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Search: a193330 -id:a193330
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Number of divisors of n^2 + 1.
+10
26
1, 2, 2, 4, 2, 4, 2, 6, 4, 4, 2, 4, 4, 8, 2, 4, 2, 8, 6, 4, 2, 8, 4, 8, 2, 4, 2, 8, 4, 4, 4, 8, 6, 8, 4, 4, 2, 8, 6, 4, 2, 6, 4, 12, 4, 4, 4, 16, 4, 4, 4, 4, 4, 8, 2, 8, 2, 16, 4, 4, 4, 4, 4, 8, 4, 4, 2, 8, 8, 4, 6, 4, 8, 16, 2, 8, 4, 8, 4, 4, 4, 8, 6, 16, 2
OFFSET
0,2
LINKS
FORMULA
a(n) = A000005(A002522(n)). - Michel Marcus, Mar 16 2018
EXAMPLE
a(7) = 6 because 7^2 + 1 = 50 and the 6 divisors are {1, 2, 5, 10, 25, 50}.
MAPLE
with(numtheory):for n from 0 to 110 do:n1:=nops(divisors(n^2+1)):s:=0:for m from 1 to n1 do: s:=s+1:od: printf(`%d, `, s):od:
MATHEMATICA
Array[DivisorSigma[0, #^2 + 1] &, 85, 0] (* Michael De Vlieger, Mar 17 2018 *)
PROG
(PARI) a(n) = numdiv(n^2+1); \\ Michel Marcus, Mar 16 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jul 28 2011
STATUS
approved
Number of distinct prime factors of n^2+1.
+10
9
1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 1, 3, 2, 2, 1, 3, 2, 3, 1, 2, 1, 3, 2, 2, 2, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 2, 3, 1, 3, 1, 3, 2, 2, 2, 2, 2, 3, 2, 2, 1, 3, 2, 2, 2, 2, 3, 4, 1, 3, 2, 3, 2, 2, 2, 3, 2, 4, 1, 2, 2, 3, 2, 3, 1, 3, 2, 3, 1, 2, 2, 3, 3, 3, 2, 2, 2
OFFSET
1,3
COMMENTS
a(n) is also the number of distinct prime factors, up to multiplication by units, of n + i in the ring of Gaussian integers. - Jason Kimberley, Dec 17 2011
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = A001221(n^2+1).
EXAMPLE
a(3) = 2 because 3^2+1 = 2*5.
MAPLE
a:= n-> nops(select(isprime, numtheory[divisors](n^2+1))):
seq(a(n), n=1..100); # Alois P. Heinz, Dec 06 2020
MATHEMATICA
a[n_]:=Length[FactorInteger[n^2 + 1]]
PROG
(PARI) a(n)=omega(n^2+1) \\ Charles R Greathouse IV, Jul 31 2011
CROSSREFS
Cf. A193330 (counted with multiplicity).
KEYWORD
nonn
AUTHOR
Kent Horvath (kenthorvath(AT)gmail.com), May 10 2007
STATUS
approved
Number of prime factors of n^4 + 1, counted with multiplicity.
+10
3
0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 2, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 2, 2, 2, 1, 3, 2, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 2, 2, 2, 3, 2, 4, 3, 3, 1, 3, 1, 3, 2, 3, 2, 3, 1, 3, 1, 2, 2, 4, 3, 2, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 2, 2, 1, 3, 3, 4, 2, 4, 1, 2, 1, 4, 2, 4, 2, 3, 1, 3, 1, 3, 2, 3, 2, 4, 3, 3, 2, 3, 2, 3, 2, 2, 3, 2, 1, 3, 2, 3, 3, 4, 2, 2, 2, 2, 2, 3, 1
OFFSET
0,4
COMMENTS
This is to A193330 as A002523(n) = n^4+1 is to A002522(n) = n^2 + 1. a(n) = 2 when n^4+1 is prime, iff n is in A037896.
LINKS
FORMULA
a(n) = A001222(A002523(n)) = bigomega(n^4+1) or Omega(n^4+1).
EXAMPLE
a(9) = 3 because 9^4+1 = 6562 = 2 * 17 * 193, which has 3 prime factors, counted with multiplicity
MATHEMATICA
Join[{0}, Table[Total[Transpose[FactorInteger[n^4 + 1]][[2]]], {n, 100}]] (* T. D. Noe, Aug 10 2011 *)
Join[{0}, Table[PrimeOmega[n^4+1], {n, 120}]] (* Harvey P. Dale, Sep 25 2012 *)
PROG
(PARI) a(n) = bigomega(n^4+1); \\ Michel Marcus, Feb 09 2020
(Magma) [0] cat [&+[p[2]: p in Factorization(n^4+1)]:n in [1..120]]; // Marius A. Burtea, Feb 09 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, Aug 09 2011
STATUS
approved
Smallest integer m such that m^2 + 1 has exactly n prime factors, counted with multiplicity.
