A193432 -id:A193432 - cp's OEIS frontend

A333167 a(n) = r_2(n^2 + 1), where r_2(k) is the number of ways of writing k as a sum of 2 squares (A004018).

4, 4, 8, 8, 8, 8, 8, 12, 16, 8, 8, 8, 16, 16, 8, 8, 8, 16, 24, 8, 8, 16, 16, 16, 8, 8, 8, 16, 16, 8, 16, 16, 24, 16, 16, 8, 8, 16, 24, 8, 8, 12, 16, 24, 16, 8, 16, 32, 16, 8, 16, 8, 16, 16, 8, 16, 8, 32, 16, 8, 16, 8, 16, 16, 16, 8, 8, 16, 32, 8, 24, 8, 32, 32

Offset: 0

Amiram Eldar, Mar 09 2020

nonn

			a(0) = r_2(0^2 + 1) = r_2(1) = A004018(1) = 4.

Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 166.

Amiram Eldar, Table of n, a(n) for n = 0..10000
E. J. Scourfield, The divisors of a quadratic polynomial, Glasgow Mathematical Journal, Vol. 5, No. 1 (1961), pp. 8-20.

Cf. A004018, A002522, A193432, A193433, A333169.

Table[SquaresR[2, k^2 + 1], {k, 0, 100}]

a(n) = A004018(A002522(n)).

A193562 Number of divisors of n^4+1.

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 4, 4, 8, 4, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 4, 4, 4, 2, 8, 4, 8, 2, 4, 4, 8, 4, 8, 2, 4, 4, 8, 4, 4, 4, 8, 4, 16, 8, 8, 2, 8, 2, 8, 4, 8, 4, 8, 2, 8, 2, 4, 4, 16, 8, 4, 4, 8, 8, 4, 8, 8, 4, 8, 8, 4, 4, 4, 2, 8, 8, 16, 4, 16, 2, 4, 2, 16, 4

Offset: 0

Jonathan Vos Post, Aug 09 2011

This is to n^4+1 as A193432 is to n^2+1.

a(n) = 2 when n^4+1 is prime, iff n is in A037896.

			a(3) = 4 because 3^4+1 = 82, whose 4 factors are {1, 2, 41, 82}.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000005, A002523, A037896, A193432 (number of divisors of n^2+1).

[NumberOfDivisors(n^4+1):n in [0..90]]; // Marius A. Burtea, Feb 09 2020

DivisorSigma[0,Range[0,90]^4+1] (* Harvey P. Dale, May 05 2013 *)

a(n) = numdiv(n^4+1); \\ Michel Marcus, Feb 09 2020

a(n) = A000005(A002523(n)) = d(n^4+1) (also called tau(n^4+1) or sigma_0(n^4+1)), the number of divisors of n^4+1.

A193929 Number of prime factors of n^4 + 1, counted with multiplicity.

Original entry on oeis.org

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

Jonathan Vos Post, Aug 09 2011

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.

			a(9) = 3 because 9^4+1 = 6562 = 2 * 17 * 193, which has 3 prime factors, counted with multiplicity

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A001222, A002523, A193432, A193562.

[0] cat [&+[p[2]: p in Factorization(n^4+1)]:n in [1..120]]; // Marius A. Burtea, Feb 09 2020

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 *)

a(n) = bigomega(n^4+1); \\ Michel Marcus, Feb 09 2020

a(n) = A001222(A002523(n)) = bigomega(n^4+1) or Omega(n^4+1).

A172438 Numbers k such that tau(k^2+1) - tau(k^2) = 1 where the function tau(k) is the number of positive divisors of k.

Original entry on oeis.org

1, 3, 5, 11, 19, 27, 29, 59, 61, 71, 79, 101, 125, 131, 139, 181, 199, 242, 243, 271, 333, 349, 379, 387, 409, 423, 449, 461, 477, 521, 569, 571, 603, 631, 641, 661, 739, 747, 751, 772, 788, 821, 881, 929, 991, 1017, 1031, 1039, 1051, 1058, 1069, 1075, 1083

Offset: 1

Michel Lagneau, Feb 02 2010

nonn

Square roots of perfect squares in A055927. [Juri-Stepan Gerasimov, Apr 06 2011]

			k=1, tau(2) - tau(1) = 2 - 1 = 1.
k=3, tau(10) - tau(9) = 4 - 3 = 1.
k=5, tau(26) - tau(25) = 4 - 3 = 1.
k=387, tau(149770)- tau(149769) = 16 - 15 = 1.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 840.
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Chap. II. (For inequalities, etc.)

Marius A. Burtea, Table of n, a(n) for n = 1..5587
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
G. E. Andrews, Some debts I owe, Séminaire Lotharingien de Combinatoire, Paper B42a, Issue 42, 2000; see (7.1).
P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., 19 (1919), 75-113.
S. Ramanujan, On The Number Of Divisors Of A Number

Cf. A055927, A000005, A002183, A002182 (records), A005237, A006558.

