A027444 -id:A027444 - cp's OEIS frontend

A058895 a(n) = n^4 - n.

0, 0, 14, 78, 252, 620, 1290, 2394, 4088, 6552, 9990, 14630, 20724, 28548, 38402, 50610, 65520, 83504, 104958, 130302, 159980, 194460, 234234, 279818, 331752, 390600, 456950, 531414, 614628, 707252, 809970, 923490, 1048544, 1185888, 1336302, 1500590, 1679580

Offset: 0

Henry Bottomley, Jan 08 2001

a(n) is the number of ways to assign 4 different students to n different dorm rooms, each of which can hold at most 3 students. In other words, a(n) is the number of functions f:[4]->[n] with the size of the pre-image set of each element of the codomain at most 3. - Dennis P. Walsh, Mar 21 2013

a(n) are the values of m that yield integer solutions to this family of equations: x = sqrt(m + sqrt(x)), which may also be viewed as an infinitely recursive radical. The real solutions for x at each m = a(n) is n^2, except at n = 1 (m = 0) where x = 0 or 1 is a solution. - Richard R. Forberg, Oct 15 2014

Harry J. Smith, Table of n, a(n) for n = 0..2000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A002061, A002378, A024001, A027444, A027482, A068601, A033562, A185065, A339605.

[n^4-n: n in [0..40]]; // Vincenzo Librandi, Feb 23 2012

seq(n*(n^3-1),n=0..25) ; # R. J. Mathar, Dec 10 2015

Table[n^4 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = { n^4-n } \\ Harry J. Smith, Jun 23 2009

(2*x^2*(7+4*x+x^2)/(1-x)^5).series(x, 37).coefficients(x, sparse=False) # Stefano Spezia, Jul 09 2021

a(n) = n*(n-1)*(n^2+n+1) = A000583(n) - n = A002061(n+1) * A002378(n-1) = (n-1) * A027444(n) = -n * A024001(n).
a(n) = 2*A027482(n). - Zerinvary Lajos, Jan 28 2008
a(n) = floor(n^7/(n^3+1)). - Gary Detlefs, Feb 11 2010
a(n)^3 = (a(n)/n)^4 + (a(n)/n)^3. - Vincenzo Librandi, Feb 23 2012
a(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3, for n > 0. - Vincenzo Librandi, Mar 13 2012
G.f.: 2*x^2*(7 + 4*x + x^2)/(1 - x)^5. - Colin Barker, Apr 23 2012
a(n) = 14*C(n,2) + 36*C(n,3) + 24*C(n,4). - Dennis P. Walsh, Mar 21 2013
Sum_{n>=2} (-1)^n/a(n) = (Pi/3)*sech(Pi*sqrt(3)/2) + 4*log(2)/3 - 1 = 0.06147271494... . - Amiram Eldar, Jul 04 2020
Sum_{n>=2} 1/a(n) = A339605. - R. J. Mathar, Jan 08 2021
E.g.f.: exp(x)*x^2*(7 + 6*x + x^2). - Stefano Spezia, Jul 09 2021
a(n) = 12*A000332(n+2) + 2*A000537(n-1). - Yasser Arath Chavez Reyes, Apr 05 2024

A098547 a(n) = n^3 + n^2 + 1.

Original entry on oeis.org

1, 3, 13, 37, 81, 151, 253, 393, 577, 811, 1101, 1453, 1873, 2367, 2941, 3601, 4353, 5203, 6157, 7221, 8401, 9703, 11133, 12697, 14401, 16251, 18253, 20413, 22737, 25231, 27901, 30753, 33793, 37027, 40461, 44101, 47953, 52023, 56317, 60841, 65601, 70603, 75853

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Oct 26 2004

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

Cf. A000578, A001093, A011379, A027444, A033431, A033562, A034262, A053698, A061317, A066023, A071568.

Cf. A000217, A081423.

