Mateusz Szymański - cp's OEIS frontend

A232007 Maximal number of moves needed by a knight to reach every square from a fixed position on an n X n chessboard, or -1 if it is not possible to reach every square.

0, -1, -1, 5, 4, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 44, 44, 45, 46, 46

Offset: 1

Mateusz Szymański, Nov 16 2013

In other words, a(n) is the graph diameter of the n X n knight graph (or -1 if the graph is disconnected). - Eric W. Weisstein, Nov 20 2019

			For a classic 8 X 8 chessboard, a knight needs at most 6 moves to reach every square starting from a fixed position, so a(8) = 6.
For a 3 X 3 chessboard, it's impossible to reach the middle square starting from any other, so a(3) = -1.

Colin Barker, Table of n, a(n) for n = 1..1000
Marco Ripà, Metric spaces in chess and international chess pieces graph diameters, arXiv:2311.00016 [math.HO], 2023. See pp. 11, 19.
Paul B. Slater, Formulas for Generalized Two-Qubit Separability Probabilities, arXiv:1609.08561 [quant-ph], 2016.
Paul B. Slater, Hypergeometric/Difference-Equation-Based Separability Probability Formulas and Their Asymptotics for Generalized Two-Qubit States Endowed with Random Induced Measure, arXiv preprint arXiv:1504.04555 [quant-ph], 2015.
Eric Weisstein's World of Mathematics, Graph Diameter
Eric Weisstein's World of Mathematics, Knight Graph
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A004523.

Replace[Table[GraphDiameter[KnightTourGraph[n, n]], {n, 20}], Infinity -> -1] (* Eric W. Weisstein, Nov 20 2019 *)
Join[{0, -1, -1, 5}, Table[Ceiling[2 n/3], {n, 5, 20}]] (* Eric W. Weisstein, Nov 20 2019 *)
Join[{0, -1, -1, 5}, LinearRecurrence[{1, 0, 1, -1}, {4, 4, 5, 6}, 20]] (* Eric W. Weisstein, Nov 20 2019 *)
CoefficientList[Series[-1 - x + 5 x^2 + x^3 (4 + x^2 - 3 x^3)/((-1 + x)^2 (1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Nov 20 2019 *)

concat(0, Vec(-x^2*(1-6*x^2+5*x^5-2*x^6)/((1-x)^2*(1+x+x^2)) + O(x^100))) \\ Colin Barker, Apr 26 2016

a(n) = ceiling(2n/3) for n > 4; see A004523. - R. J. Mathar, Nov 24 2013
From Colin Barker, Apr 26 2016: (Start)
a(n) = a(n-1)+a(n-3)-a(n-4) for n>8.
G.f.: -x^2*(1-6*x^2+5*x^5-2*x^6) / ((1-x)^2*(1+x+x^2)). (End)

More terms from Vaclav Kotesovec, Oct 21 2014

A188573 a(n) = coefficient of the sqrt(6) term in (1 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 0, 2, 6, 32, 120, 528, 2128, 8960, 36864, 153472, 635008, 2635776, 10922496, 45300736, 187800576, 778731520, 3228696576, 13387309056, 55506722816, 230146834432, 954246856704, 3956565671936, 16404954546176, 68019305840640, 282025965649920

Offset: 0

Mateusz Szymański, Dec 28 2012

nonn

From Clark Kimberling, Oct 23 2024: (Start)
Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 6,5,14,4,4,8 and these periods:
p = 2: (1, 2, 2, 1, 3, 3)
p = 3: (1, 4, 3, 8, 8)
p = 5: (1, 5, 4, 1, 1, 6, 6, 6, 6, 6, 6, 2, 4, 6)
p = 7: (1, 16, 1, 18)
p = 11: (1, 30, 1, 32)
p = 13: (1, 10, 3, 17, 10, 1, 28, 14)
See A377109 for a guide to related sequences. (End)

			a(3) = 6, because (1+sqrt(2)+sqrt(3))^3 = 16 + 14 sqrt(2) + 12 sqrt(3) + 6 sqrt(6).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188570, A188571, A188572, A377109.

