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%I #22 May 26 2023 10:51:15
%S 0,1,0,1,1,2,1,3,3,5,5,8,8,13,14,20,23,31,35,48,55,72,84,108,126,160,
%T 187,233,275,340,398,489,574,697,819,988,1158,1390,1627,1941,2271,
%U 2696,3145,3721,4335,5104,5938,6967,8088,9462,10964,12783
%N Number of partitions of n with rank 1 (the rank of a partition is the largest part minus the number of parts).
%C Column k=1 in the triangle A063995.
%D George E. Andrews, The Theory of Partitions, Addison-Wesley, Reading, Mass., 1976.
%H Seiichi Manyama, <a href="/A101198/b101198.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f. for the number of partitions of n with rank r is Sum((-1)^k*x^(r*k)*(x^((3*k^2+k)/2)-x^((3*k^2-k)/2)), k=1..infinity)/Product(1-x^k, k=1..infinity). - _Vladeta Jovovic_, Dec 20 2004
%F Also Sum(x^(2*n+r+1)*Product((1-x^(2*n+r+1-k))/(1-x^k),k=1..n),n=0..infinity). - _Vladeta Jovovic_, May 05 2008
%F a(n) ~ Pi * exp(Pi*sqrt(2*n/3)) / (3 * 2^(9/2) * n^(3/2)). - _Vaclav Kotesovec_, May 26 2023
%e a(6)=2 because the 11 partitions 6,51,42,411,33,321,3111,222,2211,21111,111111 have ranks 5,3,2,1,1,0,-1,-1,-2,-3,-5, respectively.
%p with(combinat): for n from 1 to 35 do P:=partition(n): c:=0: for j from 1 to nops(P) do if P[j][nops(P[j])]-nops(P[j])=1 then c:=c+1 else c:=c fi od: a[n]:=c: od: seq(a[n],n=1..35);
%t Table[Count[IntegerPartitions[n],_?(Max[#]-Length[#]==1&)],{n,60}] (* _Harvey P. Dale_, Nov 29 2014 *)
%Y Cf. A000041, A063995.
%Y Cf. A101198-A101200, A101707-A101709.
%K nonn
%O 1,6
%A _Emeric Deutsch_, Dec 12 2004