%I #47 Nov 20 2023 08:19:55
%S 0,1,11,91,671,4651,31031,201811,1288991,8124571,50700551,313968931,
%T 1932641711,11839990891,72260648471,439667406451,2668522016831,
%U 16163719991611,97745259402791,590286253682371
%N a(n) = 6^n - 5^n.
%C These are the numerators of a(n) = (Integral_{x=0..1/3} (1-x/2)^n dx). E.g., a(3)=671/2592. The denominators are b(n) = 3*(n+1)*6^n. E.g., b(3)=2592. the subscripts in both cases are 0. - Al Hakanson (hawkuu(AT)excite.com), Feb 22 2004
%C Number of numbers with at most n digits whose largest digit is 5. For the first 5 terms, the first differences (i.e., ...with exactly n digits...) are given in A125373. - _M. F. Hasler_, May 03 2015
%C a(n) is the number of n-digit numbers whose smallest decimal digit is 4. - _Stefano Spezia_, Nov 15 2023
%H Vincenzo Librandi, <a href="/A005062/b005062.txt">Table of n, a(n) for n = 0..300</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-30).
%F G.f.: x/((1-5*x)(1-6*x)).
%F a(n) = 11*a(n-1) - 30*a(n-2), n > 1; a(0)=0, a(1)=1. - _Philippe Deléham_, Jan 01 2009
%F E.g.f.: exp(6*x) - exp(5*x). - _Mohammad K. Azarian_, Jan 14 2009
%F a(n) = -(30)^n * a(-n) for all n in Z. - _Michael Somos_, Jul 14 2018
%e G.f. = x + 11*x^2 + 91*x^3 + 671*x^4 + 4651*x^5 + 31031*x^6 + 201811*x^7 + ... - _Michael Somos_, Jul 14 2018
%p restart:a:=n->sum(5^(n-j)*binomial(n,j),j=1..n): seq(a(n), n=0..19); # _Zerinvary Lajos_, Apr 18 2009
%t f[n_]:=6^n-5^n;f[Range[0,60]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 31 2011 *)
%t LinearRecurrence[{11,-30},{0,1},20] (* _Harvey P. Dale_, May 28 2015 *)
%o (Sage) [lucas_number1(n,11,30) for n in range(0, 20)] # _Zerinvary Lajos_, Apr 27 2009
%o (Magma) [6^n - 5^n: n in [0..25]]; // _Vincenzo Librandi_, Jun 03 2011
%o (PARI) a(n)=6^n-5^n \\ _M. F. Hasler_, May 03 2015
%o (PARI) for(d=0,9,print1(sum(n=1,10^d-1,vecmax(digits(n))==5)",")) \\ Only to illustrate the comment about "largest digit equals 5".
%Y Cf. A005060 (5^n - 4^n), A125373.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_