[(n^3+n^2+1): n in [1..60]]; // Vincenzo Librandi, Apr 06 2011

with(combinat): seq(fibonacci(3,n)+n^3, n=0..40); # Zerinvary Lajos, May 25 2008

Table[n^3+n^2+1,{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Mar 09 2011 *)

Vec(-(x^3-7*x^2+x-1)/(x-1)^4 + O(x^100)) \\ Colin Barker, Aug 29 2014

From Colin Barker, Aug 29 2014: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (1 - x + 7*x^2 - x^3)/(1-x)^4. (End)
a(n) = A081423(n) + A000217(n-1). - Bruce J. Nicholson, Jan 06 2019
E.g.f.: exp(x)*(1 + 2*x + 4*x^2 + x^3). - Elmo R. Oliveira, Apr 20 2025

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)

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

Original entry on oeis.org

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)

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)3 and containing (k+1)3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 3, 1, 7, 14, 1, 13, 81, 39, 1, 21, 304, 456, 84, 1, 31, 875, 3000, 1750, 155, 1, 43, 2106, 13875, 18500, 5265, 258, 1, 57, 4459, 50421, 128625, 84035, 13377, 399, 1, 73, 8576, 153664, 669536, 836920, 307328, 30016, 584, 1, 91, 15309, 409536, 2815344, 6001128, 4223016, 955584, 61236, 819, 1

Offset: 0

Susanne Wienand, Oct 10 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 L's increment the index of dir,
(d) consecutive R's 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 = 3.
The values in the triangle are proved by brute force for 0 <= n <= 11. The formulas are not yet proved in general. - Susanne Wienand
Let S(N,n,k) = C(n,k)^(N+1)*Sum_{j=0..N} Sum_{i=0..N} (-1)^(N-j+i)*C(N-i,j)*((n+1)/(k+1))^j. Then S(0,n,k) = A007318(n,k), S(1,n,k) = A103371(n,k), S(2,n,k) = T(n,k), S(3,n,k) = A197653(n,k), S(4,n,k) = A197654(n,k), S(5,n,k) = A197655(n,k). - Peter Luschny, Oct 21 2011
The number triangle can be calculated recursively by the number triangles A103371 and A007318. The first column of the triangle contains the central polygonal numbers A002061. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027444. Row sums are in A197657. - Susanne Wienand, Nov 24 2011
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - Susanne Wienand, Jun 24 2015

			For n = 4 and k = 2, T(3,4,2) = 456.
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(2,4,0) = 5
T(2,4,1) = 40
T(2,4,2) = 60
T(2,4,3) = 20
T(2,4,4) = 1
T(3,4,0) = T(1,4,0)^3 + T(1,4,0)*T(2,4,4-1-0) = 1^3 + 1*20 = 21
T(3,4,1) = T(1,4,1)^3 + T(1,4,1)*T(2,4,4-1-1) = 4^3 + 4*60 = 304
T(3,4,2) = T(1,4,2)^3 + T(1,4,2)*T(2,4,4-1-2) = 6^3 + 6*40 = 456
T(3,4,3) = T(1,4,3)^3 + T(1,4,3)*T(2,4,4-1-3) = 4^3 + 4*5  = 84
T(3,4,4) = 1.
Example for closed formula:
T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 L's.
  S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R
dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L
dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0
  S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L
dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0
Each value of dir occurs 15/3 = 5 times.

Susanne Wienand, Table of n, a(n) for n = 0..2015
Peter Luschny, Meanders and walks on the circle.
Susanne Wienand, Animation of a meander.
Susanne Wienand, Example of a meander.

Cf. A103371, A197653, A197654, A197655, A197657, A007318, A002061, A000012, A027444, A073254.

A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);
seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # Peter Luschny, Oct 14 2011

T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A194595(n,k) = {if(n == 1+2*k,3,(1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n))*binomial(n,k)^3} \\ Peter Luschny, Nov 24 2011

Recursive formula (conjectured):
T(n,k) = 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(2,n,k) = A103371,
T(1,n,k) = A007318 (Pascal's Triangle).
Closed formula (conjectured): T(n,k) = (C(n,k))^3 + C(n,k) * C(n,k+1) * C(n+1,k+1). - 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,2). - Peter Luschny, Oct 20 2011
T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^3, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 3. - Peter Luschny, Nov 24 2011

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

Original entry on oeis.org

2, 5, 16, 41, 86, 157, 260, 401, 586, 821, 1112, 1465, 1886, 2381, 2956, 3617, 4370, 5221, 6176, 7241, 8422, 9725, 11156, 12721, 14426, 16277, 18280, 20441, 22766, 25261, 27932, 30785, 33826, 37061, 40496, 44137, 47990, 52061, 56356, 60881, 65642, 70645

Offset: 1

Adeniji, Adenike, Apr 14 2011

The original definition was "Identity difference partial one - one transformation semigroup is a semigroup having the property that the difference between max im(alpha) and min im(alpha) is not greater than 1. This is denoted by S = IDI_n for each n." [Needs editing.]

