|
Search: id:A052379
|
|
|
| A052379 |
|
Number of integers from 1 to 10^(n+1)-1 that lack 0 and 1 as a digit. |
|
+0
8
|
|
| 8, 72, 584, 4680, 37448, 299592, 2396744, 19173960, 153391688, 1227133512, 9817068104, 78536544840, 628292358728, 5026338869832, 40210710958664, 321685687669320, 2573485501354568, 20587884010836552, 164703072086692424
(list; graph; listen)
|
|
|
OFFSET
|
0,1
|
|
|
FORMULA
|
a(n) = 8*a(n-1) + 8
Conjecture: a(n)=A023001(n+2)-1 . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 18 2007. Comment from Vim Wenders (vim(AT)gmx.li), Mar 26 2008: The conjecture is true: the g.f. leads to the closed form a(n) = -8/7*(1^n) + 64/7*(8^n) = (-8 + 64*8^n)/7 = (-8 + 8^(n+2))/7 = (8^(n+2) - 1)/7 -1 = A023001(n+2)-1.
G.f.: 8/((-1+x)*(-1+8*x)). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 19 2007
|
|
EXAMPLE
|
For n=1, the numbers from 1 to 99 which *have* 0 or 1 as a digit are the numbers 1, and 10, 20, 30, ..., 90, and 11, 12, ..., 18, 19, and 21, 31, ..., 91. So a(1) = 99 - 27 = 72.
|
|
MAPLE
|
a:=n->sum (8^j, j=1..n): seq(a(n), n=1..19); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 03 2007
|
|
CROSSREFS
|
Cf. A024101, A052386.
Sequence in context: A082141 A054615 A111919 this_sequence A062541 A057091 A055275
Adjacent sequences: A052376 A052377 A052378 this_sequence A052380 A052381 A052382
|
|
KEYWORD
|
easy,nonn,base
|
|
AUTHOR
|
Odimar Fabeny (fabeny(AT)braznet.com.br), Mar 12 2000
|
|
EXTENSIONS
|
More terms and revised description from James A. Sellers (sellersj(AT)math.psu.edu), Mar 13 2000
|
|
|
Search completed in 0.002 seconds
|
