Michel Lagneau - cp's OEIS frontend

A386516 Least k such that k^2+1 contains exactly n distinct prime factors of the form m^2+1 or 0 if no such k exists.

1, 3, 13, 183, 2843, 41323, 57109753, 1929510527, 760999599793

Offset: 1

Michel Lagneau, Jul 24 2025

			a(4)=183 because the prime factors of 183^2+1 are {2, 5, 17, 197} are of the form m^2+1 with m = 1, 2, 4 and 14.

Cf. A002522, A001912, A002496, A005574, A008784, A216784.

with(numtheory):nn:=10^6:
for n from 0 to 6 do :
 ii :=0 :
 for k from 0 to nn while(ii=0) do :
   d:=factorset(k^2+1):n0:=nops(d):it:=0:
    for i from 1 to n0 do:
      c:=d[i]-1:if sqrt(c) = floor(sqrt(c)) then it:=it+1:else fi:
    od:
     if it =n then ii :=1 :printf (`%d %d \n`,n,k):
      else
     fi :
 od:
od :

a[n_]:=Module[{k=0},Until[PrimeNu[k^2+1]==n&&AllTrue[Sqrt[First/@FactorInteger[k^2+1]-1],IntegerQ],k++];k];Array[a,6] (* James C. McMahon, Jul 25 2025 *)

a(7) from Giovanni Resta, Jul 24 2025

a(8)-a(9) from David A. Corneth, Jul 24 2025

A383644 a(n) is the number of zeros in the left half-plane of the Maclaurin polynomial of degree n for exp(z).

Original entry on oeis.org

1, 2, 3, 4, 3, 4, 5, 6, 7, 6, 7, 8, 9, 10, 11, 10, 11, 12, 13, 14, 13, 14, 15, 16, 17, 16, 17, 18, 19, 20, 19, 20, 21, 22, 23, 24, 23, 24, 25, 26, 27, 26, 27, 28, 29, 30, 29, 30, 31, 32, 33, 32, 33, 34, 35, 36, 37, 36, 37, 38, 39, 40, 39, 40, 41, 42, 43, 42, 43, 44

Offset: 1

Michel Lagneau, May 03 2025

nonn

The Maclaurin polynomial of degree n for exp(z) is P(n,z) = Sum_{i=0..n} z^i/i!

The number of zeros in the right half-plane is equal to n - a(n) because we do not observe any purely imaginary roots.

			a(4)= 4 because P(4,z) = 1 + z/1! + z^2/2! + z^3/3! + z^4/4! with 4 roots in the left half-plane:
z1 = -1.729444231-.8889743761*i,
z2 = -1.729444231+.8889743761*i,
z3 = -.2705557689-2.504775904*i,
z4 = -.2705557689+2.504775904*i

Cf. A332324, A330187, A332420.

A:=proc(n) local P, m, y, it:
  it:=0:P:=add(x^i/i!,i=0..n):
   y:=[fsolve(expand(P), x, complex)]:
    for m from 1 to nops(y) do:
     if Re(y[m])<0 then
      it:=it+1:else fi:
    od: A(n):=it:end proc:
seq(A(n), n=1..70);

A383747 Consider the polynomial P(m,z) = Sum_{k=1..r} d(k)*z^(k-1) where d(1) < d(2) < ... < d(r) are the r divisors of m. The sequence lists the numbers m such that P(m,z) contains at least three zeros of the form -1/q, i/q, -i/q, for some integer q, i = sqrt(-1).

Original entry on oeis.org

8, 27, 88, 104, 125, 128, 136, 152, 184, 232, 248, 296, 328, 343, 344, 376, 424, 472, 488, 536, 568, 584, 632, 664, 712, 776, 783, 808, 824, 837, 856, 872, 904, 968, 999, 1016, 1048, 1096, 1107, 1112, 1161, 1192, 1208, 1256, 1269, 1304, 1331, 1336, 1352, 1384, 1431

Offset: 1

Michel Lagneau, May 08 2025

nonn

Subsequence of A291127.

The corresponding integers q are in A383748.

			  n   m  P(m,z)                                  3 zeros of P(m,z)
  1   8  1+2z+4z^2+8z^3                          -1/2, -i/2, i/2
  2  27  1+3z+9z^2+27z^3                         -1/3, -i/3, i/3
  3  88  1+2z+4z^2+8z^3+11z^4+22z^5+44z^6+88z^7  -1/2, -i/2, i/2

Cf. A027750, A291127, A383748.

with(numtheory) :
A:=proc(n) local P, Q, i, q, d, ii:
d:=divisors(n):P:=add(op(i,d)*x^(i-1),i=1..nops(d)):
ii:=0:for q from 1 to 10^4 while (ii=0) do:
Q:=(x+1/q)*(x^2+1/q^2):
if divide(P,Q,'R') then ii:=1:
A(n):=n:else fi:od:end proc:
seq(A(n), n=1..2500);

A383748 a(n) = q is the smallest integer, such that the numbers -1/q, i/q, -i/q with i = sqrt(-1), are three zeros of the polynomial P(A783747(n),z) = Sum_{k=1..r} d(k)*z^(k-1) where d(1) < d(2), ..., < d(r) are the r divisors of A383747(n).

