A349960 Values taken by the function A067095 in the order of their appearance.

2, 1, 18, 181, 1817, 18175, 181757, 1817571, 18175719, 181757197, 1817571972, 18175719727, 181757197277, 1817571972772, 18175719727727, 181757197277277, 1817571972772779, 18175719727727795, 181757197277277957, 1817571972772779572, 18175719727727795720, 181757197277277957202

Offset: 1

Author

Bernard Schott, Dec 07 2021

Keywords

• nonn
• base

Comments

a(2) < a(1), but thereafter this function increases monotonically without limit (see Krusemeyer reference).

The record values > 2 of A067095(m) occur when m = 5, 50, 500, 5000, .... This happens precisely when the corresponding numerator A019520(m) goes from 2/4/6/8/10/12/....../999...98 to 2/4/6/8/10/12/....../999...98/1000...00, where here / means concatenation.

If a(n) is a k-digit number (k = A055642(a(n))), then 1.8 * 10^(k-1) < a(n) < 1.9 * 10^(k-1).

If we consider the sequence u(n) = a(n)/10^(k-1) where k = length(a(n)); we have u(n) is increasing with an upper bound 1.9; so, this sequence u(n) is convergent and, conjecture, this limit = 1.81757197277277957... found by Giorgos Kalogeropoulos; now, from this limit, it is possible to get the successive terms of this sequence here.

Examples

			Floor(A019520(5)/A019519(5)) = floor(246810/13579) = floor(18.175859...) = 18, hence, 18 is a term.
		

References

• Mark I. Krusemeyer, George T. Gilbert and Loren C. Larson, A Mathematical Orchard, Problems and Solutions, MAA, 2012, Problem 87, pp. 159-161.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..1001

Crossrefs

Cf. A019519, A019520, A055642, A067095.

Programs

Mathematica

terms=5; f[i_]:=FromDigits@Flatten[IntegerDigits/@i];
k[q_]:=f[Range[2,2q,2]]/f[Range[1,2q,2]];
DeleteDuplicates@Table[Floor[k@n],{n,10^(terms-2)/2}] (* Giorgos Kalogeropoulos, Dec 10 2021 *)

Python

def A349960(n): return 3-n if n <= 2 else int("".join(str(d) for d in range(2,10**(n-2)+1,2)))//int("".join(str(d) for d in range(1,10**(n-2),2))) # Chai Wah Wu, Dec 10 2021
from itertools import count
def A349960(n): # a more efficient implementation
    if n <= 2:
        return 3-n
    a, b = '', ''
    for i in count(1,2):
        a += str(i)
        b += str(i+1)
        ai, bi = int(a), int(b)
        if len(a)+n-2 == len(b): return bi//ai
        m = 10**(n-2-len(b)+len(a))
        lb = bi*m//(ai+1)
        ub = (bi+1)*m//ai
        if lb == ub: return lb # Chai Wah Wu, Dec 10 2021

Formula

a(n) = floor(k((n + 6)/2)*10^(n - 1 - ceiling(log_10(k((n + 6)/2))))) for k(n) = A019520(n)/A019519(n) and n >= 2 (conjectured). - Giorgos Kalogeropoulos, Dec 10 2021

Extensions

a(5)-a(7) from Michel Marcus, Dec 07 2021

a(8)-a(9) from Martin Ehrenstein, Dec 10 2021

a(10)-a(22) from Chai Wah Wu, Dec 10 2021