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A068011 Number of subsets of {1,2,3,...,n} that sum to 0 mod 5.

%I A068011 #28 May 02 2025 15:20:39
%S A068011 1,1,1,2,4,8,14,26,52,104,208,412,820,1640,3280,6560,13112,26216,
%T A068011 52432,104864,209728,419440,838864,1677728,3355456,6710912,13421792,
%U A068011 26843552,53687104,107374208,214748416,429496768,858993472,1717986944,3435973888,6871947776
%N A068011 Number of subsets of {1,2,3,...,n} that sum to 0 mod 5.
%C A068011 For n>2, a(n) = 2 * A068031(n).
%H A068011 Sophie LeBlanc, Jan 20 2002, <a href="http://groups.google.com/groups?hl=en&amp;selm=85962b5a.0201201247.4aa79c5c%40posting.google.com">sci.math posting</a>
%H A068011 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,0,0,2,-4).
%F A068011 a(k+1) = 2*a(k) if k == 2, 3, or 4 mod 5, 2*a(k)-2^(k/5) if k == 0 mod 5, 2*a(k)-2^((k-1)/5) if k == 1 mod 5.
%F A068011 G.f.: -(x^2-x+1)*(2*x^3+2*x^2-1) / ((2*x-1)*(2*x^5-1)). - _Colin Barker_, Dec 22 2012
%F A068011 If n == 0 mod 5, then a(n) = (2^n + 4*2^(n/5))/5. - _Giorgos Kalogeropoulos_, May 02 2025
%F A068011 a(n) ~ 2^n/5. - _Stefano Spezia_, May 02 2025
%p A068011 A068011_rec := proc(n); if(0 = n) then RETURN(1); fi; if(1 = (n mod 5)) then RETURN(2*A068011_rec(n-1)-2^((n-1)/5)); fi; if(2 = (n mod 5)) then RETURN(2*A068011_rec(n-1)-2^((n-2)/5)); fi; RETURN(2*A068011_rec(n-1)); end;
%p A068011 # second Maple program:
%p A068011 b:= proc(n, s) option remember; `if`(n=0, `if`(s=0, 1, 0),
%p A068011       b(n-1, s)+b(n-1, irem(s+n, 5)))
%p A068011     end:
%p A068011 a:= n-> b(n, 0):
%p A068011 seq(a(n), n=0..35);  # _Alois P. Heinz_, May 02 2025
%t A068011 LinearRecurrence[{2, 0, 0, 0, 2, -4}, {1, 1, 1, 2, 4, 8}, 40] (* _Jean-François Alcover_, Mar 06 2016 *)
%Y A068011 5th row of A068009.
%K A068011 nonn
%O A068011 0,4
%A A068011 _Antti Karttunen_, Feb 11 2002