

A300000


The sum of the first n terms of the sequence is the concatenation of the first n digits of the sequence, with a(1) = 1.


13



1, 10, 99, 999, 9990, 99900, 999000, 9990000, 99900000, 999000000, 9990000000, 99899999991, 998999999919, 9989999999190, 99899999991900, 998999999918991, 9989999999189910, 99899999991899109, 998999999918991090, 9989999999189910900, 99899999991899108991, 998999999918991089910, 9989999999189910899100
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

The sequence starts with a(1) = 1 and is always extended with the smallest integer not yet present in the sequence and not leading to a contradiction.
By definition, Sum_{k=1..n} a(k) = c(n) = concatenation of the first n digits of the sequence, therefore a(n) = c(n)  c(n1). For n > 2, this defines a(n) recursively, without the need for solving an implicit equation, as the definition might suggest.  M. F. Hasler, Feb 22 2018
From Michael S. Branicky, Dec 07 2020: (Start)
Each digit 09 eventually appears. The digit 1 appears first in a(1), 0 in a(2), 9 in a(3), 8 in a(12), 2 in a(68), 7 in a(71), 3 in a(2280), 6 in a(2283), 5 in a(2417), and 4 in a(4280).
All ten digits appear in each of a(4280) through a(121000).
Conjecture: a(n) contains all ten digits for n >= 4280. (End)
Conjecture holds through a(169000).  Michael S. Branicky, Jul 08 2022


LINKS

Michael S. Branicky, Table of n, a(n) for n = 1..1001 (terms 1..300 from JeanMarc Falcoz)


FORMULA

a(n) = c(n)  c(n1), where c(n) is the concatenation of the first n digits. c(n) ~ 1.1*10^(n1), and a(n) ~ 0.999*10^(n1).  M. F. Hasler, Feb 22 2018


EXAMPLE

1 + 10 = 11 which is the concatenation of 1 and 1.
1 + 10 + 99 = 110 which is the concatenation of 1, 1 and 0.
1 + 10 + 99 + 999 = 1109 which is the concatenation of 1, 1, 0 and 9.
Otherwise said:
a(3) = concat(1,1,0)  (1 + 10) = 110  11 = 99,
a(4) = concat(1,1,0,9)  (11 + 99) = 1109  110 = 999,
a(5) = concat(1,1,0,9,9)  1109 = 11099  1109 = 9990,
a(6) = concat(1,1,0,9,9,9)  11099 = 99900, etc.  M. F. Hasler, Feb 22 2018


MATHEMATICA

a[1]=1; a[2]=10; a[n_]:=a[n]=FromDigits[Flatten[IntegerDigits/@Table[a[k], {k, n1}]][[;; n]]]Total@Table[a[m], {m, n1}];
Table[a[l], {l, 30}] (* Giorgos Kalogeropoulos, May 20 2019 *)


PROG

(PARI) a(n, show=1, a=1, c=a, d=[c])={for(n=2, n, show&&print1(a", "); a=c+c=c*10+d[1]; d=concat(d[^1], if(n>2, digits(a)))); a} \\ M. F. Hasler, Feb 22 2018
(Python)
def a(n):
alist, c, ckm1 = [1, 10], "110", 11
for k in range(3, n+1):
ck = 10*ckm1 + int(c[k1])
ak, ckm1 = ck  ckm1, ck
c += str(ak)
alist.append(ak)
return alist[n1]
print([a(n) for n in range(1, 24)]) # Michael S. Branicky, Dec 07 2020


CROSSREFS

A299865, A299866, A299867, A299868, A299869, A299870, A299871 and A299872 show the same type of sequence but with a different start.
The partial sums (the sequence c(n) mentioned in the Comments) is A299301.
Sequence in context: A179555 A105694 A179557 * A213454 A000456 A138365
Adjacent sequences: A299997 A299998 A299999 * A300001 A300002 A300003


KEYWORD

nonn,base,nice,easy


AUTHOR

Eric Angelini, Feb 10 2018


STATUS

approved



