Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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