, -1377, 1539, -1900, 2100, -2541, 2783, -3312, 3600, -4225, 4563, -5292, 5684, -6525, 6975, -7936, 8448, -9537, 10115, -11340, 11988, -13357, 14079, -15600, 16400, -18081, 18963, -20812, 21780, -23805

Offset: 1

Roger L. Bagula, Sep 17 2006

Unsigned = row sums of triangle A143120 and Sum_{n>=1} n*A026741(n). - Gary W. Adamson, Jul 26 2008
Unsigned = partial sums of positive integers of A129194. - Omar E. Pol, Aug 22 2011
Unsigned, see A212760. - Clark Kimberling, May 29 2012

Roger G. Newton, Scattering Theory of Waves and Particles, McGraw Hill, 1966; p. 254.

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

Cf. A026741, A098023, A143120, A212760.

[(n+1)*(n+2)*(2*n+3+(-1)^n)*(-1)^(n+1)/8 : n in [0..50]]; // Wesley Ivan Hurt, Jul 22 2014

A122576:=n->(n+1)*(n+2)*(2*n+3+(-1)^n)*(-1)^(n+1)/8: seq(A122576(n), n=0..50); # Wesley Ivan Hurt, Jul 22 2014

Table[(n + 1) (n + 2) (2 n + 3 + (-1)^n) (-1)^(n + 1)/8, {n, 0, 50}] (* Wesley Ivan Hurt, Jul 22 2014 *)
CoefficientList[Series[(1 -2 x +6 x^2 -2 x^3 +x^4)/((x-1)^3 (x+1)^4), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 23 2014 *)

def A122576(n): return n*(n+1)*((2*n+1)*(-1)^n - 1)//8
[A122576(n) for n in range(1,51)] # G. C. Greubel, Nov 02 2024

a(n) = n*(n+1)/8 * ((2*n+1)*(-1)^n - 1). - Ralf Stephan, Jan 01 2014
a(n) = (n+1)*(n+2)*(2*n+3+(-1)^n)*(-1)^(n+1)/8 (with offset of 0). - Wesley Ivan Hurt, Jul 22 2014
E.g.f.: -(1/8)*x*( (6 - 9*x + 2*x^2)*exp(-x) + (2+x)*exp(x) ). - G. C. Greubel, Nov 02 2024

Edited by N. J. A. Sloane, May 20 2007. The simple generating function now used to define the sequence was found by an anonymous correspondent.

A129204 The denominator of 2/n^3.

Original entry on oeis.org

1, 1, 4, 27, 32, 125, 108, 343, 256, 729, 500, 1331, 864, 2197, 1372, 3375, 2048, 4913, 2916, 6859, 4000, 9261, 5324, 12167, 6912, 15625, 8788, 19683, 10976, 24389, 13500, 29791, 16384, 35937, 19652, 42875, 23328, 50653, 27436, 59319, 32000

Offset: 0

Paul Barry, Apr 03 2007

Take two consecutive triangular numbers t1 and t2 and create a triangle using (0,0), (t1,t2) and (t2,t1). The area of this triangle will be ((n+1)^3)/2 for t1 = n*(n+1)/2. - J. M. Bergot, May 08 2012

Multiplicative because both A000578 and A040001 are. - Andrew Howroyd, Jul 26 2018

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

Cf. A000578, A040001, A129194, A309337.

[1] cat [Denominator(2/n^3): n in [1..40]]; // Vincenzo Librandi, Jul 26 2018

Join[{1}, Table[Denominator[2 / n^3], {n, 100}]] (* Vincenzo Librandi, Jul 26 2018 *)

a(n) = if(n < 1, n==0, lcm(2, n^3)/2) \\ Andrew Howroyd, Jul 25 2018

G.f.: (1+x+23x^2+22x^4+23x^5+x^7+x^8)/(1-x^2)^4.
a(n) = 0^n + n^3*(3/4 - (-1)^n/4).
a(n+1) = A129196(n)*(5/3 + (4/3)*cos(2*Pi*(n+1)/3)).
a(2n) = 4n^3, a(2n+1) = (2n+1)^3.
a(n) = A000578(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + 9*zeta(3)/8.
Sum_{n>=0} (-1)^n/a(n) = 1 - 5*zeta(3)/8. (End)
From Peter Bala, Jan 21 2024: (Start)
For n >= 1, a(n) = n*A129194(n) = n*Sum_{d divides n} (-1)^(d+1)*J(2,n/d), where the Jordan totient function J_2(n) = A007434(n). Cf. A309337.
Dirichlet g.f. for sequence without the a(0) term: (1 - 4/2^s)*zeta(s-3). (End)

More terms from Vincenzo Librandi, Jul 26 2018

A309338 a(n) = n^4 if n odd, 7*n^4/8 if n even.

Original entry on oeis.org

0, 1, 14, 81, 224, 625, 1134, 2401, 3584, 6561, 8750, 14641, 18144, 28561, 33614, 50625, 57344, 83521, 91854, 130321, 140000, 194481, 204974, 279841, 290304, 390625, 399854, 531441, 537824, 707281, 708750, 923521, 917504, 1185921, 1169294, 1500625, 1469664, 1874161, 1824494

Offset: 0

Ilya Gutkovskiy, Jul 24 2019

Moebius transform of A284900.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-10,0,10,0,-5,0,1).

Cf. A000583, A016756, A059377, A129194, A193356, A284900, A309337.

a[n_] := If[OddQ[n], n^4, 7 n^4/8]; Table[a[n], {n, 0, 38}]
nmax = 38; CoefficientList[Series[x (1 + 14 x + 76 x^2 + 154 x^3 + 230 x^4 + 154 x^5 + 76 x^6 + 14 x^7 + x^8)/(1 - x^2)^5, {x, 0, nmax}], x]
LinearRecurrence[{0, 5, 0, -10, 0, 10, 0, -5, 0, 1}, {0, 1, 14, 81, 224, 625, 1134, 2401, 3584, 6561}, 39]
Table[n^4 (15 - (-1)^n)/16, {n, 0, 38}]

G.f.: x * (1 + 14*x + 76*x^2 + 154*x^3 + 230*x^4 + 154*x^5 + 76*x^6 + 14*x^7 + x^8)/(1 - x^2)^5.
G.f.: Sum_{k>=1} J_4(k) * x^k/(1 + x^k), where J_4() is the Jordan function (A059377).
Dirichlet g.f.: zeta(s-4) * (1 - 2^(1-s)).
a(n) = n^4 * (15 - (-1)^n)/16.
a(n) = Sum_{d|n} (-1)^(n/d + 1) * J_4(d).
Sum_{n>=1} 1/a(n) = 113*Pi^4/10080 = 1.091986834012130496797...
Multiplicative with a(2^e) = 7*2^(4*e-3), and a(p^e) = p^(4*e) for odd primes p. - Amiram Eldar, Oct 26 2020

cp's OEIS Frontend

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A122576 Expansion of g.f.: -x(1 - 2x + 6x^2 - 2x^3 + x^4)/((1-x)^3*(1+x)^4).