A053698 -id:A053698 - cp's OEIS frontend

A099721 a(n) = n^2(2n+1).

0, 3, 20, 63, 144, 275, 468, 735, 1088, 1539, 2100, 2783, 3600, 4563, 5684, 6975, 8448, 10115, 11988, 14079, 16400, 18963, 21780, 24863, 28224, 31875, 35828, 40095, 44688, 49619, 54900, 60543, 66560, 72963, 79764, 86975, 94608, 102675, 111188, 120159, 129600

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Nov 07 2004

For a right triangle with sides of lengths 8*n^3 + 12*n^2 + 8*n + 2, 4*n^4 + 8*n^3 + 4*n^2, and 4*n^4 + 8*n^3 + 12*n^2 + 8*n + 2, dividing the area by the perimeter gives a(n). - J. M. Bergot, Jul 30 2013
This sequence is the difference between the centered icosahedral (or cuboctahedral) numbers (A005902(n)) and the centered octagonal pyramidal numbers (A000447(n+1)). - Peter M. Chema, Jan 09 2016
a(n) is the sum of the integers in the closed interval (n-1)*n to n*(n+1). - J. M. Bergot, Apr 19 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Row n=3 of A229079.
Cf. A000578, A001093, A011379, A015237, A027444, A033431, A033562, A034262, A053698, A061317, A066023, A071568, A098547.
Cf. A000290, A000447, A002378, A005408, A005902, A024196.

[n^2*(2*n+1): n in [0..50]]; // Vincenzo Librandi, May 01 2011

A099721 := proc(n) n^2*(2*n+1) ; end proc:
seq(A099721(n),n=0..10) ;

a[n_]:=2*n^3+n^2; (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
LinearRecurrence[{4,-6,4,-1},{0,3,20,63},40] (* Harvey P. Dale, Aug 19 2022 *)

a(n) = ceil(sum(i=n^2-(n-1), n^2+(n-1), if(!issquare(4*i+1), (2*i+1+sqrt(4*i+1))/2, 0))); \\ Michel Marcus, Nov 14 2014, after Richard R. Forberg

G.f.: x*(3 + 8*x + x^2)/(x-1)^4.
a(n) = A024196(n) - A024196(n-1). - Philippe Deléham, May 07 2012
a(n) = ceiling(Sum_{i=n^2-(n-1)..n^2+(n-1)} s(i)), for n > 0 and integer i, where s(i) are the real solutions to x = i + sqrt(x), and the summation range excludes the integer solutions which occur where i is an oblong number (A002378). The fractional portion of the summation converges to 2/3 for large n. If s(i) is replaced with i, then the summation equals n^2*(2*n-1) = A015237. - Richard R. Forberg, Oct 15 2014
a(n) = A005902(n) - A000447(n+1). - Peter M. Chema, Jan 09 2016
From Amiram Eldar, May 17 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/6 + 4*log(2) - 4.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/12 - Pi - 2*log(2) + 4. (End)
From Elmo R. Oliveira, Aug 08 2025: (Start)
E.g.f.: x*(1 + 2*x)*(3 + x)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = A000290(n)*A005408(n). (End)

A100705 a(n) = n^3 + (n+1)^2.

Original entry on oeis.org

1, 5, 17, 43, 89, 161, 265, 407, 593, 829, 1121, 1475, 1897, 2393, 2969, 3631, 4385, 5237, 6193, 7259, 8441, 9745, 11177, 12743, 14449, 16301, 18305, 20467, 22793, 25289, 27961, 30815, 33857, 37093, 40529, 44171, 48025, 52097, 56393, 60919, 65681, 70685, 75937

Offset: 0

Giovanni Teofilatto, Jan 03 2005

The finite simple continued fraction [1;n-1,n,n+1] has a numerator in the resulting rational number (n^3+2*n+n^2+1)/(n*(n^2+1)) that is the same as a(n). - J. M. Bergot, Sep 29 2011

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

Cf. A053698.

