A193869
Smallest product of n distinct primes of the form n*k + 1.
Original entry on oeis.org
2, 15, 1729, 32045, 60551711, 85276009, 52814801041129, 1312422595226609, 1130308388231798179, 4182230628909121261, 100166053986652515419641469, 1898732717895963155960377, 1011844196551535741726366525322443
Offset: 1
a(1) = 2
a(2) = 3*5 = 15
a(3) = 7*13*19 = 1729
a(4) = 5*13*17*29 = 32045
a(5) = 11*31*41*61*71 = 60551711
a(6) = 7*13*19*31*37*43 = 85276009
-
Tj := proc(n,k) option remember: local j,p: if(k=0)then return 0:fi: for j from procname(n,k-1)+1 do if(isprime(n*j+1))then return j: fi: od: end: A193869 := proc(n) return mul(n*Tj(n,k)+1,k=1..n): end: seq(A193869(n),n=1..15); # Nathaniel Johnston, Sep 02 2011
A193873
Smallest product of three distinct primes of the form n*k+1.
Original entry on oeis.org
30, 105, 1729, 1105, 13981, 1729, 88537, 50881, 51319, 13981, 137149, 29341, 548497, 88537, 285541, 186337, 3372529, 51319, 18326641, 252601, 1152271, 137149, 1809641, 1366633, 3828001, 548497, 4814857, 645569, 4797703, 285541, 79230049, 4811297
Offset: 1
a(1) = 2*3*5 = 30
a(2) = 3*5*7 = 105
a(3) = 7*13*19 = 1729
a(4) = 5*13*17 = 1105
a(5) = 11*31*41 = 13981
-
a[n_] := Module[{s = {}, c = 0, m = n + 1}, While[c < 3, While[!PrimeQ[m], m += n]; c++; AppendTo[s, m]; m += n]; Times @@ s]; Array[a, 100] (* Amiram Eldar, Jan 17 2025 *)
-
a(n)=my(p,q,k=1);while(!isprime(k+=n),);p=k;while(!isprime(k+=n),);q=k;while(!isprime(k+=n),);p*q*k \\ Charles R Greathouse IV, Sep 03 2011
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).
Original entry on oeis.org
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
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
-
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
Showing 1-3 of 3 results.
Comments