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%I #20 Dec 08 2021 19:58:32
%S 0,1,3,6,78,686999778,9876799878,89996788896,77779987999896,
%T 589598998999878,999699998689998991,9988894989978899995,
%U 95898999989999989765,999999966989999986978996
%N Triangular numbers whose mean digit value reaches a new maximum.
%C Subsequence of A068808.
%C No triangular number ends in 9, so the mean digit value is always less than 9.
%C 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?
%C a(14) <= 999999966989999986978996. - _David A. Corneth_, Dec 05 2021
%e n a(n) digit sum #dgts mean digit value
%e -- -------------------- --------- ----- ----------------
%e 1 0 0 1 0
%e 2 1 1 1 1
%e 3 3 3 1 3
%e 4 6 6 1 6
%e 5 78 15 2 7.5
%e 6 686999778 69 9 7.66666666666...
%e 7 9876799878 78 10 7.8
%e 8 89996788896 87 11 7.90909090909...
%e 9 77779987999896 111 14 7.92857142857...
%e 10 589598998999878 120 15 8
%e 11 999699998689998991 145 18 8.05555555555...
%e 12 9988894989978899995 154 19 8.10526315789...
%e 13 95898999989999989765 163 20 8.15
%t 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 *)
%o (Python)
%o def meandigval(n): s = str(n); return sum(map(int, s))/len(s)
%o def afind(limit):
%o alst, k, t, record = [], 0, 0, -1
%o while t <= limit:
%o mdv = meandigval(t)
%o if mdv > record:
%o print(t, end=", ")
%o record = mdv
%o k += 1
%o t += k
%o afind(10**14) # _Michael S. Branicky_, Dec 03 2021
%Y Cf. A000217, A068133, A068808, A069669, A069670, A095864.
%K nonn,base,hard,more
%O 1,3
%A _Jon E. Schoenfield_, Dec 03 2021
%E a(14) verified by _Martin Ehrenstein_, Dec 06 2021