+10
3
1, 3, 7, 43, 57, 307, 1068, 2943, 12943, 45807, 110443, 670807, 2733307, 25670807, 113561432, 123327057, 657922943, 17213170807, 7200891693, 148802454193
OFFSET
1,2
COMMENTS
Or, first occurrences of n in A193330.
Is the sequence finite?
a(n) exists for arbitrarily large n, and in particular a(n+k) < A185952(n) by the Chinese Remainder Theorem and the fact that -1 is a square mod the primes in A002313, for some k >= 0. Probably a(n) exists for each n. - Charles R Greathouse IV, Apr 21 2015
From Jon E. Schoenfield, Jun 14-15 2015: (Start)
Numbers of the form m^2+1 cannot be divisible by 3; they may be divisible by 2 (but not 2^2), and the only other numbers they can have as prime factors are 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ..., i.e., the terms of A002144. This explains why 5 tends to appear with such high multiplicity in the factorizations of a(n)^2+1, as numbers with a higher multiplicity of the prime factor 5 are more likely to be small enough to be the smallest integer with n prime factors than numbers whose n prime factors (counted with multiplicity) are mostly larger than 5. Having 2 as one of the n prime factors is also an advantage, which accounts for the predominance of odd terms, yielding even values of m^2+1.
For k>1, if m^2+1 is divisible by 5^k, then there are only two possible residues of m mod 5^k; e.g., if m^2+1 is divisible by 5^4, then m mod 5^4 must be either 182 or 443. Thus it is not coincidental that the last few digits of some terms also appear as the last few digits of other terms, e.g., terms ending in 443 or 443+500 = 943, or in 182+125 = 307 or 182+625 = 807. (End)
EXAMPLE
a(1)=1 because 1^2+1=2(prime),
a(2)=3 because 3^2+1=10=2*5,
a(3)=7 because 7^2+1=50=2*2*5,
...............
a(14)=25670807 because 25670807^2+1=2*5^11*149*45289.
MATHEMATICA
Table[m = 1; While[PrimeOmega[m^2 + 1] != n, m++]; m, {n, 12}] (* Michael De Vlieger, Apr 21 2015 *)
PROG
(PARI) a(n)=my(m); while(bigomega(m++^2+1)!=n, ); m \\ Charles R Greathouse IV, Apr 21 2015
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Zak Seidov, Apr 21 2015
EXTENSIONS
a(15)-a(17) from Jon E. Schoenfield, Jun 14 2015
a(18)-a(19) from Jon E. Schoenfield, Jun 15 2015
a(20) from Jon E. Schoenfield, Jul 10 2015
STATUS
approved
Number of prime factors of n^8 + 1, counted with multiplicity.
+10
2
0, 1, 1, 3, 1, 3, 2, 3, 3, 2, 2, 3, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 4, 3, 3, 2, 6, 2, 4, 3, 3, 2, 2, 2, 4, 3, 3, 2, 4, 6, 3, 2, 2, 4, 3, 3, 2, 3, 3, 2, 2, 2, 2, 3, 3, 2, 5, 2, 3, 2, 4, 4, 4, 3, 6, 2, 5, 2, 2, 2, 5, 2, 5, 4, 4, 3, 4, 3, 5, 4, 2, 3, 4, 2, 4
OFFSET
0,4
COMMENTS
This is to A193330 as A002523(n) = n^4+1 is to A002522(n) = n^2 + 1, and as A060890(n) = n^8+1 is to A002522(n) = n^2 + 1. a(n) = 1 when n^8+1 is prime, iff n is in {1, 2, 4} unless there is a larger Fermat prime than 65537.
LINKS
FORMULA
a(n) = A001222(A060890(n)) = bigomega(n^8+1) or Omega(n^8+1)
EXAMPLE
a(10) = 2 because 10^8 + 1 = 100000001 = 17 * 5882353 has 2 prime factors.
a(40) = 6 because 40^8 + 1 = 6553600000001 = 17^2 * 113 * 337 * 641 * 929 has 6 prime factors (with multiplicity) and is the smallest example not squarefree.
MATHEMATICA
Join[{0}, Table[Total[Transpose[FactorInteger[n^8 + 1]][[2]]], {n, 50}]]
PrimeOmega[Range[0, 90]^8+1] (* Harvey P. Dale, May 27 2018 *)
PROG
(PARI) a(n) = bigomega(n^8+1); \\ Michel Marcus, Feb 09 2020
(Magma) [0] cat [&+[p[2]: p in Factorization(n^8+1)]:n in [1..90]]; // Marius A. Burtea, Feb 09 2020
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, Aug 10 2011
STATUS
approved
Numbers n such that n and n^2+1 have the same number of prime factors (including multiplicities).
