A003463 a(n) = (5^n - 1)/4. (Formerly M4209)

0, 1, 6, 31, 156, 781, 3906, 19531, 97656, 488281, 2441406, 12207031, 61035156, 305175781, 1525878906, 7629394531, 38146972656, 190734863281, 953674316406, 4768371582031, 23841857910156, 119209289550781, 596046447753906, 2980232238769531

Offset

0,3

Comments

5^a(n) is the highest power of 5 dividing (5^n)!. - Benoit Cloitre, Feb 04 2002

n such that A002294(n) is not divisible by 5. - Benoit Cloitre, Jan 14 2003

Without leading zero, i.e., sequence {a(n+1) = (5*5^n-1)/4}, this is the binomial transform of A003947. - Paul Barry, May 19 2003 [Edited by M. F. Hasler, Oct 31 2014]

Numbers n such that a(n) is prime are listed in A004061(n) = {3, 7, 11, 13, 47, 127, 149, 181, 619, 929, ...}. Corresponding primes a(n) are listed in A086122(n) = {31, 19531, 12207031, 305175781, 177635683940025046467781066894531, ...}. 3^(m+1) divides a(2*3^m*k). 31 divides a(3k). 13 divides a(4k). 11 divides a(5k). 71 divides a(5k). 7 divides a(6k). 19531 divides a(7k). 313 divides a(8k). 19 divides a(9k). 829 divides a(9k). 71 divides a(10k). 521 divides a(10k). 17 divides a(16k). p divides a(p-1) for all prime p except p = {2,5}. p^(m+1) divides a(p^m*(p-1)) for all prime p except p = {2,5}. p divides a((p-1)/2) for prime p = {11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, ...} = A045468, Primes congruent to {1, 4} mod 5. p divides a((p-1)/3) for prime p = {13, 67, 127, 163, 181, 199, 211, 241, 313, 337, 367, 379, 409, 457, ...}. p divides a((p-1)/4) for prime p = {101, 109, 149, 181, 269, 389, 401, 409, 449, 461, 521, 541, ...} = A107219, Primes of the form x^2+100y^2. p divides a((p-1)/5) for prime p = {31, 191, 251, 271, 601, 641, 761, 1091, 1861, ...}. p divides a((p-1)/6) for prime p = {181, 199, 211, 241, 379, 409, 631, 691, 739, 769, 1039, ...}. - Alexander Adamchuk, Jan 23 2007

Starting with 1 = convolution square of A026375: (1, 3, 11, 45, 195, 873, ...). - Gary W. Adamson, May 17 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 27 2010

This is the sequence A(0,1;4,5;2) = A(0,1;6,-5;0) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

It is the Lucas sequence U(6,5). - Felix P. Muga II, Mar 21 2014

a(2*n+1) is the sum of the numerators and denominators of the reduced fractions 0 < b/5^n < 1 plus 1, with b < 5^n. - J. M. Bergot, Jul 24 2015

The sequence multiplied by 10 (0, 10, 60, 310, 1560, ...) is the maximum number of coins which can be decided by n weighings on 2 balances in the counterfeit coin problem with undecided under/overweight. [Halbeisen and Hungerbuhler, Disc. Math. 147 (1995) 139 Theorem 1]. - R. J. Mathar, Sep 10 2015

Order of the rank-n projective geometry PG(n-1,5) over the finite field GF(5). - Anthony Hernandez, Oct 05 2016

Number of zeros in the substitution system {0 -> 11100, 1 -> 11110} at step n from initial string "1" (1 -> 11110 -> 1111011110111101111011100 -> ...). - Ilya Gutkovskiy, Apr 10 2017

a(n) is the numerator of Sum_{k=1..n} 1/5^k, which approaches a limit of 1/4. The denominators are 5^n. In general, Sum_{k=1..n} 1/x^k approaches a limit of 1/(x-1). It is of interest to note that as x increases, so does the rate of convergence. See Crossrefs for numerators for other values of x which have the general form (x^n-1)/(x-1). - Gary Detlefs, Aug 31 2021

References

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Joseph E. Bonin and Joseph P. S. Kung The Number of Points In A Combinatorial Geometry With No 8-Point-Line Minors, Mathematical Essays in Honor of Gian-Carlo Rota, B. Sagan and R. P. Stanley, eds., Birkhäuser, 1998, 271-284.

