A027878 a(n) = Product_{i=1..n} (10^i - 1).

1, 9, 891, 890109, 8900199891, 890011088900109, 890010198889020099891, 8900101098880002109889900109, 890010100987899112108987901010099891, 890010100097889011121088788901111989989900109

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 0..50

Crossrefs

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027879 (q=11), A027880 (q=12).

Cf. A132038, A132326.

Programs

Magma

[1] cat [&*[10^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

Mathematica

Table[Product[10^i-1,{i,n}],{n,0,10}] (* Harvey P. Dale, Aug 15 2011 *)
Abs@QPochhammer[10, 10, Range[0, 30]] (* G. C. Greubel, Nov 24 2015 *)

PARI

a(n) = prod(k=1, n, 10^k - 1) \\ Altug Alkan, Nov 25 2015

Formula

a(n) ~ c * 10^(n*(n+1)/2), where c = Product_{k>=1} (1-1/10^k) = A132038 = 0.890010099998999000000100009999999989999900000000... . - Vaclav Kotesovec, Nov 21 2015

3^n*(11)^(floor(n/2)) divides a(n) for n>=0. - G. C. Greubel, Nov 24 2015

Equals 10^(binomial(n+1,2))*(1/10;1/10){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015

G.f.: Sum_{n>=0} 10^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 10^k*x). - Ilya Gutkovskiy, May 22 2017

From Amiram Eldar, May 07 2023: (Start)

Sum_{n>=0} 1/a(n) = A132326.

Sum_{n>=0} (-1)^n/a(n) = A132038. (End)