A260615 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.

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, 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, 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

Offset

1,2

Links

Matthew Campbell, Table of n, a(n) for n = 1..116505 The first 225 rows are in the b-file.

Example

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.
From Michael De Vlieger, Aug 29 2015: (Start)
Triangle begins:
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,   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, 42,  2,  2,  7,  1,  1,  4
...
(End)

Maple

seq(op(numtheory:-cfrac(s, 'quotients')), s=ListTools:-PartialSums(map2(`/`, 1, [seq(ithprime(i), i=1..20)]))); # Robert Israel, Sep 06 2015

Mathematica

Table[ContinuedFraction[Sum[1/Prime@k, {k, n}]], {n, 11}] // Flatten (* Michael De Vlieger, Aug 29 2015 *)

Prog

(PARI) row(n) = contfrac(sum(k=1, n, 1/prime(k)));
tabf(nn) = for(n=1, nn, print(row(n))); \\ Michel Marcus, Sep 18 2015

Crossrefs

Cf. A000040.

For the continued fractions of the harmonic numbers, see A100398.

For the numerator of the sum, see A024451.

For the denominator of the sum, see A002110.

Sequence in context: A193471 A351645 A182931 * A293298 A079134 A175528

Adjacent sequences: A260612 A260613 A260614 * A260616 A260617 A260618

Keyword

nonn,tabf,cofr

Author

Matthew Campbell, Aug 29 2015

Status

approved