A061600 -id:A061600 - cp's OEIS frontend

A126420 a(n) = n^3 - n - 1.

-1, 5, 23, 59, 119, 209, 335, 503, 719, 989, 1319, 1715, 2183, 2729, 3359, 4079, 4895, 5813, 6839, 7979, 9239, 10625, 12143, 13799, 15599, 17549, 19655, 21923, 24359, 26969, 29759, 32735, 35903, 39269, 42839, 46619, 50615, 54833, 59279, 63959, 68879, 74045, 79463

Offset: 1

Artur Jasinski, Dec 26 2006

Given three consecutive numbers x=n-2, y=n-1 and z=n, the sum over all products is x*y*z + x*y + x*z + y*z + x + y + z = n^3 - n - 1 = a(n). - J. M. Bergot, Aug 25 2011

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

Cf. A002327, A116581.

[n^3-n-1: n in [1..50]]; // Vincenzo Librandi, Aug 29 2011

a = {}; Do[AppendTo[a, x^3 - x - 1], {x, 1, 100}]; a

For n > 1, a(n) = floor(n^6/(n^3+n+1)). - Gary Detlefs, Feb 10 2010
G.f.: x*(-1 + 9*x - 3*x^2 + x^3) / (x-1)^4. - R. J. Mathar, Aug 28 2011
a(-n) = -A061600(n). - Bruno Berselli, Aug 29 2011
E.g.f.: (-1 + 6*x + 6*x^2 + x^3)*exp(x) = -E(0) where E(k) = 1 - 6*x/(1 - x/(1 + x - x/(6 + x - 6/(1 - x^2*(k+1)/E(k+1) )))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 09 2013

A268457 T(n,k)=Number of length-n 0..k arrays with no adjacent pair x,x+1 followed at any distance by x+1,x.

Original entry on oeis.org

2, 3, 4, 4, 9, 7, 5, 16, 25, 11, 6, 25, 61, 67, 16, 7, 36, 121, 229, 176, 22, 8, 49, 211, 581, 852, 456, 29, 9, 64, 337, 1231, 2776, 3146, 1169, 37, 10, 81, 505, 2311, 7160, 13204, 11536, 2971, 46, 11, 100, 721, 3977, 15816, 41526, 62535, 42032, 7496, 56, 12, 121

Offset: 1

R. H. Hardin, Feb 04 2016

Table starts
..2.....3......4.......5........6.........7.........8..........9.........10
..4.....9.....16......25.......36........49........64.........81........100
..7....25.....61.....121......211.......337.......505........721........991
.11....67....229.....581.....1231......2311......3977.......6409.......9811
.16...176....852....2776.....7160.....15816.....31276......56912......97056
.22...456...3146...13204....41526....108032....245626.....504876.....959414
.29..1169..11536...62535...240170....736525...1926444....4474451....9476950
.37..2971..42032..294967..1385338...5012171..15089356...39616567...93543782
.46..7496.152254.1385969..7970326..34047931.118040270..350431909..922677334
.56.18796.548568.6488635.45742764.230889543.922247248.3096903363.9094484100

			Some solutions for n=6 k=4
..3....2....3....2....2....0....4....2....2....0....1....0....0....3....4....3
..0....2....1....2....4....0....1....1....2....4....4....2....3....2....3....3
..0....4....1....1....4....2....3....0....1....4....2....1....1....2....0....1
..3....0....4....2....3....2....1....0....2....4....4....4....3....1....0....0
..4....1....2....3....2....2....0....4....0....4....4....4....1....3....1....3
..0....1....3....4....3....1....4....0....2....0....1....1....1....2....4....2

R. H. Hardin, Table of n, a(n) for n = 1..421

Column 1 is A000124.
Row 1 is A000027(n+1).
Row 2 is A000290(n+1).
Row 3 is A061600(n+1).

Empirical for column k:
k=1: a(n) = (1/2)*n^2 + (1/2)*n + 1
k=2: a(n) = 6*a(n-1) -13*a(n-2) +14*a(n-3) -10*a(n-4) +4*a(n-5) -a(n-6)
k=3: [order 15]
k=4: [order 28]
k=5: [order 51]
k=6: [order 89]
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = n^2 + 2*n + 1
n=3: a(n) = n^3 + 3*n^2 + 2*n + 1
n=4: a(n) = n^4 + 4*n^3 + 4*n^2 + n + 1
n=5: a(n) = n^5 + 5*n^4 + 7*n^3 + n^2 + 2*n
n=6: a(n) = n^6 + 6*n^5 + 11*n^4 + 2*n^3 + 6*n - 4
n=7: a(n) = n^7 + 7*n^6 + 16*n^5 + 5*n^4 - 7*n^3 + 18*n^2 - 5*n - 5 for n>1

A269064 At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.

