OFFSET
4,1
COMMENTS
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
LINKS
Colin Barker, Table of n, a(n) for n = 4..1000
Serhat Bulut, Oktay Erkan Temizkan, Subset Sum Problem
Sela Fried, Counting r X s rectangles in nondecreasing and Smirnov words, arXiv:2406.18923 [math.CO], 2024. See p. 9.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
a(n) = 6*A000389(n+1).
G.f.: 6*x^4 / (1-x)^6. - Colin Barker, Apr 03 2015
E.g.f.: x^4*(5 + x)*exp(x)/20. - G. C. Greubel, Nov 24 2017
EXAMPLE
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}.
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).
MATHEMATICA
Drop[Plus @@ Flatten[Part[#, 1 ;; 3] & /@ Subsets[Range@ #, {4}]] & /@
Range@ 28, 3] (* Michael De Vlieger, Jan 20 2015 *)
PROG
(Magma) [6*Binomial(n+1, 5): n in [4..40]]; // Vincenzo Librandi, Feb 13 2015
(PARI) Vec(6*x^4/(1-x)^6 + O(x^100)) \\ Colin Barker, Apr 03 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Serhat Bulut, Jan 20 2015
EXTENSIONS
More terms from Vincenzo Librandi, Feb 13 2015
STATUS
approved