Cf. A048691, A193432.

[m:m in [1..1100]| #Divisors(m^2+1) - #Divisors(m^2) eq 1]; // Marius A. Burtea, Jul 12 2019

with(numtheory): for n from 1 to 100000 do; if tau(n^2+1)-tau(n^2)= 1 then print(n); else fi ; od;

dsQ[n_]:=Module[{n2=n^2},DivisorSigma[0,n2+1]-DivisorSigma[0,n2]==1]; Select[Range[1200],dsQ] (* Harvey P. Dale, May 05 2011 *)
Select[Sqrt[#]&/@Flatten[Position[Partition[DivisorSigma[0,Range[1200000]],2,1],?(#[[2]]-#[[1]]==1&),1,Heads->False]],IntegerQ] (* _Harvey P. Dale, Apr 09 2022 *)

A194003 Number of prime factors of n^8 + 1, counted with multiplicity.

Original entry on oeis.org

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

Jonathan Vos Post, Aug 10 2011

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.

			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.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A001222, A060890, A002523, A193432, A193562, A193929.

[0] cat [&+[p[2]: p in Factorization(n^8+1)]:n in [1..90]]; // Marius A. Burtea, Feb 09 2020

Join[{0}, Table[Total[Transpose[FactorInteger[n^8 + 1]][[2]]], {n, 50}]]
PrimeOmega[Range[0,90]^8+1] (* Harvey P. Dale, May 27 2018 *)

a(n) = bigomega(n^8+1); \\ Michel Marcus, Feb 09 2020

a(n) = A001222(A060890(n)) = bigomega(n^8+1) or Omega(n^8+1)

A209877 a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).

Original entry on oeis.org

1, 4, 2, 6, 3, 16, 15, 25, 23, 17, 11, 5, 38, 49, 50, 22, 14, 40, 81, 56, 7, 61, 72, 32, 8, 41, 30, 114, 69, 144, 57, 74, 68, 21, 52, 137, 167, 10, 133, 196, 127, 191, 174, 24, 104, 143, 26, 59, 43, 12, 258, 238, 289, 97, 77, 252, 53, 29, 13, 283, 48, 190, 335, 361, 31, 228, 291, 159, 263, 123, 260, 325, 363, 247, 162

Offset: 1

M. F. Hasler, Mar 14 2012

nonn

Also: Square root of -1/4 in Z/pZ, for Pythagorean primes p=A002144(n).
Also: Least m>0 such that the Pythagorean prime p=A002144(n) divides 4(kp +/- m)^2+1 for all k>=0.
In practice these can also be determined by searching the least N^2+1 whose least prime factor is p=A002144(n): For given p, all of these N will have a(n) or p-a(n) as remainder mod 2p.

			a(1)=1 since A002144(1)=5 and 4*1^2+1 is divisible by 5; as a consequence 4*(5k+/-1)^2+1 = 100k^2 +/- 40k + 5 is divisible by 5 for all k.
a(2)=4 since A002144(2)=13 and 4*4^2+1 = 65 is divisible by 13, while 4*1^1+1=5, 4*2^2+1=17 and 4*3^2+1=37 are not. As a consequence, 4*(13k+/-4)^2+1 = 13(...)+4*4^1+1 is divisible by 13 for all k.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A209874.
Cf. A002496, A014442, A085722, A144255, A089120, A193432.
Cf. A002314, A177979.

f:= proc(p) local m;
   if not isprime(p) then return NULL fi;
   m:= numtheory:-msqrt(-1/4, p);
   min(m,p-m);
end proc:
map(f, [seq(i,i=5..1000,4)]); # Robert Israel, Mar 13 2018

f[p_] := Module[{r}, r /. Solve[4 r^2 == -1, r, Modulus -> p] // Min];
f /@ Select[4 Range[300] + 1, PrimeQ] (* Jean-François Alcover, Jul 27 2020 *)

apply(p->lift(sqrt(Mod(-1,p)/4)), A002144)

A333173 a(n) = r_4(n^2 + 1), where r_4(k) is the number of ways of writing k as a sum of 4 squares (A000118).

Original entry on oeis.org

8, 24, 48, 144, 144, 336, 304, 744, 672, 1008, 816, 1488, 1440, 2592, 1584, 2736, 2064, 4320, 3472, 4368, 3216, 6048, 4704, 7776, 4624, 7536, 5424, 10656, 7584, 10128, 7776, 12768, 10416, 15840, 10080, 14736, 10384, 19872, 14736, 18288, 12816, 20904, 16992, 28272

Offset: 0

Amiram Eldar, Mar 09 2020

nonn

			a(0) = r_4(0^2 + 1) = r_4(1) = A000118(1) = 8.

Amiram Eldar, Table of n, a(n) for n = 0..10000
R. Sitaramachandrarao and P. V. Krishnaiah, On the sums Sigma_{n<=x} A(f(n)) and Sigma_{p<=x} A(f(p)), Journal of Number Theory, Vol. 23, No. 2 (1986), pp. 149-168.

