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%I #32 Oct 02 2015 13:25:40
%S 0,2,0,1,5,1,30,1,5,1,2,12,1,3,1,2,1,9,1,1,7,1,2,1,9,1,2,1,2,12,7,1,2,
%T 2,13,1,1,1,8,13,5,4,1,2,5,8,1,2,6,1,1,4,10,1,2,3,1,3,1,2,238,1,28,1,
%U 42,2,2,7,1,1,4,1,1,1,6,1,41,3,1,1,51,1,9,2,3,2,5,1,2,1,6,1,1,1,3,3,3,1,1,1,3,3,1,2,19,1,13,1,1,3,4,7,1,1,3,2,1,10
%N Irregular triangle read by rows: the n-th row is the continued fraction expansion of the sum of the reciprocals of the first n primes.
%H Matthew Campbell, <a href="/A260615/b260615.txt">Table of n, a(n) for n = 1..116505</a> The first 225 rows are in the b-file.
%e For row 3, the sum of the first three prime reciprocals equals 1/2 + 1/3 + 1/5 = 31/30. The continued fraction expansion of 31/30 is 1 + (1/30). Because of this, the terms in row 3 are 1 and 30.
%e From _Michael De Vlieger_, Aug 29 2015: (Start)
%e Triangle begins:
%e 0, 2
%e 0, 1, 5
%e 1, 30
%e 1, 5, 1, 2, 12
%e 1, 3, 1, 2, 1, 9, 1, 1, 7
%e 1, 2, 1, 9, 1, 2, 1, 2, 12, 7
%e 1, 2, 2, 13, 1, 1, 1, 8, 13, 5, 4
%e 1, 2, 5, 8, 1, 2, 6, 1, 1, 4, 10, 1, 2, 3, 1, 3
%e 1, 2, 238, 1, 28, 1, 42, 2, 2, 7, 1, 1, 4
%e ...
%e (End)
%p seq(op(numtheory:-cfrac(s,'quotients')),s=ListTools:-PartialSums(map2(`/`,1,[seq(ithprime(i),i=1..20)]))); # _Robert Israel_, Sep 06 2015
%t Table[ContinuedFraction[Sum[1/Prime@k, {k, n}]], {n, 11}] // Flatten (* _Michael De Vlieger_, Aug 29 2015 *)
%o (PARI) row(n) = contfrac(sum(k=1, n, 1/prime(k)));
%o tabf(nn) = for(n=1, nn, print(row(n))); \\ _Michel Marcus_, Sep 18 2015
%Y Cf. A000040.
%Y For the continued fractions of the harmonic numbers, see A100398.
%Y For the numerator of the sum, see A024451.
%Y For the denominator of the sum, see A002110.
%K nonn,tabf,cofr
%O 1,2
%A _Matthew Campbell_, Aug 29 2015
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