For all n >= 3, a(n) expressed in base n has the three digits n-2, 2, and 1; for example, a(16) in hexadecimal is "E21". For all n >= 3, a(n+1) expressed in base n is "1112". For all n >= 7, a(n+2) expressed in base n is "1465". - Mathew Englander, Jan 07 2021

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

Cf. A027444, A053698, A056106 (first differences), A060354, A162607, A188377, A188716.

[n^3 - 2*n^2 + 2*n + 1: n in [1..30]]; // Vincenzo Librandi, May 01 2011

A188947[n_] := n^3 - 2*n^2 + 2*n + 1; Table[A188947[n], {n, 1, 42}] (* Robert P. P. McKone, Jan 31 2021 *)

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

a(n) = (n+1) + n*(n-1)^2 = n^3 - 2*n^2 + 2*n + 1 = 1 + A053698(n-1).
G.f.: ( -x*(-2 + 3*x - 8*x^2 + x^3) ) / ( (x-1)^4 ). - R. J. Mathar, Apr 14 2011
a(n) = A060354(n) + A162607(n+1). - Lechoslaw Ratajczak, Sep 24 2020
E.g.f.: exp(x)*(1 + x)*(1 + x^2) - 1. - Stefano Spezia, Apr 10 2022

Edited by N. J. A. Sloane, Apr 23 2011

A270109 a(n) = n^3 + (n+1)*(n+2).

Original entry on oeis.org

2, 7, 20, 47, 94, 167, 272, 415, 602, 839, 1132, 1487, 1910, 2407, 2984, 3647, 4402, 5255, 6212, 7279, 8462, 9767, 11200, 12767, 14474, 16327, 18332, 20495, 22822, 25319, 27992, 30847, 33890, 37127, 40564, 44207, 48062, 52135, 56432, 60959, 65722, 70727, 75980, 81487, 87254

Offset: 0

Bruno Berselli, Mar 11 2016, at the suggestion of Giuseppe Amoruso in BASE Cinque forum

For n>1, many consecutive terms of the sequence are generated by floor(sqrt(n^2 + 2)^3) + n^2 + 2.
It appears that this is a subsequence of A000037 (the nonsquares).
The primes in the sequence belong to A045326.
Inverse binomial transform is 2, 5, 8, 6, 0, 0, 0, ... (0 continued).

Bruno Berselli, Table of n, a(n) for n = 0..1000
MathsSmart, Number pattern and Puzzle - 7, 20, 47, 94, 167.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Subsequence of A001651, A047212.
Cf. A000037, A045326.
Cf. A027444: numbers of the form n^3+n*(n+1); A085490: numbers of the form n^3+(n-1)*n.
Cf. A008865: numbers of the form n+(n+1)*(n+2); A130883: numbers of the form n^2+(n+1)*(n+2).

Magma
```
[n^3+(n+1)*(n+2): n in [0..50]];
```

Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]

Maxima
```
makelist(n^3+(n+1)*(n+2), n, 0, 50);
```
PARI
```
vector(50, n, n--; n^3+(n+1)*(n+2))
```
Sage
```
[n^3+(n+1)*(n+2) for n in (0..50)]
```

O.g.f.: (2 - x + 4*x^2 + x^3)/(1 - x)^4.
E.g.f.: (2 + x)*(1 + x)^2*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
a(n+h) - a(n) + a(n-h) = n^3 + n^2 + (6*h^2+3)*n + (2*h^2+2) for any h. This identity becomes a(n) = n^3 + n^2 + 3*n + 2 if h=0.
a(h*a(n) + n) = (h*a(n))^3 + (3*n+1)*(h*a(n))^2 + (3*n^2+2*n+3)*(h*a(n)) + a(n) for any h, therefore a(h*a(n) + n) is always a multiple of a(n).
a(n) + a(-n) = 2*A059100(n) = A255843(n).
a(n) - a(-n) = 4*A229183(n).

