A262222 Decimal representations of hexadecimal numbers that can be misinterpreted as decimal numbers in scientific E notation.

480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 736, 737, 738, 739, 740, 741, 742, 743, 744, 745, 992, 993, 994, 995, 996, 997, 998, 999, 1000, 1001, 1248, 1249, 1250, 1251, 1252, 1253, 1254, 1255, 1256, 1257, 1504, 1505, 1506, 1507, 1508, 1509, 1510, 1511

Offset: 1

Author

Adam J.T. Partridge, Sep 16 2015

Keywords

• nonn
• base

Comments

Hexadecimal numbers containing no alpha digits other than a single "e", where the "e" is not the first or last digit (e.g., 3e12), may be misinterpreted as numbers in scientific E notation.

These numbers are especially troublesome when importing hexadecimal numbers from CSV files into Microsoft Excel.

These numbers can be written as 16^(i+1)a + 14*16^i + b where a and b are members of A102489 with a>0, b<16^i and i>0.

Numbers whose hexadecimal representation matches the regular expression [1-9][0-9]*e[0-9]+. - Eric M. Schmidt, Sep 28 2023

Examples

			480_10 = 1e0_16 and 1e0 = 1 in E notation.
739_10 = 2e3_16 and 2e3 = 2000_10 in E notation.
		

Links

• Christian Perfect, Table of n, a(n) for n = 1..99999
• Microsoft Community, Unable to disable scientific notation in Microsoft Excel.
• Wikipedia, Scientific E notation.

Crossrefs

A subset of A102490.

Programs

C

#include 
#define DIGIT_E 14
bool isA262222(int k)
{
    if (k <= 0 || k % 16 == DIGIT_E) return false;
    bool foundE = false;
    int digit;
    while (k > 0) {
        digit = k % 16;
        if (digit == DIGIT_E) {
            if (foundE) return false;
            foundE = true;
        }
        else if (digit > 9) return false;
        k /= 16;
    }
    return foundE && digit != DIGIT_E;
} // Eric M. Schmidt, Sep 28 2023

Python

from itertools import count,product
# every string of d characters with exactly one 'e' in it, and all the other characters digits 0-9, in ascending lexicographic order
def mids(d):
    if d<1:
        raise Exception("d<1")
    if d==1:
        yield 'e'
        return
    for i in range(0,10):
        for m in mids(d-1):
            yield str(i)+m
    for i in range(10**(d-1)):
        yield 'e'+str(i).zfill(d-1)
def a_generator():
    for d in count(1):
        for start in range(1,10): # for each leading digit 1-9
            for mid in mids(d): # for all possible middles made of d characters, containing exactly one 'e'
                for end in range(10): #for each possible final digit, 0-9
                    s = '{}{}{}'.format(start,''.join(mid),end)
                    i = int(s,16)
                    yield i
a262222 = a_generator()
[next(a262222) for  in range(48)] # _Christian Perfect, Oct 20 2015