This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A319515 #36 Jun 19 2025 09:19:58
%S A319515 1,4,5,8,10,13,16,19,21,24,27,30,32,35,38,41,43,46,49,51,54,57,60,62,
%T A319515 65,68,70,73,76,79,81,84,87,90,92,95,98,100,103,106,109,111,114,117,
%U A319515 119,122,125,128,130,133,136,138,141,144,147,149,152,155,158,160,163,166,168,171,174,177
%N A319515 Second term of the simple continued fraction of (Sum_{k=1..n} k^k)/(n^n).
%C A319515 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....
%H A319515 G. W. Wishard and F. Underwood, <a href="https://www.jstor.org/stable/2306261">Problem 4155</a>, Amer. Math. Monthly, 53 (1946), 471.
%e A319515 a(1)=1 because the continued fraction of 1 is written as 0 + 1/1 = (0;1).
%e A319515 a(3)=5 because (1 + 4 + 27)/27 = 1 + 5/27 = 1 + 1/(5 + 1/(2 + 1/2)) = (1;5,2,2).
%t A319515 a[1]=1; a[n_] := ContinuedFraction[Sum[k^k, {k, 1, n}]/n^n][[2]]; Array[a, 100]
%o A319515 (PARI) a(n) = if (n==1, 1, contfrac(sum(k=1, n, k^k)/n^n)[2]); \\ _Michel Marcus_, Sep 23 2018
%Y A319515 Cf. A001923, A022843.
%K A319515 nonn,easy
%O A319515 1,2
%A A319515 _Greg Huber_, Sep 21 2018