A027445 -id:A027445 - cp's OEIS frontend

A016861 a(n) = 5*n + 1.

1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, 91, 96, 101, 106, 111, 116, 121, 126, 131, 136, 141, 146, 151, 156, 161, 166, 171, 176, 181, 186, 191, 196, 201, 206, 211, 216, 221, 226, 231, 236, 241, 246, 251, 256, 261, 266, 271, 276, 281

Offset: 0

N. J. A. Sloane, Dec 11 1996

Numbers ending in 1 or 6.
Apart from initial terms, same as 5n-14.
Complement of A047203; A027445(a(n)) mod 10 = 4. - Reinhard Zumkeller, Oct 23 2006
Campbell reference shows: "A graph on n vertices with at least 4n-9 edges is intrinsically linked. A graph on n vertices with at least 5n-14 edges is intrinsically knotted." - Jonathan Vos Post, Jan 18 2007
Central terms of the triangle in A153125: a(n) = A153125(2*n+1, n+1). - Reinhard Zumkeller, Dec 20 2008
For n > 2, also the number of (not necessarily maximal) cliques in the n-Moebius ladder graph. - Eric W. Weisstein, Nov 29 2017
For n > 3, also the number of (not necessarily maximal) cliques in the n-prism graph. - Eric W. Weisstein, Nov 29 2017
For n >= 1, a(n) is the size of any hexagonal chain graph with n cells. - Christian Barrientos, Sarah Minion, Mar 07 2018
For n >= 1, a(n) is the number of possible outcomes of the summation when using n dice. - Bram Kole, Dec 24 2018
Numbers congruent to 1 (mod 5). - Muniru A Asiru, Jan 01 2019
Numbers k such that the k-th Fibonacci number, A000045(k), and the k-th Lucas number, A000032(k), end with the same decimal digit. - Amiram Eldar, Apr 15 2023

Ivan Panchenko, Table of n, a(n) for n = 0..1000
J. Campbell, T. W. Mattman, R. Ottman, J. Pyzer, M. Rodrigues and S. Williams, Intrinsic knotting and linking of almost complete graphs, arXiv:math/0701422 [math.GT], Jan 15 2007.
Tanya Khovanova, Recursive Sequences.
Eric Weisstein's World of Mathematics, Clique.
Eric Weisstein's World of Mathematics, Moebius Ladder.
Eric Weisstein's World of Mathematics, Prism Graph.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093562 ((5, 1) Pascal, column m=1).
Cf. A000566 (partial sums).
Cf. A000032, A000045, A001622, A131843, A342819.

a:=List([0..60],n->5*n+1);; Print(a); # Muniru A Asiru, Jan 01 2019

a016861 = (+ 1) . (* 5)
a016861_list = [1, 6 ..]  -- Reinhard Zumkeller, Jun 16 2013

A016861:=n->5*n+1; seq(A016861(n), n=0..100); # Wesley Ivan Hurt, May 03 2014

Range[1, 500, 5] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
LinearRecurrence[{2, -1}, {6, 11}, {0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)
CoefficientList[Series[(1 + 4 x)/(-1 + x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Nov 29 2017 *)

a(n)=5*n+1 \\ Charles R Greathouse IV, Jul 10 2016

G.f.: (1+4*x)/(1-x)^2.
Row sums of triangle A131843. - Gary W. Adamson, Jul 21 2007
a(n) = 2*a(n-1) - a(n-2) with a(0)=1, a(1)=6. - Vincenzo Librandi, Aug 01 2010
a(n) = A017293(n)/2 = A008587(n)+1. - Wesley Ivan Hurt, May 03 2014
E.g.f.: exp(x)*(1 + 5*x). - Stefano Spezia, Mar 23 2021
Sum_{n>=0} (-1)^n/a(n) = sqrt(2+2/sqrt(5))*Pi/10 + log(phi)/sqrt(5) + log(2)/5, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023

More terms from Reinhard Zumkeller, Oct 23 2006

A228275 A(n,k) = Sum_{i=1..k} n^i; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 6, 3, 0, 0, 4, 14, 12, 4, 0, 0, 5, 30, 39, 20, 5, 0, 0, 6, 62, 120, 84, 30, 6, 0, 0, 7, 126, 363, 340, 155, 42, 7, 0, 0, 8, 254, 1092, 1364, 780, 258, 56, 8, 0, 0, 9, 510, 3279, 5460, 3905, 1554, 399, 72, 9, 0

