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A363271
Vertical sum of n in base 10.
0
1, 2, 3, 4, 5, 6, 7, 8, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 6, 7, 8, 9, 10, 11, 12
OFFSET
1,2
COMMENTS
The Sum_{n>=1} a(n)/10^n = 10/81, it is the vertical sum of each integer. The pattern is easy to see but apparently impossible for a program to find any closed form or recurrence. The sequence is generated by adding each integer with an offset of 1 at each step.
If you sum integers with each term divided by 10^n, at n = 9 there are 2 terms in the column 9 + 1 = 10 which is a(10).
Here is the actual sum:
.100000000000000000000
.020000000000000000000
.003000000000000000000
.000400000000000000000
.000050000000000000000
.000006000000000000000
.000000700000000000000
.000000080000000000000
.000000009000000000000
.000000001000000000000
.000000000110000000000
.000000000012000000000
.000000000001300000000
.000000000000140000000
.000000000000015000000
.000000000000001600000
.000000000000000170000
.000000000000000018000
.000000000000000001900
.000000000000000000200
.000000000000000000021
.000000000000000000002
...
By adding each column we get a(n), which explains why a(9) = 10.
EXAMPLE
The original sequence is 1 2 3 4 5 6 7 8 9 10 11 12 ... but when we sum digit per digit (in base 10) the sequence is not a rational fraction.
MAPLE
p:=proc(v) local n, aa, nn, s, k, t;
aa := v;
nn := nops(aa);
s := [seq(1 + aa[k]/10^k,
k = 1 .. nops(aa))];
[seq(sum(trunc(10*frac(10^t*s[k])),
k = 1 .. nops(aa)),
t = 0 .. nops(aa))]
end;
# enter a sequence like a(n) = [1, 2, 3, 4, ...] it will return a sequence r such that sum(r(n)/10^n) is equal to sum(a(n)/10^n).
CROSSREFS
Cf. A021085 (10/81), A089400 (binary analog).
Sequence in context: A110370 A249517 A353907 * A031044 A367306 A194978
KEYWORD
nonn,base
AUTHOR
Simon Plouffe, May 24 2023
STATUS
approved