A083289 Least k such that 10^n+k is a brilliant number (cf. A078972).

3, 0, 21, 3, 201, 13, 18081, 43, 140049, 81, 600009, 147, 6000009, 73, 380000361, 3, 1400000049, 831, 14000000049, 49, 380000000361, 987, 600000000009, 691, 78000000001521, 183, 740000000001369, 4153, 6200000000000961, 279

Offset: 0

Author

Jason Earls, Jun 03 2003

Keywords

• base
• nonn

Comments

If n is an even positive exponent, then a(n) is the first prime greater than 10^(n/2) squared less 10^n.

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..42
• Dario Alejandro Alpern, Brilliant numbers

Crossrefs

Cf. A078972, A084475, A084476.

Programs

Mathematica

NextPrim[n_] := Block[{k = n + 1}, While[ !PrimeQ[k], k++ ]; k]; LengthBase10[n_] := Floor[ Log[10, n] + 1]; f[n_] := Block[{k = 0}, If[ EvenQ[n] && n > 1, NextPrim[ 10^(n/2)]^2 - 10^n, While[fi = FactorInteger[10^n + k]; Plus @@ Flatten[ Table[ # [[2]], {1}] & /@ fi] != 2 || Length[ Union[ LengthBase10 /@ Flatten[ Table[ # [[1]], {1}] & /@ fi]]] != 1, k++ ]; k]]; Table[ f[n], {n, 0, 30}]

Python

from sympy import nextprime, factorint
def A083289(n):
    a, b = divmod(n,2)
    c, d = 10**n, 10**a
    if b == 0: return nextprime(d)**2-c
    k = 0
    while True:
        fs = factorint(c+k,multiple=True)
        if len(fs) == 2 and min(fs) >= d:
            return k
        k += 1 # Chai Wah Wu, Sep 28 2021

Extensions

Edited and extended by Robert G. Wilson v, Jun 27 2003