[n^3 + (n+1)^2: n in [0..50]]; // Vincenzo Librandi, Sep 30 2011

Table[n^3 + (n + 1)^2, {n, 0, 45}]
LinearRecurrence[{4,-6,4,-1},{1,5,17,43},50] (* Harvey P. Dale, Dec 03 2014 *)

a(n)=n^3+(n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1+x+3*x^2+x^3)/(x-1)^4. - R. J. Mathar, Sep 29 2011
a(n) = A053698(n) + n. - Bruno Berselli, Sep 30 2011
From Elmo R. Oliveira, Apr 19 2025: (Start)
E.g.f.: exp(x)*(1 + x)*(1 + 3*x + x^2).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)

More terms from Mark Hudson and Farideh Firoozbakht, Jan 04 2005

A102909 a(n) = Sum_{j=0..8} n^j.

Original entry on oeis.org

1, 9, 511, 9841, 87381, 488281, 2015539, 6725601, 19173961, 48427561, 111111111, 235794769, 469070941, 883708281, 1589311291, 2745954241, 4581298449, 7411742281, 11668193551, 17927094321, 26947368421, 39714002329, 57489010371, 81870575521, 114861197401

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Mar 01 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A001016, A060890, A059839.

Cf. similar sequences of the type a(n) = Sum_{j=0..m} n^j: A000027 (m=1), A002061 (m=2), A053698 (m=3), A053699 (m=4), A053700 (m=5), A053716 (m=6), A053717 (m=7), this sequence (m=8), A103623 (m=9), A060885 (m=10), A105067 (m=11), A060887 (m=12), A104376 (m=13), A104682 (m=14), A105312 (m=15), A269442 (m=16), A269446 (m=18).

[(&+[n^j: j in [0..8]]): n in [0..30]]; // G. C. Greubel, Feb 13 2018

1 + Sum[Range[0, 30]^j, {j, 1, 8}] (* G. C. Greubel, Feb 13 2018 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,9,511,9841,87381,488281,2015539,6725601,19173961},30] (* Harvey P. Dale, Feb 01 2025 *)

a(n)=n^8+n^7+n^6+n^5+n^4+n^3+n^2+n+1 \\ Charles R Greathouse IV, Oct 07 2015

[sum(n^j for j in (0..8)) for n in (0..30)] # G. C. Greubel, Apr 14 2019

a(n) = (n^2+n+1) * (n^6+n^3+1) and so is never prime. - Jonathan Vos Post, Dec 21 2012
G.f.: (x^8 + 162*x^7 + 3418*x^6 + 14212*x^5 + 16578*x^4 + 5482*x^3 + 466*x^2 + 1)/(1-x)^9. - Colin Barker, Nov 05 2012, edited by M. F. Hasler, Dec 31 2012
a(n) = (n^9-1)/(n-1) with a(1) = 9. - L. Edson Jeffery and M. F. Hasler, Dec 30 2012
E.g.f.: exp(x)*(1 + 8*x + 247*x^2 + 1389*x^3 + 2127*x^4 + 1206*x^5 + 288*x^6 + 29*x^7 + x^8). - Stefano Spezia, Oct 03 2024

Offset corrected by N. J. A. Sloane, Dec 30 2012

A104376 a(n) = Sum_{j=0..13} n^j.

Original entry on oeis.org

1, 14, 16383, 2391484, 89478485, 1525878906, 15672832819, 113037178808, 628292358729, 2859599056870, 11111111111111, 37974983358324, 116719860413533, 328114698808274, 854769755812155, 2085209001813616, 4803839602528529, 10523614159962558, 22047845151887119

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Apr 16 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Cf. similar sequences of the type a(n) = Sum_{j=0..m} n^j are: A000027 (m=1), A002061 (m=2), A053698 (m=3), A053699 (m=4), A053700 (m=5), A053716 (m=6), A053717 (m=7), A102909 (m=8), A103623 (m=9), A060885 (m=10), A105067 (m=11), A060887 (m=12), this sequence (m=13), A104682 (m=14), A105312 (m=15), A269442 (m=16), A269446 (m=18).

[(&+[n^j: j in [0..13]]): n in [0..20]]; // Vincenzo Librandi, May 01 2011

Table[1+Sum[n^j, {j,1,13}], {n,0,20}] (* G. C. Greubel, Apr 14 2019 *)
LinearRecurrence[{14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1},{1,14,16383,2391484,89478485,1525878906,15672832819,113037178808,628292358729,2859599056870,11111111111111,37974983358324,116719860413533,328114698808274},20] (* Harvey P. Dale, Sep 04 2023 *)

a(n)=sum(j=0,13, n^j) \\ Charles R Greathouse IV, Oct 07 2015

[sum(n^j for j in (0..13)) for n in (0..20)] # G. C. Greubel, Apr 14 2019

a(n) = n^13 + n^12 + n^11 + n^10 + n^9 + n^8 + n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n^1 + 1.

