A340287 Numbers k for which there are A000005(k+1)-1 bases b such that k in base b contains digit b-1.

1, 3, 5, 9, 11, 17, 27, 35, 37, 39, 41, 65, 81, 83, 85, 89, 131, 149, 179, 203, 255, 257, 263, 407, 419, 455, 539, 739, 811, 899, 1031, 1109, 1385, 1619, 1631, 1883, 2819, 3299, 3527, 4133, 4139, 4151, 4919, 5669, 5939, 6299, 7055, 7307, 8303, 9829, 9839, 10661

Offset: 1

Author

Devansh Singh, Jan 03 2021

Keywords

• nonn
• base

Comments

If k written in some base b has b-1 as its least significant digit, then b is a divisor of k+1 (because k = high*b + b-1 means k+1 = b*(high+1)). The present sequence is those k where such divisors are in fact the only bases b where digit b-1 occurs.

In terms of the count of bases in A337496, the divisors give a lower bound A337496(k) >= tau(k+1)-1 (number of divisors except 1). The present sequence is those k at this lower bound.

There are 147 terms to k <= 10^8.

Examples

			1 is a term because A000005(1+1)-1 = 1 such that 1 in base 2 (2|2) contains digit 1 and there are no such bases of 1 which are non-divisors of 1+1.
5 is a term because A000005(5+1)-1 = 3 such that 5 in base 2,3,6 (2|6,3|6 and 6|6) contains digit 1,2,5 respectively and there are no such bases of 5 which are non-divisors of 5+1.
		

Links

• Kevin Ryde, C code searching for terms

Crossrefs

Cf. A000005, A337496.

Programs

Mathematica

baseQ[n_, b_] := MemberQ[IntegerDigits[n, b], b - 1]; q[n_] := Count[Range[2, n + 1], ?(baseQ[n, #] &)] == DivisorSigma[0, n + 1] - 1; Select[Range[1000], q] (* _Amiram Eldar, Jan 03 2021 *)

PARI

isok(k) = sum(b=2, k+1, (#select(x->(x==(b-1)), digits(k, b)))>0) == numdiv(k+1)-1; \\ Michel Marcus, Jan 03 2021

Python

def is_n_with_no_nondivisor_baseb(N):
    return list(filter(n_with_no_nondivisor_baseb,range(1,N+1,2)))
def n_with_no_nondivisor_baseb(n):
    main_base_counter=0
    for b in range(3,((n+1)//2) +1):
        if (n+1)%b!=0:
            main_base_counter=main_base_check(n//b,b)+main_base_counter
            if main_base_counter==1:
                break
    return main_base_counter==0
def main_base_check(m,b):
    while m!=0:
        if m%b == b-1:
            return 1
        m = m//b
    if m==0:
        return 0
print(is_n_with_no_nondivisor_baseb(int(input())))