login
A253945
a(n) = 6*binomial(n+1,5).
3
6, 36, 126, 336, 756, 1512, 2772, 4752, 7722, 12012, 18018, 26208, 37128, 51408, 69768, 93024, 122094, 158004, 201894, 255024, 318780, 394680, 484380, 589680, 712530, 855036, 1019466, 1208256, 1424016, 1669536, 1947792, 2261952, 2615382, 3011652, 3454542
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
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.
FORMULA
a(n) = 6*A000389(n+1).
G.f.: 6*x^4 / (1-x)^6. - Colin Barker, Apr 03 2015
a(n) = Sum_{i=1..n-2} A000217(i-1)*A000217(i+1) with a(3)=0. [Bruno Berselli, Jul 20 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 *)
6 Binomial[Range[5, 29], 5] (* Michael De Vlieger, Feb 13 2015, after Alonso del Arte at A253946 *)
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
Sequence in context: A083374 A264366 A061707 * A331576 A056375 A360588
KEYWORD
nonn,easy
AUTHOR
Serhat Bulut, Jan 20 2015
EXTENSIONS
More terms from Vincenzo Librandi, Feb 13 2015
STATUS
approved