Original entry on oeis.org

0, 1, 4, 10, 26, 48, 87, 135, 208, 293, 410, 542, 714, 904, 1141, 1399, 1712, 2049, 2448, 2874, 3370, 3896, 4499, 5135, 5856, 6613, 7462, 8350, 9338, 10368, 11505, 12687, 13984, 15329, 16796, 18314, 19962, 21664, 23503, 25399, 27440, 29541, 31794, 34110, 36586, 39128, 41837, 44615, 47568

Offset: 0

Luce ETIENNE, Feb 18 2016

At stage n, we count (6*n^2-6*n+5-3*(2*n-1)*(-1)^n)/8 unit up-pointing triangles and 3*(2*n^2-2*n+1+(2*n-1)*(-1)^n)/8 unit down-pointing triangles.
At stage n, the total number of unit triangles is (3*n^2-3*n+2)/2 = A005448(n). It is the same total as for A064412. Note also that A064412 gives number of triangles in a geometrical structure according to expansion side-side (mode S-S).
The edges of several unit triangles can form larger size triangles, and these are also up- or down-pointing. The number of all such larger is given by :(14*n^3-9*n^2+11*n+18-(9*n^2+3*n+14)*(-1)^n-4*((-1)^((2*n+1-(-1)^n)/4)))/64 up-pointing triangles and (14*n^3-15*n^2+35*n-6+(9*n^2+21*n+2)*(-1)^n+4*((-1)^((2*n-1+(-1)^n)/4)))/64 down-pointing triangles.
As for A265282 we observe that starting with n = 4 we can see and count hexagonal and dodecagonal forms for example in a reticular system (incomplete with hexagonal holes) by opposition to a compact shape A064412.

			a(0)= 0, a(1) = 1, a(2) = 4, a(3) = 7+3 = 10, a(4) = 19 + 6 + 1 = 26, a(5) = 31 + 12 + 4 + 1 = 48.

Colin Barker, Table of n, a(n) for n = 0..1000
Luce ETIENNE, Illustration of initial terms
Kival Ngaokrajang, Illustration of triangles expansion
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Cf. A003215, A005448, A005993, A033428, A064412, A061600, A006003, A039623, A102214, A265282.

[(14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n) div 4)-(-1)^((6*n-1+(-1)^n) div 4)))/32: n in [0..50]]; // Vincenzo Librandi, Feb 19 2016

Table[(14 n^3 - 12 n^2 + 23 n + 6 + 3 (3 n - 2) (-1)^n + 2 ((-1)^((2*n - 1 + (-1)^n) / 4) - (-1)^((6 n - 1 + (-1)^n) / 4))) / 32, {n, 0, 45}] (* Vincenzo Librandi, Feb 19 2016 *)

concat(0, Vec(x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6)/((1-x)^4*(1+x)^2*(1+x^2)) + O(x^50))) \\ Colin Barker, Feb 24 2016

a(n) = (7*n^3-3*n^2+4*n)/2 for n even.
a(n) = (28*n^3+30*n^2+16*n+7+(-1)^n)/8 for n odd.
a(n) = (14*n^3-12*n^2+23*n+6+3*(3*n-2)*(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)-(-1)^((6*n-1+(-1)^n)/4)))/32.
G.f.: x*(1+2*x+2*x^2+8*x^3+2*x^4+5*x^5+x^6) / ((1-x)^4*(1+x)^2*(1+x^2)). - Colin Barker, Feb 24 2016

A100698 Primes of the form k^3 - k + 1.

Original entry on oeis.org

7, 61, 211, 337, 991, 1321, 2731, 3361, 6841, 9241, 10627, 15601, 17551, 29761, 42841, 59281, 68881, 74047, 91081, 124951, 140557, 157411, 185137, 238267, 421801, 456457, 592621, 614041, 658417, 728911, 778597, 857281, 970201, 1030201, 1061107, 1190911, 1367521

Offset: 1

Giovanni Teofilatto, Dec 09 2004

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Marie Euler and Christophe Petit, Expanding the use of quasi-subfield polynomials, arXiv:1909.11326 [cs.CR], 2019.

Cf. A061600, A236477.

[ a: n in [0..400] | IsPrime(a) where a is n^3 - n +1]; // Vincenzo Librandi, Nov 17 2010

Select[Table[n^3-n+1,{n,0, 1500}],PrimeQ] (* Vincenzo Librandi, Jul 18 2012 *)

a(n) = A061600(A236477(n)). - Elmo R. Oliveira, Apr 19 2025

Inaccurate comment removed by D. S. McNeil, Nov 17 2010

A193871 Square array T(n,k) = k^n - k + 1 read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 25, 13, 1, 1, 31, 79, 61, 21, 1, 1, 63, 241, 253, 121, 31, 1, 1, 127, 727, 1021, 621, 211, 43, 1, 1, 255, 2185, 4093, 3121, 1291, 337, 57, 1, 1, 511, 6559, 16381, 15621, 7771, 2395, 505, 73, 1, 1, 1023, 19681, 65533, 78121, 46651, 16801, 4089, 721, 91, 1

Offset: 1

Omar E. Pol, Aug 21 2011

The columns give 1^n-0, 2^n-1, 3^n-2, 4^n-3, 5^n-4, etc.

