%I #19 Jul 04 2019 03:34:18
%S 1,2,3,5,9,37,195,8584,1281621,1325419784,40182098746967,
%T 203448501599750774078,4275655952199444141114482835180,
%U 10920781877316031992615629928696178128586477545
%N a(1)=1; for each n > 1, a(n) is the smallest number such that Sum_{i=1..n} 1/a(i)^2 < sqrt(2).
%C In other words, the sequence is the lexicographically first infinite sequence of positive integers whose squared reciprocals sum to less than sqrt(2). After a(1)=1, each term is the smallest number that will not cause the sum of the squares of the reciprocals to exceed the square root of 2.
%F a(n+1) = ceiling(1/sqrt(sqrt(2) - Sum_{i=1..n} 1/a(i)^2)). - _R. J. Mathar_, Jul 11 2010
%e a(1)=1; 1/1^2 = 1;
%e a(2)=2; 1 + 1/2^2 = 5/4 = 1.25;
%e a(3)=3; 5/4 + 1/3^2 = 49/36 = 1.3611111111...;
%e a(4)=5; 49/36 + 1/5^2 = 1261/900 = 1.4011111111...;
%e a(5)=9; 1261/900 + 1/9^2 = 11449/8100 = 1.4134567901...;
%e (sums approach sqrt(2) = 1.4142135623...).
%p Digits := 200 : A179332 := proc(n) option remember; if n = 1 then 1; else sqrt(2)-add( 1/procname(i)^2,i=1..n-1) ; ceil( 1/sqrt(%)) ; end if; end proc: seq(A179332(n),n=1..14) ; # _R. J. Mathar_, Jul 11 2010
%Y Cf. A216245.
%K easy,nonn
%O 1,2
%A _Ben Paul Thurston_, Jul 10 2010
%E More terms from _R. J. Mathar_, Jul 11 2010
%E Name changed, comments expanded, and example corrected and expanded by _Jon E. Schoenfield_, Feb 28 2014
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