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a(1)=1, a(n) = 5*a(n-1) + n.
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%I #37 Jun 07 2024 17:32:28

%S 1,7,38,194,975,4881,24412,122068,610349,3051755,15258786,76293942,

%T 381469723,1907348629,9536743160,47683715816,238418579097,

%U 1192092895503,5960464477534,29802322387690,149011611938471,745058059692377,3725290298461908,18626451492309564

%N a(1)=1, a(n) = 5*a(n-1) + n.

%H Vincenzo Librandi, <a href="/A014827/b014827.txt">Table of n, a(n) for n = 1..1000</a>

%H László Tóth, <a href="https://arxiv.org/abs/2002.06584">On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers</a>, arXiv:2002.06584 [math.NT], 2020. See also <a href="https://doi.org/10.1090/proc/14863">Proc. Amer. Math. Soc.</a> 148 (2020), 461-469.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-11,5).

%F a(n) = (5^(n+1) - 4*n - 5)/16.

%F G.f.: x/((1-5*x)*(1-x)^2).

%F a(n) = Sum_{k=0..n} (n-k)*5^k = Sum_{k=0..n} k*5^(n-k). - _Paul Barry_, Jul 30 2004

%F a(n) = Sum_{k=0..n} binomial(n+2, k+2)*4^k [Offset 0]. - _Paul Barry_, Jul 30 2004

%p a:=n->sum((5^(n-j)-1^(n-j))/4,j=0..n): seq(a(n), n=1..21); # _Zerinvary Lajos_, Jan 04 2007

%t Join[{a=1,b=7},Table[c=6*b-5*a+1;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 06 2011 *)

%o (Sage) [(gaussian_binomial(n,1,5)-n)/4 for n in range(2,23)] # _Zerinvary Lajos_, May 29 2009

%o (Magma) [(5^(n+1)-4*n-5)/16: n in [1..30]]; // _Vincenzo Librandi_, Aug 23 2011

%Y Cf. A016218, A016208, A000392, A000225, A003462, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A016256.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_