Cf. A000118, A002522 A193432, A193433, A333167, A333169.

Table[SquaresR[4, k^2 + 1], {k, 0, 100}]

a(n) = A000118(A002522(n)).

A261609 Number of composite divisors of n^2+1.

Original entry on oeis.org

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

Michel Lagneau, Aug 26 2015

			a(7) = 3 because the composite divisors of 7^2+1 are 10, 25, 50.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A002522, A055212, A128428, A193330, A193432.

Table[ Count[ PrimeQ[ Divisors[n^2+1] ], False] - 1, {n, 1, 105} ]
Table[Count[Divisors[n^2+1],?CompositeQ],{n,100}] (* _Harvey P. Dale, Dec 16 2024 *)

a(n) = A055212(A002522(n)).

A333171 a(n) = Sum_{k=0..n} d(k^2 + 1), where d(k) is the number of divisors of k (A000005).

Original entry on oeis.org

1, 3, 5, 9, 11, 15, 17, 23, 27, 31, 33, 37, 41, 49, 51, 55, 57, 65, 71, 75, 77, 85, 89, 97, 99, 103, 105, 113, 117, 121, 125, 133, 139, 147, 151, 155, 157, 165, 171, 175, 177, 183, 187, 199, 203, 207, 211, 227, 231, 235, 239, 243, 247, 255, 257, 265, 267, 283

Offset: 0

Amiram Eldar, Mar 09 2020

nonn

			a(0) = d(0^1 + 1) = d(1) = 1.
a(1) = d(0^1 + 1) + d(1^1 + 1) = d(1) + d(2) = 1 + 2 = 3.

Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 166.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Christopher Hooley, On the number of divisors of quadratic polynomials, Acta Mathematica, Vol. 110 (1963), pp. 97-114.
James McKee, On the average number of divisors of quadratic polynomials, Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 117. No. 3 (1995), pp. 389-392, alternative link.
James McKee, The average number of divisors of an irreducible quadratic polynomial, Mathematical Proceedings of the Cambridge Philosophical Society. Vol. 126. No. 1. (1999), pp. 17-22.

Partial sums of A193432.

Cf. A000005, A002522.

Accumulate @ Table[DivisorSigma[0, k^2 + 1], {k, 0, 100}]

a(n) = sum(k=0, n, numdiv(k^2+1)); \\ Michel Marcus, Mar 10 2020

a(n) ~ (3/Pi) * n * log(n).

A353008 a(n) is the smallest positive k such that k^2 + 1 has 2*n divisors, or -1 if no such k exists.

Original entry on oeis.org

1, 3, 7, 13, 182, 43, 1068, 47, 268, 443, 15905182, 157, 1832311432, 14557, 16432, 307, 255250280182, 1407, 355101282318, 3307, 92682, 3626068, 21346690797155182, 993, 313932, 120813568, 51982, 16693, 982692130687379186432, 2943, 2444574943897581751068, 2163

Offset: 1

Jon E. Schoenfield, May 15 2022

nonn

From Jon E. Schoenfield, Jun 14 2024: (Start)
For integers k, neither 3 nor 4 ever divides k^2 + 1, so there exists no prime p < 5 such that p^2 divides k^2 + 1.
For n <= 32, the only n for which the 5-adic valuation of a(n)^2 + 1 is not gpf(n) - 1 is n = 16 (see Examples).
Conjecture: a(n) is never -1. (End)

			From _Jon E. Schoenfield_, Jun 14 2024: (Start)
From a(5) = 182 because 182 is the smallest positive integer k such that k^2 + 1 has 2*5 divisors: 182^2 + 1 = 33125 = 5^4 * 53.
a(16) = 307 because 307 is the smallest positive integer k such that k^2 + 1 has 2*16 divisors: 307^2 + 1 = 94250 = 2 * 5^3 * 377.
a(31) = 2444574943897581751068: 2444574943897581751068^2 + 1 = 5975946656331864965715445578098297119140625 = 5^30 * 6416623862896477837609. (End)

Cf. A006530, A112765, A193432, A347193.

a(26), a(29), and a(31) corrected by Jon E. Schoenfield, Jun 14 2024

cp's OEIS Frontend

A333167 a(n) = r_2(n^2 + 1), where r_2(k) is the number of ways of writing k as a sum of 2 squares (A004018).

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A193562 Number of divisors of n^4+1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A193929 Number of prime factors of n^4 + 1, counted with multiplicity.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A172438 Numbers k such that tau(k^2+1) - tau(k^2) = 1 where the function tau(k) is the number of positive divisors of k.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

A194003 Number of prime factors of n^8 + 1, counted with multiplicity.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A209877 a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A333173 a(n) = r_4(n^2 + 1), where r_4(k) is the number of ways of writing k as a sum of 4 squares (A000118).

Views

Author

Keywords