+10
2
2, 9, 15, 18, 22, 25, 27, 34, 35, 39, 46, 49, 51, 58, 62, 63, 65, 69, 70, 75, 85, 86, 95, 98, 105, 106, 121, 125, 132, 138, 141, 145, 147, 148, 153, 158, 159, 166, 169, 172, 174, 178, 194, 201, 202, 205, 209, 212, 214, 219, 221, 226, 254, 262, 274, 282, 285, 289, 298, 299
OFFSET
1,1
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
FORMULA
Numbers n such that bigomega(n) = bigomega(n^2+1).
MAPLE
with(numtheory): A272044:=n->`if`(bigomega(n)=bigomega(n^2+1), n, NULL): seq(A272044(n), n=1..500); # Wesley Ivan Hurt, Apr 19 2016
MATHEMATICA
Select[Range@ 300, PrimeOmega[#^2 + 1] == PrimeOmega@ # &] (* Michael De Vlieger, Apr 19 2016 *)
PROG
(PARI) is(n)=bigomega(n)==bigomega(n^2+1) \\ Charles R Greathouse IV, Apr 18 2016
CROSSREFS
Cf. A001222 (bigomega), A193330.
KEYWORD
nonn
AUTHOR
Benjamin Przybocki, Apr 18 2016
STATUS
approved
Number of composite divisors of n^2+1.
+10
1
0, 0, 1, 0, 1, 0, 3, 1, 1, 0, 1, 1, 4, 0, 1, 0, 4, 3, 1, 0, 4, 1, 4, 0, 1, 0, 4, 1, 1, 1, 4, 3, 4, 1, 1, 0, 4, 3, 1, 0, 3, 1, 8, 1, 1, 1, 11, 1, 1, 1, 1, 1, 4, 0, 4, 0, 12, 1, 1, 1, 1, 1, 4, 1, 1, 0, 4, 5, 1, 3, 1, 4, 11, 0, 4, 1, 4, 1, 1, 1, 4, 3, 11, 0, 1, 1
OFFSET
1,7
LINKS
FORMULA
a(n) = A055212(A002522(n)).
EXAMPLE
a(7) = 3 because the composite divisors of 7^2+1 are 10, 25, 50.
MATHEMATICA
Table[ Count[ PrimeQ[ Divisors[n^2+1] ], False] - 1, {n, 1, 105} ]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michel Lagneau, Aug 26 2015
STATUS
approved
Array, by antidiagonals, A(k,n) is the number of prime factors of n^(2^k) + 1, counted with multiplicity.
+10
0
0, 0, 1, 0, 1, 1, 0, 1, 1, 2, 0, 1, 1, 2, 1, 0, 1, 1, 3, 1, 2, 0, 1, 2, 2, 1, 2, 1, 0, 1, 2, 2, 2, 3, 1, 3, 0, 1, 2, 2, 2, 3, 2, 2, 2, 0, 1, 2, 6, 2, 4, 3, 3, 2, 2, 0, 1, 3, 5, 2, 4, 3, 3, 3, 3, 1, 3, 0, 1, 4, 7, 3, 4, 3, 4, 3, 2, 2, 2, 1, 0, 1, 5
OFFSET
0,10
COMMENTS
The main diagonal A(n,n) = number of prime factors of n^(2^n) + 1, counted with multiplicity, begins 0, 1, 1, 3, 2, 4, 3, 6, 6.
EXAMPLE
A(4,5) = 3 because 1+5^16 = 152587890626 = 2 * 2593 * 29423041, which has 3 prime factors. The array begins:
================================================================
....|n=0|n=1|n=2|n=3|n=4|n=5|n=6|n=7|n=8|n=9|.10|.11|comment
====|===|===|===|===|===|===|===|===|===|===|===|===|===========
k=0.|.0.|.1.|.1.|.2.|.1.|.2.|.1.|.3.|.2.|.2.|.1.|.3.|A001222
k=1.|.0.|.1.|.1.|.2.|.1.|.2.|.1.|.3.|.2.|.2.|.1.|.2.|A193330
k=2.|.0.|.1.|.1.|.2.|.1.|.2.|.1.|.2.|.2.|.3.|.2.|.2.|A193929
k=3.|.0.|.1.|.1.|.3.|.1.|.3.|.2.|.3.|.3.|.2.|.2.|.3.|A194003
k=4.|.0.|.1.|.1.|.2.|.2.|.3.|.3.|.3.|.3.|.2.|.5.|.3.|not in OEIS
k=5.|.0.|.1.|.2.|.2.|.2.|.4.|.3.|.4.|.3.|.2.|.4.|.4.|not in OEIS
================================================================
KEYWORD
nonn,hard,tabl
AUTHOR
Jonathan Vos Post, Aug 11 2011
EXTENSIONS
Edited by Alois P. Heinz, Aug 11 2011
More terms from Max Alekseyev, Sep 09 2011
STATUS
approved

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