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A023000
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a(n) = (7^n - 1)/6.
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71
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0, 1, 8, 57, 400, 2801, 19608, 137257, 960800, 6725601, 47079208, 329554457, 2306881200, 16148168401, 113037178808, 791260251657, 5538821761600, 38771752331201, 271402266318408, 1899815864228857, 13298711049602000
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OFFSET
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0,3
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COMMENTS
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Apart form a(0), numbers of the form 11...11 (i.e., repunits) in base 7.
7^(floor(7^n/6)) is the highest power of 7 dividing (7^n)!. - Benoit Cloitre, Feb 04 2002
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=7, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). [Milan Janjic, Feb 21 2010]
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=8, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^n*charpoly(A,1). [Milan Janjic, Feb 21 2010]
This is the sequence A(0,1;6,7;2) = A(0,1;8,-7;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. [Wolfdieter Lang, Oct 18 2010]
From Wolfdieter Lang, May 02 2012 (Start)
6*a(n) =: z(n) gives the approximation up to 7^n for one of the three 7-adic integers (-1)^(1/3), i.e. z(n)^3 + 1 == 0 (mod 7^n), n>=0, and z(n) == 6 (mod 7) == -1 (mod 7), n>=1. The companion sequences are x(n)=A210852(n) and y(n)=A212153(n). This leads to a(n) == 1 (mod 7) for n>=1 (this is also clear from some of the formulas given below). Also 216*a(n)^3 + 1 == 0 (mod 7^n), n>=0, as well as 3*216*a(n)^2 + A212156(n) == 0 (mod 7^n), n>=0. a(n) = 6^(7^(n-1)-1) (mod 7^n), n>=1. A recurrence is a(n) = a(n-1) + 7^(n-1), with a(0)=0, for n>=1.
Also a(n) = (1/6)*(6*a(n-1))^7 (mod 7^n) with a(1)=1 for n>=1. Finally, 6^3*a(n-1)*a(n)^2 + 1 == 0 (mod 7^(n-1)), n>=1.
(End)
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Roger B. Eggleton, Maximal Midpoint-Free Subsets of Integers, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Eric Weisstein's World of Mathematics, Repunit
Index entries for linear recurrences with constant coefficients, signature (8,-7).
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FORMULA
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a(n) = ((4+sqrt(9))^n-(4-sqrt(9))^n)/6. [Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009]
a(n) = 8*a(n-1)-7*a(n-2). G.f.: x/((1-x)*(1-7*x)). [R. J. Mathar, Jun 21 2009]
From Wolfdieter Lang, Oct 18 2010: (Start)
a(n) = 6*a(n-1) + 7*a(n-2) + 2, a(0)=0, a(1)=1.
a(n) = 7*a(n-1) + a(n-2) - 7*a(n-3) = 9*a(n-1) - 15*a(n-2) + 7*a(n-3), a(0)=0, a(1)=1, a(2)= 8. Observation by G. Detlefs. See the W. Lang comment and link. (End)
a(n) = 7*a(n-1)+1 (with a(0)=0). [Vincenzo Librandi, Nov 19 2010]
a(n) = a(n-1) + 7^(n-1), with a(0)=0, n>=1. - See a Wolfdieter Lang comment above, May 02 2012
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MATHEMATICA
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LinearRecurrence[{8, -7}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
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PROG
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(Sage)
def a(n): return (7**n-1)//6
[a(n) for n in range(66)] # show terms
# Joerg Arndt, May 28 2012
(PARI) a(n)=(7^n-1)/6; /* Joerg Arndt, May 28 2012 */
(Maxima) A023000(n):=floor((7^n-1)/6)$ makelist(A023000(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(MAGMA) [n le 2 select n-1 else 8*Self(n-1) - 7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012
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CROSSREFS
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Sequence in context: A295711 A164031 A297369 * A331792 A097114 A022038
Adjacent sequences: A022997 A022998 A022999 * A023001 A023002 A023003
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KEYWORD
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easy,nonn
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AUTHOR
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David W. Wilson
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STATUS
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approved
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