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

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

Author

N. J. A. Sloane

Keywords

• nonn
• easy

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

Examples

			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
		

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).
• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 282.

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.
• Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
• 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).

Crossrefs

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

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

Programs

Magma

[(5^n-1)/4 : n in [0..30]]; // Wesley Ivan Hurt, Sep 25 2014

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 *)

Maxima

A003463(n):=floor((5^n-1)/4)$ makelist(A003463(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

PARI

a(n)=5^n\4; \\ Charles R Greathouse IV, Jul 15 2011

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

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)