a[n_] := Sum[Sum[2^(Floor[n/2] - j - 1 - k) 3^j Multinomial[2 k + n - 2 Floor[n/2], 2 j + 1, 2 Floor[n/2] - 2 k - 1 - 2 j], {j, 0, Floor[n/2] - k - 1}], {k, 0, Floor[n/2] - 1}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[6]]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)

From G. C. Greubel, Apr 10 2018: (Start)
Empirical: a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4).
Empirical: G.f.: 2*x^2*(1-x)/(1 - 4*x - 4*x^2 + 16*x^3 - 8*x^4). (End)
The conjectures by Greubel are true. See link. - Sela Fried, Jan 01 2025

Keyword tabl removed by Michel Marcus, Apr 11 2018

Edited by Clark Kimberling, Oct 23 2024

A188571 a(n) = coefficient of sqrt(2) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

0, 1, 2, 14, 48, 224, 880, 3760, 15360, 64192, 265088, 1101440, 4561920, 18925568, 78447616, 325313536, 1348730880, 5592420352, 23187169280, 96141172736, 398624489472, 1652807303168, 6852965761024, 28414229807104, 117812861337600, 488483370827776

Offset: 0

Mateusz Szymański, Dec 28 2012

From Clark Kimberling, Oct 23 2024: (Start)
Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 4,8,14,3,3,5 and these periods:
p = 2: (4)
p = 3: (6, 1, 1, 3, 1, 4, 2, 6)
p = 5: (6, 6, 6, 2, 4, 6, 1, 5, 4, 1, 1, 6, 6, 6)
p = 7: (9, 18, 9)
p = 11: (43, 10, 11)
p = 13: (21, 7, 28, 7, 21)
See A377109 for a guide to related sequences. (End)

			a(3) = 14 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188570, A377109.

a[n_] := Sum[Sum[2^(Floor[(n - 1)/2] - k - j) 3^j Multinomial[2 Floor[(n - 1)/2] + 1 - 2 j - 2 k, 2 j, 2 k + 1 - n + 2 Floor[n/2]], {j, 0, Floor[(n - 1)/2] - k + 1}], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[2]] /. Sqrt[3] -> 0; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)

Conjectures from Colin Barker, Jan 08 2013: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4).
G.f.: -x*(2*x^2-2*x+1) / (8*x^4-16*x^3+4*x^2+4*x-1). (End)
The conjectures by Barker are true. See link. - Sela Fried, Jan 01 2025

Edited by Clark Kimberling, Oct 20 2024

A188570 a(n) = coefficient of the term that is independent of sqrt(2) and sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Original entry on oeis.org

1, 1, 6, 16, 80, 296, 1296, 5216, 21952, 90304, 375936, 1555456, 6456320, 26754560, 110963712, 460015616, 1907494912, 7908659200, 32792076288, 135963148288, 563742310400, 2337417887744, 9691567030272, 40183767891968, 166612591968256, 690819710058496

Offset: 0

Mateusz Szymański, Dec 28 2012

nonn

From Clark Kimberling, Oct 23 2024: (Start)
Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 1,5,5,3,3,5 and these periods:
p = 2: (4)
p = 3: (8, 1, 4, 3, 8)
p = 5: (9, 10, 1, 20, 20)
p = 7: (9, 9, 18)
p = 11: (10, 11, 43)
p = 13: (7, 21, 21, 7, 28)
See A377109 for a guide to related sequences. (End)

			a(3) = 16 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188571, A188572, A188573, A377109.

a[n_] := Sum[Sum[2^(Floor[n/2] - k - j) 3^j Multinomial[2 k + n - 2 Floor[n/2], 2 j, 2 Floor[n/2] - 2 k - 2 j], {j, 0, Floor[n/2] - k}], {k, 0, Floor[n/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Expand[(1 + Sqrt[2] + Sqrt[3])^n] /. Sqrt[] -> 0; Table[a[n], {n, 0, 25}] (* _Jean-François Alcover, Jan 08 2013 *)
LinearRecurrence[{4,4,-16,8},{1,1,6,16},30] (* Harvey P. Dale, Jan 25 2019 *)

Recurrence: a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4). - Vaclav Kotesovec, Aug 13 2013

a(n) ~ (1+sqrt(3)+sqrt(2))^n/4. - Vaclav Kotesovec, Aug 13 2013

Edited by Clark Kimberling, Oct 20 2024

A188572 a(n) = coefficient of sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n sequence.

