A319515 Second term of the simple continued fraction of (Sum_{k=1..n} k^k)/(n^n).

1, 4, 5, 8, 10, 13, 16, 19, 21, 24, 27, 30, 32, 35, 38, 41, 43, 46, 49, 51, 54, 57, 60, 62, 65, 68, 70, 73, 76, 79, 81, 84, 87, 90, 92, 95, 98, 100, 103, 106, 109, 111, 114, 117, 119, 122, 125, 128, 130, 133, 136, 138, 141, 144, 147, 149, 152, 155, 158, 160, 163, 166, 168, 171, 174, 177

Offset

1,2

Comments

Ignoring the first two terms, a(n+1) is very similar to floor(n*e) (the Beatty sequence for e, A022843). This breaks down at n=11, where a(12)=30 and floor(11*e)=29. The successive terms (or partial denominators) in the sequence differ by 3 except for semi-regular "glitches". For example, the pattern of differences is 313233323332333233233323323332333233233323323332332333233323323332... A natural conjecture is that the mean of these differences is asymptotic to e. The glitches are positioned between terms n and n+1, for values of n which begin 2,4,8,12,16,19,23,26,30,34,37,41,44,48,.... If one counts the 3's immediately preceding the 2's in the pattern of differences, one gets the "derivative" sequence 233323233232323323....

Links

Table of n, a(n) for n=1..66.

G. W. Wishard & F. Underwood, Problem 4155, Amer. Math. Monthly, 53 (1946), 471.

Example

a(1)=1 because the continued fraction of 1 is written as 0 + 1/1 = (0;1).
a(3)=5 because (1 + 4 + 27)/27 = 1 + 5/27 = 1 + 1/(5 + 1/(2 + 1/2)) = (1;5,2,2).

Mathematica

a[1]=1; a[n_] := ContinuedFraction[Sum[k^k, {k, 1, n}][[2]]; Array[a, 100]

Prog

(PARI) a(n) = if (n==1, 1, contfrac(sum(k=1, n, k^k)/n^n)[2]); \\ Michel Marcus, Sep 23 2018

Crossrefs

Cf. A001923, A022843.

Sequence in context: A310574 A059659 A140459 * A139132 A295068 A022435

Adjacent sequences: A319512 A319513 A319514 * A319516 A319517 A319518

Keyword

nonn,easy

Author

Greg Huber, Sep 21 2018

Status

approved