Roger B. Eggleton, Maximal Midpoint-Free Subsets of Integers, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 374

Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992

Eric Weisstein's World of Mathematics, Repunit

Index entries for linear recurrences with constant coefficients, signature (6,-5).

Formula

Second binomial transform of A015518; binomial transform of A000302 (preceded by 0). - Paul Barry, Mar 28 2003

a(n) = Sum_{k=1..n} binomial(n,k)*4^(k-1). - Paul Barry, Mar 28 2003

a(n) = (-1)^n times the (i, j)-th element of M^n (for all i and j such that i is not equal to j), where M = ((1, -1, 1, -2), (-1, 1, -2, 1), (1, -2, 1, -1), (-2, 1, -1, 1)). - Simone Severini, Nov 25 2004

a(n) = A125118(n,4) for n>3. - Reinhard Zumkeller, Nov 21 2006

a(n) = ((3+sqrt(4))^n - (3-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008

a(n) = 6*a(n-1) - 5*a(n-2) n>1, a(0)=0, a(1)=1. - Philippe Deléham, Jan 01 2009

From Wolfdieter Lang, Oct 18 2010: (Start)

O.g.f.: x/((1-5*x)*(1-x)).

a(n) = 4*a(n-1) + 5*a(n-2) + 2, a(0)=0, a(1)=1.

a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3), a(0)=0, a(1)=1, a(2)=6. Observation by G. Detlefs. See the W. Lang comment and link. (End)

a(n) = 5*a(n-1) + 1 with n>0, a(0)=0. - Vincenzo Librandi, Nov 17 2010

a(n) = a(n-1) + A000351(n-1) n>0, a(0)=0. - Felix P. Muga II, Mar 19 2014

a(n) = a(n-1) + 20*a(n-2) + 5 for n > 1, a(0)=0, a(1)=1. - Felix P. Muga II, Mar 19 2014

a(n) = A060458(n)/2^(n+2), for n > 0. - R. J. Cano, Sep 25 2014

From Ilya Gutkovskiy, Oct 05 2016: (Start)

E.g.f.: (exp(4*x) - 1)*exp(x)/4.

Convolution of A000351 and A057427. (End)

Example

Base 5...........decimal
0......................0
1......................1
11.....................6
111...................31
1111.................156
11111................781
111111..............3906
1111111............19531
11111111...........97656
111111111.........488281
1111111111.......2441406
etc. ...............etc.
- Zerinvary Lajos, Jan 14 2007

Maple

a:=n->sum(5^(n-j), j=1..n): seq(a(n), n=0..23); # Zerinvary Lajos, Jan 04 2007
A003463:=1/(5*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=6*a[n-1]-5*a[n-2]od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 21 2008

Mathematica

lst={}; Do[p=(5^n-1)/4; AppendTo[lst, p], {n, 0, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Sep 29 2008 *)
Table[((5^n-1)/4), {n, 0, 25}] (* Vincenzo Librandi, Aug 20 2012 *)
NestList[5 # + 1 &, 0, 23] (* Bruno Berselli, Feb 06 2013 *)
LinearRecurrence[{6, -5}, {0, 1}, 30] (* Harvey P. Dale, Sep 20 2023 *)

Prog

(Sage) [lucas_number1(n, 6, 5) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009
(Sage) [gaussian_binomial(n, 1, 5) for n in range(0, 24)] # Zerinvary Lajos, May 28 2009
(PARI) a(n)=5^n\4; \\ Charles R Greathouse IV, Jul 15 2011
(Maxima) A003463(n):=floor((5^n-1)/4)$ makelist(A003463(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Magma) [(5^n-1)/4 : n in [0..30]]; // Wesley Ivan Hurt, Sep 25 2014

Crossrefs

Cf. A004061, A026375, A045468, A060458, A074479, A086122, A107219.

Cf. A003462, A002450, A003464, A023000, A023001, A002452, A002275.

Sequence in context: A227505 A026705 A243874 * A026771 A289788 A065096

Adjacent sequences: A003460 A003461 A003462 * A003464 A003465 A003466

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved