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A253942
a(n) = 3*binomial(n+1, 5).
2
3, 18, 63, 168, 378, 756, 1386, 2376, 3861, 6006, 9009, 13104, 18564, 25704, 34884, 46512, 61047, 79002, 100947, 127512, 159390, 197340, 242190, 294840, 356265, 427518, 509733, 604128, 712008, 834768, 973896, 1130976, 1307691, 1505826, 1727271, 1974024, 2248194
OFFSET
4,1
COMMENTS
For a set of integers {1,2,...,n}, a(n) is the sum of the 2 smallest elements of each subset with 4 elements, which is 3*binomial(n+1, 5) (for n >= 4), hence a(n) = 3*binomial(n+1, 5) = 3*A000389(n+1).
LINKS
Serhat Bulut and Oktay Erkan Temizkan, Subset Sum Problem [Dead link]
FORMULA
a(n) = 3*A000389(n+1).
a(n) = (n-3)*(n-2)*(n-1)*n*(1+n)/40. - Colin Barker, Jan 20 2015
G.f.: 3*x^4 / (x-1)^6. - Colin Barker, Jan 20 2015
E.g.f.: x^4*(x+5)*exp(x)/40. - G. C. Greubel, Nov 25 2017
a(n) = Sum_{k=3..n-1} A050534(k). - Ivan N. Ianakiev, Oct 08 2023
EXAMPLE
For A={1,2,3,4}, the only subset with 4 elements is {1,2,3,4}; sum of 2 minimum elements of this subset: a(4) = 1+2 = 3 = 3*binomial(4+1, 5).
For A={1,2,3,4,5}, the 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 2 smallest elements of each subset: a(5) = (1+2)+(1+2)+(1+2)+(1+3)+(2+3) = 18 = 3*binomial(5+1, 5).
MATHEMATICA
a253942[n_] := Drop[Plus @@ Flatten[Part[#, 1 ;; 2] & /@ Subsets[Range@ #, {4}]] & /@ Range@ n, 3]; a253942[28] (* Michael De Vlieger, Jan 20 2015 *)
Table[3 Binomial[n + 1, 5], {n, 4, 35}] (* Vincenzo Librandi, Feb 14 2015 *)
PROG
(PARI) a(n) = 3*binomial(n+1, 5); \\ Michel Marcus, Jan 20 2015
(PARI) Vec(3*x^4/(x-1)^6 + O(x^100)) \\ Colin Barker, Jan 20 2015
(Magma) [3*Binomial(n+1, 5): n in [4..40]]; Vincenzo Librandi, Feb 14 2015
CROSSREFS
Cf. A000389.
Sequence in context: A210366 A000648 A235988 * A317404 A110689 A027333
KEYWORD
nonn,easy
AUTHOR
Serhat Bulut, Jan 20 2015
STATUS
approved