A131835 Numbers starting with 1.
1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142
Offset: 1
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- Bryan Brown, Michael Dairyko, Stephan Ramon Garcia, Bob Lutz and Michael Someck, Four quotient set gems, The American Mathematical Monthly, Vol. 121, No. 7 (2014), pp. 590-598; arXiv preprint, arXiv:1312.1036 [math.NT], 2013.
- Index entries for 10-automatic sequences.
Crossrefs
Programs
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Haskell
a131835 n = a131835_list !! (n-1) a131835_list = concat $ iterate (concatMap (\x -> map (+ 10 * x) [0..9])) [1] -- Reinhard Zumkeller, Jul 16 2014 -
Maple
isA131835 := proc(n) if op(-1,convert(n,base,10)) = 1 then true; else false ; fi ; end: for n from 1 to 300 do if isA131835(n) then printf("%d, ",n) ; fi ; od : # R. J. Mathar, Jul 24 2007
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Mathematica
Select[Range[150], IntegerDigits[#][[1]] == 1 &] (* Amiram Eldar, Feb 27 2021 *)
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PARI
a(n, {base=10}) = my (o=1); while (n>o, n-=o; o*=base); return (o+n-1) \\ Rémy Sigrist, Jun 23 2017 -
PARI
a(n) = n--; s = #digits(9*n+1); n + 8 * (10^(s-1))/9 + 1/9 \\ David A. Corneth, Jun 23 2017
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PARI
nxt(n) = my(d = digits(n+1)); if(d[1]==1, n+1, 10^#d) \\ David A. Corneth, Jun 23 2017
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Python
def A131835(n): return n+(10**(len(str(9*n-8))-1)<<3)//9 # Chai Wah Wu, Dec 07 2024
Formula
A000030(a(n)) = 1. - Reinhard Zumkeller, Jul 16 2014
a(A002275(n)+1) = 10^n for any n >= 0. - Rémy Sigrist, Jun 23 2017
a(n) = n + (8*10^floor(log_10(9*n-8))-8)/9. - Alan Michael Gómez Calderón, May 16 2023
Extensions
More terms from R. J. Mathar, Jul 24 2007
Comments