A260615 - OEIS

%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