%I #10 Jun 17 2022 14:31:44
%S 1,2,8,10,60,75,131,195,988,1120,1130,1232,1345,1347,1953,2933,3549,
%T 9956,13797,13970,14586,14652,14903,17166,19176,19634,22584,24354,
%U 24842,26488,29388,30840,31409,34934,36059,45149,49793,52690,59413,61063
%N a(1)=1, a(n) is the smallest integer > a(n-1) such that the sum of elements of the simple continued fraction for S(n)=1/a(1)+1/a(2)+...+1/a(n) equals n*(n+1)/2 the n-th triangular number.
%e 1/a(1)+1/a(2)+1/a(3)+1/a(4) = (1+1/2+1/8+1/10) which continued fraction is {1, 1, 2, 1, 1, 1, 3} and 1+1+2+1+1+1+3 = 10 = 4*5/2 the fourth triangular number.
%o (PARI) s=1; t=1; for(n=2,22,s=s+1/t; while(abs(n*(n+1)/2+1-sum(i=1,length(contfrac(s+1/t)), component(contfrac(s+1/t),i)))>0,t++); print1(t,","))
%K easy,nonn
%O 1,2
%A _Benoit Cloitre_, Jun 10 2002
%E More terms from Lambert Klasen (lambert.klasen(AT)gmx.de), Dec 18 2004