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

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

Author

Carlos Alves, Jan 21 2003

Keywords

• base
• nonn
• easy

Comments

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.

Examples

			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.
		

Links

• 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).

Crossrefs

Cf. A061600.

Programs

GAP

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

Magma

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

Maple

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

Mathematica

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 *)

PARI

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

Sage

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

Formula

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)

Extensions

Edited by T. D. Noe, Jun 24 2009