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A121544 Sum of all proper base 4 numbers with n digits (those not beginning with 0).

%I A121544 #21 Nov 07 2024 08:48:41
%S A121544 6,114,1896,30624,491136,7862784,125822976,2013241344,32212156416,
%T A121544 515395682304,8246335635456,131941389041664,2111062300164096,
%U A121544 33776997104615424,540431954881806336,8646911282940739584,138350580546379186176,2213609288819376390144
%N A121544 Sum of all proper base 4 numbers with n digits (those not beginning with 0).
%C A121544 Sum of the first 3 * 4^(n-1) integers starting with 4^(n-1).
%C A121544 Sum of the integers from 4^(n-1) to 4^n -1.
%C A121544 First differences of A026337.
%H A121544 G. C. Greubel, <a href="/A121544/b121544.txt">Table of n, a(n) for n = 1..820</a>
%H A121544 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-64).
%F A121544 a(n) = 3 * 4^(n-1) * (4^(n-1) + 4^n - 1)/2.
%F A121544 G.f.: 6*x*(1-x) / ((1-4*x)*(1-16*x)). - _Colin Barker_, Apr 30 2013
%F A121544 From _G. C. Greubel_, Nov 07 2024: (Start)
%F A121544 a(n) = (3/5)*binomial(5*4^(n-1), 2).
%F A121544 E.g.f.: (3/32)*(-1 - 4*exp(4*x) + 5*exp(16*x)). (End)
%e A121544 a(1) = 6 = 1 + 2 + 3.
%e A121544 a(2) = 114 = 10_4 + 11_4 + 12_4 + 13_4 + 20_4 + 21_4 + 22_4 + 23_4 + 30_4 + 31_4 + 32_4 + 33_4 = (4+5+6+7+8+9+10+11+12+13+14+15)_10.
%t A121544 Table[3*4^(n-1)*(5*4^(n-1) - 1)/2, {n,20}] (* _James C. McMahon_, Oct 19 2024 *)
%o A121544 (Magma) [3*Binomial(5*4^(n-1), 2)/5: n in [1..20]]; // _G. C. Greubel_, Nov 07 2024
%o A121544 (SageMath)
%o A121544 def A121544(n): return 3*binomial(5*4^(n-1), 2)//5
%o A121544 [A121544(n) for n in range(1,21)] # _G. C. Greubel_, Nov 07 2024
%Y A121544 Cf. A007090, A010036, A026121, A026337, A101291.
%K A121544 easy,nonn,base
%O A121544 1,1
%A A121544 _Jonathan Vos Post_, Sep 08 2006
%E A121544 More terms from _Colin Barker_, Apr 30 2013
%E A121544 Edited by _Michel Marcus_, Apr 15 2024