G.f.: (5461*x^12 + 1119288*x^11 + 37443654*x^10 + 372458048*x^9 + 1409085783*x^8 + 2263446576*x^7 + 1598944452*x^6 + 484853760*x^5 + 57484467*x^4 + 2163032*x^3 + 16278*x^2 + 1)/(1-x)^14. - Colin Barker, Nov 04 2012

Name changed by G. C. Greubel, Apr 14 2019

A104682 a(n) = Sum_{j=0..14} n^j.

Original entry on oeis.org

1, 15, 32767, 7174453, 357913941, 7629394531, 94036996915, 791260251657, 5026338869833, 25736391511831, 111111111111111, 417724816941565, 1400638324962397, 4265491084507563, 11966776581370171, 31278135027204241, 76861433640456465, 178901440719363487

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Apr 22 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).

Cf. similar sequences of the type a(n) = Sum_{j=0..m} n^j are: A000027 (m=1), A002061 (m=2), A053698 (m=3), A053699 (m=4), A053700 (m=5), A053716 (m=6), A053717 (m=7), A102909 (m=8), A103623 (m=9), A060885 (m=10), A105067 (m=11), A060887 (m=12), A104376 (m=13), this sequence (m=14), A105312 (m=15), A269442 (m=16), A269446 (m=18).

[(&+[n^j: j in [0..14]]): n in [0..20]]; // Vincenzo Librandi, May 01 2011

With[{f=Total[n^Range[0,14]]},Table[f,{n,0,20}]] (* Harvey P. Dale, Jun 11 2011 *)

a(n) = sum(j=0, 14, n^j) \\ Charles R Greathouse IV, Oct 07 2015

[sum(n^j for j in (0..14)) for n in (0..20)] # G. C. Greubel, Apr 15 2019

a(n) = n^14 + n^13 + n^12 + n^11 + n^10 + n^9 + n^8 + n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n^1 + 1.
a(n) = (n^2 + n + 1) * (n^4 + n^3 + n^2 + n + 1) * (n^8 - n^7 + n^5 - n^4 + n^3 - n + 1). - Jonathan Vos Post, Apr 23 2005
G.f.: (x^14 +10908*x^13 +3423487*x^12 +162086420*x^11 +2236727781*x^10 +11806635128*x^9 +27116815299*x^8 +28635678216*x^7 +13957353555*x^6 +2999111468*x^5 +253732221*x^4 +6684068*x^3 +32647*x^2 +1)/(1-x)^15. - Colin Barker, Nov 04 2012

More terms from Harvey P. Dale, Jun 11 2011

Name changed by G. C. Greubel, Apr 15 2019

A105067 a(n) = Sum_{j=0..11} n^j.

Original entry on oeis.org

1, 12, 4095, 265720, 5592405, 61035156, 435356467, 2306881200, 9817068105, 35303692060, 111111111111, 313842837672, 810554586205, 1941507093540, 4361070182715, 9267595563616, 18764998447377, 36413889826860

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Apr 05 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Cf. similar sequences of the type a(n) = Sum_{j=0..m} n^j: A000027 (m=1), A002061 (m=2), A053698 (m=3), A053699 (m=4), A053700 (m=5), A053716 (m=6), A053717 (m=7), A102909 (m=8), A103623 (m=9), A060885 (m=10), this sequence (m=11), A060887 (m=12), A104376 (m=13), A104682 (m=14), A105312 (m=15), A269442 (m=16), A269446 (m=18).

