A338784 a(n) is the smallest number with exactly n divisors such that all its divisors end with the same digit (which is necessarily 1).

1, 11, 121, 341, 14641, 3751, 1771561, 13981, 116281, 453871, 25937424601, 153791, 3138428376721, 54918391, 14070001, 852841, 45949729863572161, 4767521, 5559917313492231481, 18608711, 1702470121, 804060162631, 81402749386839761113321, 9381251, 13521270961, 97291279678351, 195468361

Offset: 1

Author

Bernard Schott, Nov 09 2020

Keywords

• nonn
• base

Comments

As 1 is a divisor for each number, all the divisors must end with 1.

Examples

			121 is the smallest number whose 3 divisors (1, 11, 121) end with 1, hence a(3) = 121.
3751 is the smallest number whose 6 divisors (1, 11, 31, 121, 341, 3751) end with 1, hence a(6) = 121.
a(18) = 4767521 = 11^2 * 31^2 * 41 as it has 18 divisors all of which end in 1. - _David A. Corneth_, Nov 09 2020
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..966
• Project Euler, Problem 474: Last digits of divisors.

Crossrefs

Cf. A001020, A030430, A092609, A330348.

Subsequence of A004615.

Programs

PARI

a(n) = {my(pr); if(n==1, return(1)); if(isprime(n), return(11^(n-1))); forstep(i = 1, oo, 10, f = factor(i); if(numdiv(f) == n, pr = 1; for(j = 1, #f~, if(f[j, 1]%10 != 1, pr = 0; next(2) ) ) ); if(pr, return(i)); ) } \\ David A. Corneth, Nov 09 2020

Formula

If n is prime p, then a(p) = 11^(p-1) = A001020(p-1).

For k>=1, a(2^k) = {Product_m=1..k} A030430(m) = A092609(k).

Extensions

Data corrected by David A. Corneth, Nov 09 2020