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a(n) = 6*binomial(n+1,7).
1

%I #57 Sep 08 2022 08:46:11

%S 6,48,216,720,1980,4752,10296,20592,38610,68640,116688,190944,302328,

%T 465120,697680,1023264,1470942,2076624,2884200,3946800,5328180,

%U 7104240,9364680,12214800,15777450,20195136,25632288,32277696,40347120,50086080,61772832,75721536

%N a(n) = 6*binomial(n+1,7).

%C For a set of integers {1,2,...,n}, a(n) is the sum of the 3 smallest elements of each subset with 6 elements, which is 6*binomial(n+1,7) (for n>=6), hence a(n) = 6*binomial(n+1,7) = 6*A000580(n+1).

%H Colin Barker, <a href="/A253947/b253947.txt">Table of n, a(n) for n = 6..1000</a>

%H Serhat Bulut, Oktay Erkan Temizkan, <a href="http://matematikproje.com/dosyalar/7e1cdSubset_smallest_elements_Sum.pdf">Subset Sum Problem</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).

%F a(n) = 6*binomial(n+1,7) = 6*A000580(n+1).

%F G.f.: 6*x^6 / (1-x)^8. - _Colin Barker_, Apr 03 2015

%F a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8). - _Wesley Ivan Hurt_, Sep 03 2022

%e For A={1,2,3,4,5,6,7}, the subsets with 6 elements are {1,2,3,4,5,6}, {1,2,3,4,5,7}, {1,2,3,4,6,7}, {1,2,3,5,6,7}, {1,2,4,5,6,7}, {1,3,4,5,6,7}, and {2,3,4,5,6,7}.

%e Sum of 3 smallest elements of each subset: a(7) = (1+2+3) + (1+2+3) + (1+2+3) + (1+2+3) + (1+2+4) + (1+3+4) + (2+3+4) = 48 = 6*binomial(7+1,7) = 6*A000580(7+1).

%t Drop[Plus @@ Flatten[Part[#, 1 ;; 3] & /@ Subsets[Range@ #, {6}]] & /@

%t Range@ 30, 5] (* _Michael De Vlieger_, Jan 20 2015 *)

%t 6 Binomial[Range[7, 31], 7] (* _Michael De Vlieger_, Feb 13 2015, after _Alonso del Arte_ at A253946 *)

%o (Magma) [6*Binomial(n+1, 7): n in [6..40]]; // _Vincenzo Librandi_, Feb 13 2015

%o (PARI) Vec(6*x^6/(1-x)^8 + O(x^100)) \\ _Colin Barker_, Apr 03 2015

%Y Cf. A000580 (binomial(n, 7)).

%K nonn,easy

%O 6,1

%A _Serhat Bulut_, Jan 20 2015

%E More terms from _Vincenzo Librandi_, Feb 13 2015