Offset: 0

Alois P. Heinz, Aug 19 2013

A(n,k) is the total sum of lengths of longest ending contiguous subsequences with the same value over all s in {1,...,n}^k:
A(4,1) = 4 = 1+1+1+1: [1], [2], [3], [4].
A(1,4) = 4: [1,1,1,1].
A(3,2) = 12 = 2+1+1+1+2+1+1+1+2: [1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3].
A(2,3) = 14 = 3+1+1+2+2+1+1+3: [1,1,1], [1,1,2], [1,2,1], [1,2,2], [2,1,1], [2,1,2], [2,2,1], [2,2,2].

			Square array A(n,k) begins:
  0, 0,  0,   0,    0,     0,      0,      0, ...
  0, 1,  2,   3,    4,     5,      6,      7, ...
  0, 2,  6,  14,   30,    62,    126,    254, ...
  0, 3, 12,  39,  120,   363,   1092,   3279, ...
  0, 4, 20,  84,  340,  1364,   5460,  21844, ...
  0, 5, 30, 155,  780,  3905,  19530,  97655, ...
  0, 6, 42, 258, 1554,  9330,  55986, 335922, ...
  0, 7, 56, 399, 2800, 19607, 137256, 960799, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A000004, A001477, A002378, A027444, A027445, A152031, A228290, A228291, A228292, A228293, A228294.
Rows n=0-11 give: A000004, A001477, A000918(k+1), A029858(k+1), A080674, A104891, A105281, A104896, A052379(k-1), A052386, A105279, A105280.
Main diagonal gives A031972.
Lower diagonal gives A226238.
Cf. A228250.

A:= (n, k)-> `if`(n=1, k, (n/(n-1))*(n^k-1)):
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[0, 0] = 0; a[1, k_] := k; a[n_, k_] := n*(n^k-1)/(n-1); Table[a[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 16 2013 *)

A(1,k) = k, else A(n,k) = n/(n-1)*(n^k-1).
A(n,k) = Sum_{i=1..k} n^i.
A(n,k) = Sum_{i=1..k+1} binomial(k+1,i)*A(n-i,k)*(-1)^(i+1) for n>k, given values A(0,k), A(1,k),..., A(k,k). - Yosu Yurramendi, Sep 03 2013

A047203 Numbers that are congruent to {0, 2, 3, 4} mod 5.

Original entry on oeis.org

0, 2, 3, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 20, 22, 23, 24, 25, 27, 28, 29, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 87, 88, 89

Offset: 1

N. J. A. Sloane

Complement of A016861. - Reinhard Zumkeller, Oct 23 2006

Melvyn B. Nathanson, On the fractional parts of roots of positive real numbers, Amer. Math. Monthly, Vol. 120, No. 5 (2013), pp. 409-429 [see p. 417].
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001068, A001622, A002265, A016861, A027445, A047207, A047211, A047218, A177704.

[Floor((5*n-2)/4) : n in [1..100]]; // Wesley Ivan Hurt, May 14 2016

seq(floor(5*n-2)/4), n=1..72); # Gary Detlefs, Mar 06 2010
seq(floor((15*n-5)/12), n=1..72); # Gary Detlefs, Mar 07 2010

Flatten[Table[5*n + {0, 2, 3, 4}, {n, 0, 20}]] (* T. D. Noe, Nov 12 2013 *)

a(n)=(5*n-2)\4 \\ Charles R Greathouse IV, Jun 11 2015

A027445(a(n)) mod 10 = 0. - Reinhard Zumkeller, Oct 23 2006
a(n) = floor((5n-2)/4). - Gary Detlefs, Mar 06 2010
a(n) = floor((15n-5)/12). - Gary Detlefs, Mar 07 2010
G.f.: x^2*(2+x+x^2+x^3)/((1+x)*(1+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, May 14 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (10*n-7+(-1)^n+2*(-1)^((2*n+3+(-1)^n)/4))/8.
a(2n) = A047211(n), a(2n-1) = A047218(n).
a(n) = A047207(n+1) - 1.
a(n+2) = n + 2 + A002265(n) for n>0.
a(n+3)-a(n+2) = A177704(n) for n>0.
a(1-n) = - A001068(n). (End)
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 - sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021

More terms from Reinhard Zumkeller, Oct 23 2006

A100019 a(n) = n^4 + n^3 + n^2.