[n^11 + n^10 + n^9 + n^8 + n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n + 1: n in [0..20]]; // Vincenzo Librandi, May 01 2011

1+Sum[Range[0,20]^j, {j,1,11}] (* G. C. Greubel, Apr 13 2019 *)

a(n)=polcyclo(11,n)+n^11 \\ Charles R Greathouse IV, Sep 03 2011

[sum(n^j for j in (0..11)) for n in (0..20)] # G. C. Greubel, Apr 13 2019

Factorization of the polynomial into irreducible components over integers: n^11 + n^10 + n^9 + n^8 + n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n + 1 = +- (n + 1) * (n^2 - n + 1) * (n^2 + 1) * (n^2 + n + 1) * (n^4 - n^2 + 1). - Jonathan Vos Post, Apr 06 2005

G.f.: (1365*x^10 + 116480*x^9 + 1851213*x^8 + 8893248*x^7 + 15593370*x^6 + 10568064*x^5 + 2671890*x^4 + 217152*x^3 + 4017*x^2 + 1)/(x - 1)^12. - Colin Barker, Oct 29 2012

Signature changed by Georg Fischer, Apr 13 2019

A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)4 and containing (k+1)4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 4, 1, 15, 30, 1, 40, 324, 120, 1, 85, 2080, 3120, 340, 1, 156, 9375, 40000, 18750, 780, 1, 259, 32886, 328125, 437500, 82215, 1554, 1, 400, 96040, 1959216, 6002500, 3265360, 288120, 2800, 1

Offset: 0

Susanne Wienand, Oct 17 2011

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 4.
The values in the triangle are proved by brute force for 0 <= n <= 8. The formulas are not yet proved in general.
The number triangle can be calculated recursively by the number triangles A007318, A103371 and A194595. The first column of the triangle seems to be A053698. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027445. Row sums are in A198256.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 43. - Susanne Wienand, Jun 29 2015

			For n = 4 and k = 2, T(4,4,2) = 3120.
Recursive example:
T(1,4,0) = 1,
T(1,4,1) = 4,
T(1,4,2) = 6,
T(1,4,3) = 4,
T(1,4,4) = 1,
T(3,4,0) = 21,
T(3,4,1) = 304,
T(3,4,2) = 456,
T(3,4,3) = 84,
T(3,4,1) = 1,
T(4,4,2) = 6^4 + 6*304 = 3120.
Example for closed formula:
T(4,2) = 6^4 + 6^3 * 4 + 6^2 * 4^2 + 6 * 4^3 = 3120.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*4 = 20 and S contains (2+1)*4 = 12 Ls.
  S: L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R
dir: 1,2,3,0,1,2,3,0,1,2,3,0,0,3,2,1,0,3,2,1
  S: L,L,L,R,L,L,R,L,L,R,R,L,L,L,R,L,L,R,R,R
dir: 1,2,3,3,3,0,0,0,1,1,0,0,1,2,2,2,3,3,2,1
  S: L,R,L,L,L,L,R,R,R,L,L,R,R,L,L,L,R,L,L,R
dir: 1,1,1,2,3,0,0,3,2,2,3,3,2,2,3,0,0,0,1,1
Each value of dir occurs 20/4 = 5 times.
Triangle begins:
   1;
   4,    1;
  15,   30,    1;
  40,  324,  120,   1;
  85, 2080, 3120, 340, 1;
  ...

Susanne Wienand, Table of n, a(n) for n = 0..989
Peter Luschny, Meanders and walks on the circle.

Cf. A000012, A007318, A027445, A053698, A103371, A194595, A197654, A197655, A198063, A198256.

A197653 := (n,k) -> binomial(n,k)^4*(n+1)*(n^2-2*n*k+1+2*k+2*k^2)/((1+k)^3);
seq(print(seq(A197653(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011

T[n_, k_] := Binomial[n, k]^4 (n+1)(n^2 - 2n*k + 1 + 2k + 2k^2)/((1+k)^3);
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A197653(n,k) = {if(n==1+2*k,4,(1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n))*binomial(n,k)^4} \\ Peter Luschny, Nov 24 2011

def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
def A197653(n,k) : return S(3,n,k)
for n in (0..5) : print([A197653(n,k) for k in (0..n)])  ## Peter Luschny, Oct 24 2011