The main diagonal gives A006091, which is a sequence related to the famous "coconuts" problem.

			Array begins:
  1,   1,    1,     1,     1,    1,    1,   1,   1,   1
  1,   3,    7,    13,    21,   31,   43,  57,  73
  1,   7,   25,    61,   121,  211,  337, 505
  1,  15,   79,   253,   621, 1291, 2395
  1,  31,  241,  1021,  3121, 7771
  1,  63,  727,  4093, 15621
  1, 127, 2185, 16381
  1, 255, 6559
  1, 511
  1

M. B. Richardson, A Needlessly Complicated and Unhelpful Solution to Ben Ames Williams' Coconuts Problem, The Winnower, 3 (2016), e147175.52128. doi: 10.15200/winn.147175.52128

Row 1: A000012. Rows 2,3: A002061, A061600 but both without repetitions.
Columns 1..10: A000012, positives A000225, A058481, A141725, A164785, A164784, A164783, A164786, A177095, A170955.
Cf. A276135.

Table[k^# - k + 1 &[n - k + 1], {n, 11}, {k, n}] // Flatten (* Michael De Vlieger, Nov 16 2016 *)

A079505 The last number for which a determinant of base-n numbers is nonzero.

Original entry on oeis.org

14, 25, 61, 121, 211, 337, 505, 721, 991, 1321, 1717, 2185, 2731, 3361, 4081, 4897, 5815, 6841, 7981, 9241, 10627, 12145, 13801, 15601, 17551, 19657, 21925, 24361, 26971, 29761, 32737, 35905, 39271, 42841, 46621, 50617, 54835, 59281, 63961, 68881

Offset: 2

Carlos Alves, Jan 21 2003

Suppose the number k written in base b requires N digits. We build A_k, a square N X N matrix with the digits of k, k-1,...,k-N+1 in base b. The number Det[A_k] is 0 for k greater than b^3-b+1 (except if b=2).

|Det[A_k]| is at most (b-1)^2. The last nonzero value is 1-b, which occurs for k = b^3-b+1 (cf. A061600) except for b=2, though I did not prove it.

			a(3)=25 because the determinant sequence in base 3 is 1, 2, 2, -1, -1, 4, -2, -2, -2, 2, 0, 1, -1, 0, 1, -1, 0, -4, 4, 0, 2, -2, 0, 2, -2, 0, 0, 0, 0, 0, 0, 0, .... and Det[A_k]=0 for k > 25.

Muniru A Asiru, Table of n, a(n) for n = 2..5000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A061600.

Concatenation([14], List([3..50], n -> n^3-n+1)); # G. C. Greubel, Jan 18 2019

[14] cat [n^3-n+1: n in [3..50]]; // Vincenzo Librandi, Dec 01 2018

seq(coeff(series(x^2*(7*x^4-29*x^3+45*x^2-31*x+14)/(x-1)^4,x,n+1), x, n), n = 2 .. 35); # Muniru A Asiru, Nov 30 2018

Table[ls = {}; Do[nt = Length[IntegerDigits[k, b]]; Ak = Table[IntegerDigits[k - i, b, nt], {i, 0, nt - 1}]; AppendTo[ls, Det[Ak]], {k, 1, b^4}]; Position[ls, _?(#!=0&)][[ -1, 1]], {b, 2, 10}]
CoefficientList[Series[(7x^4 -29x^3 +45x^2 -31x +14)/(x-1)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Dec 01 2018 *)
LinearRecurrence[{4,-6,4,-1},{14,25,61,121,211},40] (* Harvey P. Dale, Feb 12 2023 *)

vector(50, n, n++; if(n==2, 14, n^3-n+1)) \\ G. C. Greubel, Jan 18 2019

[14] + [n^3-n+1 for n in (3..50)] # G. C. Greubel, Jan 18 2019

a(n) = n^3 - n + 1 (except for n=2, a(2)=14).
From Chai Wah Wu, Nov 30 2018: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 6.
G.f.: x^2*(7*x^4 - 29*x^3 + 45*x^2 - 31*x + 14)/(x - 1)^4. (End)

Edited by T. D. Noe, Jun 24 2009

cp's OEIS Frontend

A126420 a(n) = n^3 - n - 1.

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A268457 T(n,k)=Number of length-n 0..k arrays with no adjacent pair x,x+1 followed at any distance by x+1,x.

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A269064 At stage 1, start with a unit equilateral triangle. At each successive stage add 3*(n-1) new triangles around outside with vertex-to-vertex contacts. Sequence gives number of triangles at n-th stage.

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A100698 Primes of the form k^3 - k + 1.

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A193871 Square array T(n,k) = k^n - k + 1 read by antidiagonals.

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A079505 The last number for which a determinant of base-n numbers is nonzero.

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