OFFSET
1,3
COMMENTS
Subsequence of A068808.
No triangular number ends in 9, so the mean digit value is always less than 9.
Is this sequence finite? Or does the mean digit value approach some upper limit arbitrarily closely without ever reaching it exactly, and, if so, what is that limit?
a(14) <= 999999966989999986978996. - David A. Corneth, Dec 05 2021
EXAMPLE
n a(n) digit sum #dgts mean digit value
-- -------------------- --------- ----- ----------------
1 0 0 1 0
2 1 1 1 1
3 3 3 1 3
4 6 6 1 6
5 78 15 2 7.5
6 686999778 69 9 7.66666666666...
7 9876799878 78 10 7.8
8 89996788896 87 11 7.90909090909...
9 77779987999896 111 14 7.92857142857...
10 589598998999878 120 15 8
11 999699998689998991 145 18 8.05555555555...
12 9988894989978899995 154 19 8.10526315789...
13 95898999989999989765 163 20 8.15
MATHEMATICA
seq = {}; max = -1; Do[If[(m = Mean @ IntegerDigits[(t = n*(n + 1)/2)]) > max, max = m; AppendTo[seq, t]], {n, 0, 10^6}]; seq (* Amiram Eldar, Dec 03 2021 *)
PROG
(Python)
def meandigval(n): s = str(n); return sum(map(int, s))/len(s)
def afind(limit):
alst, k, t, record = [], 0, 0, -1
while t <= limit:
mdv = meandigval(t)
if mdv > record:
print(t, end=", ")
record = mdv
k += 1
t += k
afind(10**14) # Michael S. Branicky, Dec 03 2021
CROSSREFS
KEYWORD
nonn,base,hard,more
AUTHOR
Jon E. Schoenfield, Dec 03 2021
EXTENSIONS
a(14) verified by Martin Ehrenstein, Dec 06 2021
STATUS
approved