Recursive formula (conjectured):
T(n,k)   = T(4,n,k)
T(4,n,k) = T(1,n,k)^4 + T(1,n,k)*T(3,n,n-1-k), 0 <= k < n
T(4,n,n) = 1                                        k = n
T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1                                        k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k)*T(1,n,n-1-k), 0 <= k < n
T(2,n,n) = 1                                        k = n
T(3,n,k) = A194595
T(2,n,k) = A103371
T(1,n,k) = A007318 (Pascal's Triangle)
closed formula (conjectured): T(n,n) = 1,                         k = n
                T(n,k) = A + B + C + D,             k < n
                     A = (C(n,k))^4
                     B = (C(n,k))^3 * C(n,n-1-k)
                     C = (C(n,k))^2 *(C(n,n-1-k))^2
                     D =  C(n,k)    *(C(n,n-1-k))^3
[Susanne Wienand]
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,3). - Peter Luschny, Oct 20 2011
T(n,k) = A198063(n+1,k+1)*C(n,k)^4/(k+1)^3. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^4, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n) if 1+2*k-n <> 0, otherwise h(n,k) = 4. - Peter Luschny, Nov 24 2011

A237627 Semiprimes of the form n^3 + n^2 + n + 1.

Original entry on oeis.org

4, 15, 85, 259, 1111, 4369, 47989, 65641, 291919, 2016379, 2214031, 3397651, 3820909, 5864581, 9305311, 13881841, 15687751, 16843009, 19756171, 22030681, 28746559, 62256349, 64160401, 74264821, 79692331, 101412319, 117889591, 172189309, 185518471, 191435329

Offset: 1

K. D. Bajpai, Apr 22 2014

nonn

All the terms in the sequence, except a(1), are odd.

Since n^3 + n^2 + n + 1 = (n^2 + 1)(n + 1), it is a necessary condition for n^2 + 1 to be a prime, and n + 1 as well. - Alonso del Arte, Apr 22 2014

			85 is in the sequence since 4^3 + 4^2 + 4 + 1 = 85 = 5 * 17, which is a semiprime.
259 is in the sequence since 6^3 + 6^2 + 6 + 1 = 259 = 7 * 37 which is a semiprime.
585 is not in the sequence, because, although it is 8^3 + 8^2 + 8 + 1, it has more than two prime factors.

K. D. Bajpai, Table of n, a(n) for n = 1..2539

Cf. A070689 (associated n).

Cf. A053698, A001358, A005898, A046315, A046388, A240859, A240884, A241060, A241102.

IsSemiprime:=func; [s: n in [1..1000] | IsSemiprime(s) where s is n^3+n^2+n+1]; // Bruno Berselli, Apr 23 2014

Maple

select(x-> numtheory[bigomega](x)=2, [n^3+n^2+n+1$n=1..1500])[];

Mathematica

A237627 = {}; Do[t = n^3 + n^2 + n + 1; If[PrimeOmega[t] == 2, AppendTo[A237627, t]], {n, 1500}]; A237627 (* K. D. Bajpai *) (* For the b-file: *) n = 0; Do[t = k^3 + k^2 + k + 1; If[PrimeOmega[t] == 2, n++; Print[n, " ", t]], {k, 300000}] (* K. D. Bajpai *) Select[Table[n^3 + n^2 + n + 1, {n, 500}], PrimeOmega[#] == 2 &] (* Alonso del Arte, Apr 22 2014 *)

PARI

is(n)=isprime(n^2+1) && isprime(n+1) \\ Charles R Greathouse IV, Aug 25 2014

Python

from itertools import islice from sympy import isprime, nextprime def A237627_gen(): # generator of terms p = 1 while (p:=nextprime(p)): if isprime((p-1)**2+1): yield p*((p-1)**2+1) A237627_list = list(islice(A237627_gen(),20)) # Chai Wah Wu, Feb 27 2023

Sage

A237627 = list(n^3 + n^2 + n + 1 for n in (1..1000) if is_prime(n^2+1) and is_prime(n+1)); print(A237627) # Bruno Berselli, Apr 23 2014 - see comment by Alonso del Arte

Formula

Union of {4} and the members of A176070. - R. J. Mathar, Oct 04 2018

cp's OEIS Frontend

A099721 a(n) = n^2*(2*n+1).

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A100705 a(n) = n^3 + (n+1)^2.

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A102909 a(n) = Sum_{j=0..8} n^j.

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A104376 a(n) = Sum_{j=0..13} n^j.

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A104682 a(n) = Sum_{j=0..14} n^j.

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A105067 a(n) = Sum_{j=0..11} n^j.

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A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*4 and containing (k+1)*4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

A099721 a(n) = n^2(2n+1).

A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)4 and containing (k+1)4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

A188947 a(n) = n^3 - 2n^2 + 2n + 1.