A247237 Triangle read by rows: T(n,k) is the coefficient in the transformation Sum_{k=0..n} (k+1)x^k = Sum_{k=0..n} T(n,k)(x-k)^k.

Original entry on oeis.org

1, 3, 2, 3, 14, 3, 3, 50, 39, 4, 3, 130, 279, 84, 5, 3, 280, 1479, 984, 155, 6, 3, 532, 6519, 8544, 2675, 258, 7, 3, 924, 25335, 61464, 34035, 6138, 399, 8, 3, 1500, 89847, 388056, 356595, 106938, 12495, 584, 9, 3, 2310, 297207, 2225136, 3259635, 1524438, 284655, 23264, 819, 10

Offset: 0

Derek Orr, Nov 27 2014

Consider the transformation 1 + 2x + 3x^2 + 4x^3 + ... + (n+1)*x^n = T(n,0)*(x-0)^0 + T(n,1)*(x-1)^1 + T(n,2)*(x-2)^2 + ... + T(n,n)*(x-n)^n, for n >= 0.

			From _Wolfdieter Lang_, Jan 14 2015: (Start)
The triangle T(n,k) starts:
n\k 0    1      2       3       4       5      6     7   8  9 ...
0:  1
1:  3    2
2:  3   14      3
3:  3   50     39       4
4:  3  130    279      84       5
5:  3  280   1479     984     155       6
6:  3  532   6519    8544    2675     258      7
7:  3  924  25335   61464   34035    6138    399     8
8:  3 1500  89847  388056  356595  106938  12495   584   9
9:  3 2310 297207 2225136 3259635 1524438 284655 23264 819 10
...
-----------------------------------------------------------------
n = 3: 1 + 2*x + 3*x^2 + 4*x^3 = 3*(x-0)^0 +  50*(x-1)^1 + 39*(x-2)^2 + 4*(x-3)^3.
(End)

Cf. A247236, A153978, A027444, A253381.

T(n,k)=(k+1)-sum(i=k+1,n,(-i)^(i-k)*binomial(i,k)*T(n,i))
for(n=0,10,for(k=0,n,print1(T(n,k),", ")))

T(n,n) = n+1, n >= 0.
T(n,1) = n(n+1)(n+2)(3*n+1)/12 (A153978), for n >= 1.
T(n,n-1) = n^3 + n^2 + n (A027444), for n >= 1.
T(n,n-2) = (n-1)^2 (n^3-2)/2, for n >= 2.

Edited by Wolfdieter Lang, Jan 14 2015

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).

A164939 These are prime numbers p such that p^3 + p^2 + p + 2 is also prime.

Original entry on oeis.org

3, 5, 7, 13, 23, 29, 43, 79, 89, 113, 139, 163, 193, 197, 199, 229, 233, 277, 283, 317, 367, 379, 389, 503, 619, 727, 769, 797, 829, 839, 953, 967, 977, 997, 1063, 1229, 1297, 1307, 1399, 1409, 1483, 1607, 1619, 1637, 1697, 1759, 1777, 1877, 1979, 1987, 1999

Offset: 1

B. R. Becker (bbecker(AT)panda3.phys.unm.edu), Sep 01 2009

			3 is prime as is 3^3 + 3^2 + 3 + 2 = 41

Harvey P. Dale, Table of n, a(n) for n = 1..1000

A027444, A053698

Select[Prime[Range[500]],PrimeQ[ #^3 + #^2 + # + 2]&]

cp's OEIS Frontend

A058895 a(n) = n^4 - n.

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A098547 a(n) = n^3 + n^2 + 1.

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

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

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A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*3 and containing (k+1)*3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

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

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A270109 a(n) = n^3 + (n+1)*(n+2).

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

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)3 and containing (k+1)3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

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

A247237 Triangle read by rows: T(n,k) is the coefficient in the transformation Sum_{k=0..n} (k+1)x^k = Sum_{k=0..n} T(n,k)(x-k)^k.