Original entry on oeis.org

0, 1, 2, 12, 40, 184, 720, 3072, 12544, 52416, 216448, 899328, 3724800, 15452672, 64052224, 265617408, 1101234176, 4566192128, 18932244480, 78498938880, 325475532800, 1349511512064, 5595423113216, 23200121487360, 96193798471680, 398845002121216

Offset: 0

Mateusz Szymański, Dec 28 2012

nonn

From Clark Kimberling, Oct 23 2024: (Start)
Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 6,8,5,4,4,8 and these periods:
p = 2: (3, 3, 1, 2, 2, 1)
p = 3: (4, 2, 6, 6, 1, 1, 3, 1)
p = 5: (20, 20, 9, 10, 1)
p = 7: (18, 1, 16, 1)
p = 11: (32, 1, 30, 1)
p = 13: (28, 14, 1, 10, 3, 17, 10, 1)
See A377109 for a guide to related sequences. (End)
Cf. A377109.

			a(3) = 12 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Cf. A188570, A188571, A188573, A377109.

a[n_] := Sum[Sum[3^(Floor[(n - 1)/2] - k - j) 2^j Multinomial[2 Floor[(n - 1)/2] + 1 - 2 j - 2 k, 2 j, 2 k + 1 - n + 2 Floor[n/2]], {j, 0, Floor[(n - 1)/2] - k + 1}], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[3]] /. Sqrt[2] -> 0; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)

Conjectures from R. J. Mathar, Jan 09 2013: (Start)
a(n) = +4*a(n-1) +4*a(n-2) -16*a(n-3) +8*a(n-4).
G.f.: x*(-1+2*x)/( -1+4*x+4*x^2-16*x^3+8*x^4 ). (End)
The conjectures by Mathar are true. See link. - Sela Fried, Jan 01 2025

Edited by Clark Kimberling, Oct 20 2024

A182262 Least prime p that 6^n - p is prime.

Original entry on oeis.org

3, 5, 5, 5, 17, 7, 17, 7, 7, 7, 59, 19, 17, 13, 7, 19, 137, 13, 19, 7, 23, 97, 19, 89, 17, 223, 29, 109, 5, 19, 5, 59, 197, 5, 17, 307, 59, 83, 109, 157, 19, 23, 43, 109, 103, 7, 23, 19, 7, 269, 43, 13, 5, 67, 89, 83, 479, 53, 53, 383, 7, 83, 113, 37, 5, 23

Offset: 1

Mateusz Szymański, Apr 21 2012

nonn

			For n=3 p=5 is the least prime that 6^3-p is prime (211).

Robert Israel, Table of n, a(n) for n = 1..2035

Cf. A013607, A059614 (n such that a(n)=5).

f:= proc(n) local t,p;
  t:= 6^n;
  p:= 2;
  do
    p:= nextprime(p);
  until isprime(t-p);
  p
end proc:
map(f, [$1..100]); # Robert Israel, Nov 05 2020

f[n_] := Block[{p = 2}, While[! PrimeQ[6^n - p], p = NextPrime[p]];
  p]; Array[f, 60]

a(n) = my(p = 2); while(!isprime(6^n-p), p = nextprime(p+1)); p; \\ Michel Marcus, Mar 23 2016

cp's OEIS Frontend

User: Mateusz Szymański

A232007 Maximal number of moves needed by a knight to reach every square from a fixed position on an n X n chessboard, or -1 if it is not possible to reach every square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A188573 a(n) = coefficient of the sqrt(6) term in (1 + sqrt(2) + sqrt(3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A188571 a(n) = coefficient of sqrt(2) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A188570 a(n) = coefficient of the term that is independent of sqrt(2) and sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A188572 a(n) = coefficient of sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n sequence.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A182262 Least prime p that 6^n - p is prime.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI