A363271 Vertical sum of n in base 10.

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

Author

Simon Plouffe, May 24 2023

Keywords

• nonn
• base

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.

Examples

			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.
		

Crossrefs

Cf. A021085 (10/81), A089400 (binary analog).

Programs

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).