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%I #14 Mar 09 2024 13:00:54
%S 0,7,45,186,630,1905,5355,14308,36828,92115,225225,540606,1277874,
%T 2981797,6881175,15728520,35651448,80215911,179306325,398458690,
%U 880803630,1937768217,4244635395,9261022956,20132658900,43620761275
%N Sum of binary numbers with n 1's and two (possibly leading) 0's.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (9,-33,63,-66,36,-8).
%F a(n) = (2^(n+2) - 1)*n*(n+1)/2 = A059672(n) + A059938(n) = a(n-1)*2*(n+1)/(n-1) + n(n+1)/2.
%F G.f.: x*(12*x^2-18*x+7) / ((x-1)^3*(2*x-1)^3). - _Colin Barker_, Sep 13 2014
%e a(2) = 45 since binary sum of 1100 + 1010 + 1001 + 0110 + 0101 + 0011 is 12 + 10 + 9 + 6 + 5 + 3 = 45.
%o (PARI) concat(0, Vec(x*(12*x^2-18*x+7)/((x-1)^3*(2*x-1)^3) + O(x^100))) \\ _Colin Barker_, Sep 13 2014
%Y Cf. A059672, A059673, A059938.
%K nonn,easy
%O 0,2
%A _Henry Bottomley_, Feb 13 2001
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