A180278
Smallest nonnegative integer k such that k^2 + 1 has exactly n distinct prime factors.
Original entry on oeis.org
0, 1, 3, 13, 47, 447, 2163, 24263, 241727, 2923783, 16485763, 169053487, 4535472963, 36316463227, 879728844873, 4476534430363, 119919330795347, 1374445897718223, 106298577886531087
Offset: 0
a(2) = 3 because the 2 distinct prime factors of 3^2 + 1 are {2, 5};
a(10) = 16485763 because the 10 distinct prime factors of 16485763^2 + 1 are {2, 5, 13, 17, 29, 37, 41, 73, 149, 257}.
-
a[n_] := a[n] = Module[{k = 1}, If[n == 0, Return[0]]; Monitor[While[PrimeNu[k^2 + 1] != n, k++]; k, {n, k}]]; Table[a[n], {n, 0, 8}] (* Robert P. P. McKone, Sep 13 2023 *)
-
a(n)=for(k=0, oo, if(omega(k^2+1) == n, return(k))) \\ Andrew Howroyd, Sep 12 2023
-
from itertools import count
from sympy import factorint
def A180278(n):
return next(k for k in count() if len(factorint(k**2+1)) == n) # Pontus von Brömssen, Sep 12 2023
a(9), a(10) and example corrected; a(11) added by
Donovan Johnson, Aug 27 2012
A258929
a(n) is the unique even-valued residue modulo 5^n of a number m such that m^2+1 is divisible by 5^n.
Original entry on oeis.org
2, 18, 68, 182, 1068, 1068, 32318, 280182, 280182, 3626068, 23157318, 120813568, 1097376068, 1097376068, 11109655182, 49925501068, 355101282318, 355101282318, 15613890344818, 15613890344818, 365855836217682, 2273204469030182, 2273204469030182, 49956920289342682
Offset: 1
If m^2+1 is divisible by 5, then m mod 5 is either 2 or 3; the even value is 2, so a(1)=2.
If m^2+1 is divisible by 5^2, then m mod 5^2 is either 7 or 18; the even value is 18, so a(2)=18.
If m^2+1 is divisible by 5^3, then m mod 5^3 is either 57 or 68; the even value is 68, so a(3)=68.
A259266
a(n) is the unique odd-valued residue modulo 5^n of a number m such that m^2+1 is divisible by 5^n.
Original entry on oeis.org
3, 7, 57, 443, 2057, 14557, 45807, 110443, 1672943, 6139557, 25670807, 123327057, 123327057, 5006139557, 19407922943, 102662389557, 407838170807, 3459595983307, 3459595983307, 79753541295807, 110981321985443, 110981321985443, 9647724486047943, 9647724486047943
Offset: 1
If m^2+1 is divisible by 5, then m mod 5 is either 2 or 3; the odd value is 3, so a(1)=3.
If m^2+1 is divisible by 5^2, then m mod 5^2 is either 7 or 18; the odd value is 7, so a(2)=7.
If m^2+1 is divisible by 5^3, then m mod 5^3 is either 57 or 68; the odd value is 57, so a(3)=57.
A138769
a(n) = least positive integer k such that k^2+3 is divisible by at least n distinct primes.
Original entry on oeis.org
1, 3, 9, 33, 201, 1125, 5259, 98481, 1176579, 4970985, 83471355, 607500315, 20298622815, 302065005093, 2979977447571, 46728566085441, 541457057096937, 13094093041014057, 231069516389617197, 5992213273680818217
Offset: 1
a(3)=9 because 1^2+3=2*2, 2^2+3=7, 3^2+3=2*2*3, 4^2+3=19, 5^2+3=2*2*7, 6^2+3=3*13, 7^2+3=2*2*13, 8^2+3=67 have at most 2 distinct prime divisors, while 9^2+3=2*2*3*7 has 3 distinct prime divisors.
- Elgin H. Johnston, Problem 1792, Matematikolimpiyatokulu, Solution, 2009, p. 149. [broken link]
- H. A. ShahAli, Problem 1792, Math. Magazine, vol. 81, No. 2, 2008, p. 155.
-
n:=7: with(numtheory): for k while nops(factorset(k^2+3)) < n do end do: a[n]:=k; A[n]:=factorset(k^2+3); # yields a(7) as well as its 7 prime divisors; change the value of n to obtain other terms.
-
a[n_] := Block[{k=1}, While[PrimeNu[k^2 + 3] != n, k++]; k]; Array[a, 8] (* Giovanni Resta, Nov 29 2019 *)
Showing 1-4 of 4 results.
Comments