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%I #66 Feb 17 2024 10:33:22
%S 0,4,5,9,10,14,15,19,20,24,25,29,30,34,35,39,40,44,45,49,50,54,55,59,
%T 60,64,65,69,70,74,75,79,80,84,85,89,90,94,95,99,100,104,105,109,110,
%U 114,115,119,120,124,125,129,130,134,135,139,140,144,145,149
%N Numbers that are congruent to {0, 4} mod 5.
%C Also solutions to 3^x + 5^x == 2 (mod 11). - _Cino Hilliard_, May 18 2003
%H G. C. Greubel, <a href="/A047208/b047208.txt">Table of n, a(n) for n = 1..5000</a>
%H Cino Hilliard, <a href="https://web.archive.org/web/20080906234512/http://groups.msn.com/BC2LCC/3x5x211k.msnw">solutions to 3^x + 5^x == 2 mod 11/</a>, June 2003.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F From _R. J. Mathar_, Jan 24 2009: (Start)
%F G.f.: x^2*(4+x)/((1-x)^2*(1+x)).
%F a(n) = a(n-2) + 5. (End)
%F a(n) = 5*n - 6 - a(n-1) (with a(1)=0). - _Vincenzo Librandi_, Nov 18 2010
%F a(n+1) = Sum_{k>=0} A030308(n,k)*b(k), with b(0)=4 and b(k) = A020714(k-1) = 5*2^(k-1) for k>0. - _Philippe Deléham_, Oct 17 2011
%F a(n) = ceiling((5/3)*ceiling(3*n/2)). - _Clark Kimberling_, Jul 04 2012
%F a(n) = (5*(n-1) + 3*(n-1 mod 2))/2 = (5*(n-1) + A010674(n-1))/2. - _G. C. Greubel_, Nov 23 2021
%F Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + log(phi)/(2*sqrt(5)) - sqrt(1+2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - _Amiram Eldar_, Dec 07 2021
%F E.g.f.: 1 + ((5*x - 7/2)*exp(x) + (3/2)*exp(-x))/2. - _David Lovler_, Aug 23 2022
%t {#,#+4}&/@(5*Range[0,30])//Flatten (* _Harvey P. Dale_, Apr 05 2019 *)
%o (PARI) forstep(n=0,200,[4,1],print1(n", ")) \\ _Charles R Greathouse IV_, Oct 17 2011
%o (Magma) [(5*(n-1) + 3*((n-1) mod 2))/2: n in [1..100]]; // _G. C. Greubel_, Nov 23 2021
%o (Sage) [(5*(n-1) +3*((n-1)%2))/2 for n in (1..100)] # _G. C. Greubel_, Nov 23 2021
%Y Cf. A001622, A010674, A010685 (first differences), A274406.
%K nonn,easy
%O 1,2
%A _N. J. A. Sloane_
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