Original entry on oeis.org

0, 3, 28, 117, 336, 775, 1548, 2793, 4672, 7371, 11100, 16093, 22608, 30927, 41356, 54225, 69888, 88723, 111132, 137541, 168400, 204183, 245388, 292537, 346176, 406875, 475228, 551853, 637392, 732511, 837900, 954273, 1082368, 1222947

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Nov 19 2004

a(n) are the numbers m such that: j^2 = j + m + sqrt(j*m) with corresponding numbers j given by A002061(n+1), and with sqrt(j*m) = A027444(n) = n* A002061(n+1). - Richard R. Forberg, Sep 03 2013.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000583, A002523, A091940, A071253, A059826, A027445, A053699.

[n^4+n^3+n^2: n in [0..50]]; // Vincenzo Librandi, Jun 09 2011

A100019:=n->n^4+n^3+n^2: seq(A100019(n), n=0..50); # Wesley Ivan Hurt, Apr 14 2017

f[n_]:=n^4+n^3+n^2;Array[f,50,0] (* Vladimir Joseph Stephan Orlovsky, Apr 12 2011 *)

a(n) = n^4 + n^3 + n^2 \\ Indranil Ghosh, Apr 15 2017

From Indranil Ghosh, Apr 15 2017: (Start)
G.f.: -x(3 + 13x + 7x^2 + x^3)/(x - 1)^5
E.g.f.: exp(x)*x*(3 + 11x + 7x^2 + x^3)
(End)

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

A071237 a(n) = n(n+1)(n^2+1)/2.

Original entry on oeis.org

0, 2, 15, 60, 170, 390, 777, 1400, 2340, 3690, 5555, 8052, 11310, 15470, 20685, 27120, 34952, 44370, 55575, 68780, 84210, 102102, 122705, 146280, 173100, 203450, 237627, 275940, 318710, 366270, 418965, 477152, 541200, 611490, 688415, 772380, 863802, 963110

Offset: 0

N. J. A. Sloane, Jun 12 2002

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A027445, A037270.

[n*(n+1)*(n^2+1)/2: n in [0..40] ]; // Vincenzo Librandi, May 23 2011

Table[(n^4 + n^3 + n^2 + n)/2, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)
LinearRecurrence[{5,-10,10,-5,1},{0,2,15,60,170},40] (* Harvey P. Dale, Feb 08 2025 *)

def A071237(n): return (n^2+1)*binomial(n+1,2)
[A071237(n) for n in range(51)] # G. C. Greubel, Aug 05 2024

From Arkadiusz Wesolowski, Apr 01 2012: (Start)
a(n) = A000217(n)*A002522(n).
a(0) = 0, a(1) = 2; for n >= 2, a(n) = ceiling(n^5/(2*n-2)) - 1.
G.f.: x*(2 + 5*x*(1 + x))/(1 - x)^5. (End)
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5) for n>4, a(0)=0, a(1)=2, a(2)=15, a(3)=60, a(4)=170. - Yosu Yurramendi, Sep 03 2013
E.g.f.: (1/2)*x*(4 + 11*x + 7*x^2 + x^3)*exp(x). - G. C. Greubel, Aug 05 2024

A124260 a(n) = Sum_{k=1..A124259(n)} n^k.

Original entry on oeis.org

4, 126, 12, 4, 780, 94036996914, 56, 8, 9, 1111111110, 132, 12, 30940, 8108730, 240, 16, 306, 18, 380, 20, 204204, 11154, 552, 24, 25, 702, 27, 28, 732540, 27930, 992, 32, 1222980, 62556901638174, 1260, 36, 1926220, 2141490, 1560, 40, 2896404

Offset: 1

Reinhard Zumkeller, Oct 23 2006

nonn

Amiram Eldar, Table of n, a(n) for n = 1..1589

Cf. A002378, A013929, A027444, A027445, A124259.

a[n_] := Module[{k = 1, s = n}, While[SquareFreeQ[s], k++; s += n^k]; s]; Array[a, 100] (* Amiram Eldar, Dec 26 2020  *)

a(n) = my(k=1, s); while (issquarefree(s=sum(i=1, k, n^i)), k++); s; \\ Michel Marcus, Dec 26 2020

a(A013929(n)) = A013929(n).

