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A253945 a(n) = 6*binomial(n+1,5). 3

%I #56 Sep 08 2022 08:46:11

%S 6,36,126,336,756,1512,2772,4752,7722,12012,18018,26208,37128,51408,

%T 69768,93024,122094,158004,201894,255024,318780,394680,484380,589680,

%U 712530,855036,1019466,1208256,1424016,1669536,1947792,2261952,2615382,3011652,3454542

%N a(n) = 6*binomial(n+1,5).

%C For a set of integers {1,2,...,n}, a(n) is the sum of the 3 smallest elements of each subset with 4 elements, which is 6*binomial(n+1,5) for n>=4, hence a(n) = 6*binomial(n+1,5) = 6*A000389(n+1). - _Serhat Bulut_, Oktay Erkan Temizkan, Jan 20 2015

%H Colin Barker, <a href="/A253945/b253945.txt">Table of n, a(n) for n = 4..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_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F a(n) = 6*A000389(n+1).

%F G.f.: 6*x^4 / (1-x)^6. - _Colin Barker_, Apr 03 2015

%F a(n) = Sum_{i=1..n-2} A000217(i-1)*A000217(i+1) with a(3)=0. [_Bruno Berselli_, Jul 20 2015]

%F E.g.f.: x^4*(5 + x)*exp(x)/20. - _G. C. Greubel_, Nov 24 2017

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

%e Sum of 3 smallest elements of each subset: a(5) = (1+2+3) + (1+2+3) + (1+2+4) + (1+3+4) + (2+3+4) = 36 = 6*binomial(5+1,5) = 6*A000389(5+1).

%t Drop[Plus @@ Flatten[Part[#, 1 ;; 3] & /@ Subsets[Range@ #, {4}]] & /@

%t Range@ 28, 3] (* _Michael De Vlieger_, Jan 20 2015 *)

%t 6 Binomial[Range[5, 29], 5] (* _Michael De Vlieger_, Feb 13 2015, after _Alonso del Arte_ at A253946 *)

%o (Magma) [6*Binomial(n+1, 5): n in [4..40]]; // _Vincenzo Librandi_, Feb 13 2015

%o (PARI) Vec(6*x^4/(1-x)^6 + O(x^100)) \\ _Colin Barker_, Apr 03 2015

%Y Cf. A000389, A000217,

%K nonn,easy

%O 4,1

%A _Serhat Bulut_, Jan 20 2015

%E More terms from _Vincenzo Librandi_, Feb 13 2015

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