A338784 - OEIS

%I #33 Nov 10 2020 03:50:17

%S 1,11,121,341,14641,3751,1771561,13981,116281,453871,25937424601,

%T 153791,3138428376721,54918391,14070001,852841,45949729863572161,

%U 4767521,5559917313492231481,18608711,1702470121,804060162631,81402749386839761113321,9381251,13521270961,97291279678351,195468361

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

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

%H David A. Corneth, <a href="/A338784/b338784.txt">Table of n, a(n) for n = 1..966</a>

%H Project Euler, <a href="https://projecteuler.net/problem=474">Problem 474: Last digits of divisors</a>.

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

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

%e 121 is the smallest number whose 3 divisors (1, 11, 121) end with 1, hence a(3) = 121.

%e 3751 is the smallest number whose 6 divisors (1, 11, 31, 121, 341, 3751) end with 1, hence a(6) = 121.

%e 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

%o (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

%Y Cf. A001020, A030430, A092609, A330348.

%Y Subsequence of A004615.

%K nonn,base

%O 1,2

%A _Bernard Schott_, Nov 09 2020

%E Data corrected by _David A. Corneth_, Nov 09 2020