A253915 Primes p such that the polynomial k^4 + k^3 + k^2 + k + p yields primes for k = 0..8, but not for k = 9.

Original entry on oeis.org

43, 967, 11923, 213943, 2349313, 3316147, 30637567, 33421159, 39693817, 49978447, 105963769, 143405887, 148248949, 153756073, 156871549, 172981279, 187310803, 196726693, 203625283, 211977523, 220825453, 268375879, 350968543, 357834283, 414486697, 427990369

Offset: 1

K. D. Bajpai, Jan 18 2015

nonn

All the terms in this sequence are congruent to 1 (mod 3).

			a(1) = 43:
0^4 + 0^3 + 0^2 + 0 + 43 =   43;
1^4 + 1^3 + 1^2 + 1 + 43 =   47;
2^4 + 2^3 + 2^2 + 2 + 43 =   73;
3^4 + 3^3 + 3^2 + 3 + 43 =  163;
4^4 + 4^3 + 4^2 + 4 + 43 =  383;
5^4 + 5^3 + 5^2 + 5 + 43 =  823;
6^4 + 6^3 + 6^2 + 6 + 43 = 1597;
7^4 + 7^3 + 7^2 + 7 + 43 = 2843;
8^4 + 8^3 + 8^2 + 8 + 43 = 4723;
all nine are primes, and
9^4 + 9^3 + 9^2 + 9 + 43 = 7423 = 13*571 is composite.
The next prime for p=43 appears for k=13, namely 30983.

Jon E. Schoenfield, Table of n, a(n) for n = 1..155 (terms < 2*10^10)

Cf. A027445, A144051, A187057, A187058, A187060, A190800, A191456, A191457, A191458, A247949, A247966, A248206.

Select[Prime[Range[118*10^5]],AllTrue[#+{0,4,30,120,340,780,1554,2800,4680},PrimeQ]&&CompositeQ[#+7380]&] (* Harvey P. Dale, Sep 10 2021 *)

forprime(p=1, 1e10, if(isprime(p+4)&& isprime(p+30)&& isprime(p+120)&& isprime(p+340)&& isprime(p+780)&& isprime(p+1554)&& isprime(p+2800)&& isprime(p+4680) && !isprime(p+7380), print1(p,", ")))

Edited by Wolfdieter Lang, Feb 20 2015

Corrected and extended by Harvey P. Dale, Sep 10 2021

A100606 a(n) = n^4 + n^3 + n.

Original entry on oeis.org

0, 3, 26, 111, 324, 755, 1518, 2751, 4616, 7299, 11010, 15983, 22476, 30771, 41174, 54015, 69648, 88451, 110826, 137199, 168020, 203763, 244926, 292031, 345624, 406275, 474578, 551151, 636636, 731699, 837030, 953343, 1081376, 1221891, 1375674, 1543535, 1726308

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Nov 30 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A002523, A027445, A053699, A059826, A071253, A091940, A100019.

[n^4+n^3+n: n in [0..50]]; // Vincenzo Librandi, Jun 09 2011

Table[n^4+n^3+n,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,3,26,111,324},40] (* Harvey P. Dale, Apr 25 2015 *)

a(n)=n^4+n^3+n \\ Charles R Greathouse IV, Oct 21 2022

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=3, a(2)=26, a(3)=111, a(4)=324. - Harvey P. Dale, Apr 25 2015
From Elmo R. Oliveira, Aug 29 2025: (Start)
G.f.: x*(3 + 11*x + 11*x^2 - x^3)/(1-x)^5.
E.g.f.: x*(3 + 10*x + 7*x^2 + x^3)*exp(x). (End)

cp's OEIS Frontend

A016861 a(n) = 5*n + 1.

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A228275 A(n,k) = Sum_{i=1..k} n^i; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A047203 Numbers that are congruent to {0, 2, 3, 4} mod 5.

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A100019 a(n) = n^4 + n^3 + n^2.

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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.

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A071237 a(n) = n*(n+1)*(n^2+1)/2.

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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.

A071237 a(n) = n(n+1)(n^2+1)/2.