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%I M4209 #174 Sep 20 2023 11:54:04
%S 0,1,6,31,156,781,3906,19531,97656,488281,2441406,12207031,61035156,
%T 305175781,1525878906,7629394531,38146972656,190734863281,
%U 953674316406,4768371582031,23841857910156,119209289550781,596046447753906,2980232238769531
%N a(n) = (5^n - 1)/4.
%C 5^a(n) is the highest power of 5 dividing (5^n)!. - _Benoit Cloitre_, Feb 04 2002
%C n such that A002294(n) is not divisible by 5. - _Benoit Cloitre_, Jan 14 2003
%C 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]
%C 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
%C Starting with 1 = convolution square of A026375: (1, 3, 11, 45, 195, 873, ...). - _Gary W. Adamson_, May 17 2009
%C 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
%C 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
%C It is the Lucas sequence U(6,5). - _Felix P. Muga II_, Mar 21 2014
%C 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
%C 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
%C Order of the rank-n projective geometry PG(n-1,5) over the finite field GF(5). - _Anthony Hernandez_, Oct 05 2016
%C 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
%C 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
%D Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A003463/b003463.txt">Table of n, a(n) for n = 0..200</a>
%H Joseph E. Bonin and Joseph P. S. Kung <a href="http://home.gwu.edu/~jbonin/u6web.pdf">The Number of Points In A Combinatorial Geometry With No 8-Point-Line Minors</a>, Mathematical Essays in Honor of Gian-Carlo Rota, B. Sagan and R. P. Stanley, eds., Birkhäuser, 1998, 271-284.
%H Roger B. Eggleton, <a href="http://dx.doi.org/10.1155/2015/216475">Maximal Midpoint-Free Subsets of Integers</a>, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=374">Encyclopedia of Combinatorial Structures 374</a>
%H Wolfdieter Lang, <a href="/A003463/a003463.pdf">Notes on certain inhomogeneous three term recurrences.</a>
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Repunit.html">Repunit</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-5).
%F Second binomial transform of A015518; binomial transform of A000302 (preceded by 0). - _Paul Barry_, Mar 28 2003
%F a(n) = Sum_{k=1..n} binomial(n,k)*4^(k-1). - _Paul Barry_, Mar 28 2003
%F 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
%F a(n) = A125118(n,4) for n>3. - _Reinhard Zumkeller_, Nov 21 2006
%F a(n) = ((3+sqrt(4))^n - (3-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
%F a(n) = 6*a(n-1) - 5*a(n-2) n>1, a(0)=0, a(1)=1. - _Philippe Deléham_, Jan 01 2009
%F From _Wolfdieter Lang_, Oct 18 2010: (Start)
%F O.g.f.: x/((1-5*x)*(1-x)).
%F a(n) = 4*a(n-1) + 5*a(n-2) + 2, a(0)=0, a(1)=1.
%F 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)
%F a(n) = 5*a(n-1) + 1 with n>0, a(0)=0. - _Vincenzo Librandi_, Nov 17 2010
%F a(n) = a(n-1) + A000351(n-1) n>0, a(0)=0. - _Felix P. Muga II_, Mar 19 2014
%F 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
%F a(n) = A060458(n)/2^(n+2), for n > 0. - _R. J. Cano_, Sep 25 2014
%F From _Ilya Gutkovskiy_, Oct 05 2016: (Start)
%F E.g.f.: (exp(4*x) - 1)*exp(x)/4.
%F Convolution of A000351 and A057427. (End)
%e Base 5...........decimal
%e 0......................0
%e 1......................1
%e 11.....................6
%e 111...................31
%e 1111.................156
%e 11111................781
%e 111111..............3906
%e 1111111............19531
%e 11111111...........97656
%e 111111111.........488281
%e 1111111111.......2441406
%e etc. ...............etc.
%e - _Zerinvary Lajos_, Jan 14 2007
%p a:=n->sum(5^(n-j),j=1..n): seq(a(n), n=0..23); # _Zerinvary Lajos_, Jan 04 2007
%p A003463:=1/(5*z-1)/(z-1); # _Simon Plouffe_ in his 1992 dissertation
%p 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
%t lst={}; Do[p=(5^n-1)/4; AppendTo[lst, p], {n, 0, 5!}]; lst (* _Vladimir Joseph Stephan Orlovsky_, Sep 29 2008 *)
%t Table[((5^n-1)/4),{n,0,25}] (* _Vincenzo Librandi_, Aug 20 2012 *)
%t NestList[5 # + 1 &, 0, 23] (* _Bruno Berselli_, Feb 06 2013 *)
%t LinearRecurrence[{6,-5},{0,1},30] (* _Harvey P. Dale_, Sep 20 2023 *)
%o (Sage) [lucas_number1(n,6,5) for n in range(0, 24)] # _Zerinvary Lajos_, Apr 22 2009
%o (Sage) [gaussian_binomial(n,1,5) for n in range(0,24)] # _Zerinvary Lajos_, May 28 2009
%o (PARI) a(n)=5^n\4; \\ _Charles R Greathouse IV_, Jul 15 2011
%o (Maxima) A003463(n):=floor((5^n-1)/4)$ makelist(A003463(n),n,0,30); /* _Martin Ettl_, Nov 05 2012 */
%o (Magma) [(5^n-1)/4 : n in [0..30]]; // _Wesley Ivan Hurt_, Sep 25 2014
%Y Cf. A004061, A026375, A045468, A060458, A074479, A086122, A107219.
%Y Cf. A003462, A002450, A003464, A023000, A023001, A002452, A002275.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
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