|
%I #4 Sep 03 2014 14:39:28
%S 1,1,3,1533,4870483401,10632494904416274948861848751148863,
%T 442778652527729430645666843207235634221292901,
%U 8594831104112238244501123836952492157088005557663896974587707618787108,970692073484990407927190417652798419153
%N Integers of the form (2^(k+1) - 1)/C(k+2,2).
%C The numbers k for which (2^(k+1) - 1)/C(k+2,2) is an integer are given by A246636. For each such k, (2^(k+1) - 1)/C(k+2,2) is the mean of the numbers in all the rows of Pascal's triangle, from row 0 through row k.
%e The sum of the numbers in Pascal's triangle, from row 0 through row 17, is 2^18 - 1 = 262143; the number of such numbers is C(19,2) = 171, and 262143/171 = 1533; thus is in A246637 and 17 is in A246636.
%t z = 3000; t = Select[Range[0, z], IntegerQ[(2^(# + 1) - 1)/Binomial[# + 2, 2]] &] (* A246636 *)
%t Table[(2^(t[[n]] + 1) - 1)/Binomial[t[[n]] + 2, 2], {n, 1, 10}] (*A246637*)
%Y Cf. A246636, A007318.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Sep 01 2014
|