Original entry on oeis.org

2, 3, 2, 2, 5, 2, 2, 2, 2, 2, 2, 2, 2, 7, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, 2, 11, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 13, 2, 3, 2, 2, 2, 2, 3, 2, 2

Offset: 1

Michel Lagneau, May 08 2025

nonn

			 n  q  m = A783747(n)         P(m,z)          3 zeros of P(m,z)
 1  2       8          1+2z+4z^2+8z^3         -1/2, -i/2, i/2
 2  3      27          1+3z+9z^2+27z^3        -1/3, -i/3, i/3
 3  2      88          1+2z+4z^2+8z^3+11z^4+  -1/2, -i/2, i/2
                       22z^5+44z^6+88z^7

Cf. A027750, A291127, A383747.

with(numtheory) :
A:=proc(n) local P, Q, i, q, d, ii:
d:=divisors(n):P:=add(op(i,d)*x^(i-1),i=1..nops(d)):
ii:=0:for q from 1 to n while (ii=0) do:
Q:=(x+1/q)*(x^2+1/q^2):
if divide(P,Q,'R') then ii:=1:
A(n):=q:else fi:od:end proc:
seq(A(n), n=1..2500);

A383109 Consider the isosceles triangle whose vertices are the Gaussian integers z1=0, z2 = x+iy, z3=x-iy. The sequence lists the pairs of positive integer foci (f_i, f_j), f_i < f_j of Steiner inellipse for some z2, z3.

Original entry on oeis.org

3, 5, 6, 10, 9, 15, 13, 15, 12, 20, 15, 25, 18, 30, 15, 37, 15, 41, 17, 39, 21, 35, 26, 30, 24, 40, 27, 45, 25, 51, 30, 50, 39, 45, 33, 55, 36, 60, 29, 75, 30, 74, 39, 65, 51, 53, 30, 82, 34, 78, 42, 70, 52, 60, 45, 75, 39, 85, 48, 80, 51, 85, 65, 75, 54, 90, 61

Offset: 1

Michel Lagneau, Apr 16 2025

nonn

With z1=0, z2 = x+i*y, z3=x-i*y and P(z)=(z-z1)*(z-z2)*(z-z3) we obtain:
(1) P(z) = z^3 -2x*z^2 + (x^2+y^2)*z.
(2) P’(z)=3*z^2-4*x*z+ x^2+y^2.
The two zeros of P’(z) are: (2*x-sqrt(x^2-3*y^2))/3 and (2*x+sqrt(x^2-3*y^2))/3.

			(a(1),a(2)) = (3,5) because from (2) with (x,y) = (6,3), the two zeros of P’(z) are: (2*6-sqrt(6^2-3*3^2))/3 = (12 - sqrt(9))/3 = 3 and (2*6+sqrt(6^2-3*3^2))/3 = (12 + sqrt(9))/3 = 5. The two foci are integers.
(a(7),a(8)) = (13,15) because from (2) with (x,y) = (21,12), the two zeros of P’(z) are (2*21-sqrt(21^2-3*12^2))/3 = (42 - sqrt(9))/3 = 13 and (2*21+sqrt(21^2-3*12^2))/3 = (12 + sqrt(9))/3 = 15. The two foci are integers.

Beniamin Bogosel, A Geometric Proof of the Siebeck-Marden Theorem, Amer. Math. Monthly, vol. 125, no 4, 2017, p. 459-463.
A. Eydelzon, On a New Property of the Steiner Inellipse, Amer. Math. Monthly, vol. 127, no 10, 2020, p. 933-935.

Wikipedia, Steiner inellipse
Wikipedia, Marden's theorem

nn:=200:
for x from 1 to nn do:
for y from 1 to nn while(x^2>3*y^2) do:
 u:=sqrt(x^2+y^2):v:=2*b:s:=sqrt(x^2+y^2)+y:
  z1:=(2*x-sqrt(x^2-3*y^2))/3:z2:=(2*x+sqrt(x^2-3*y^2))/3:
  if z1=floor(z1) and z2=floor(z2) then printf(`%d, `,z1):
   printf(`%d, `,z2):
  else fi:
  od:
od:

A382108 Number of zeros (counted with multiplicity) on the unit circle of the polynomial P(n,z) = Sum_{k=0..n} T(n,k)*z^k where T(n,k) = A214292(n,k) is the first differences of rows in Pascal's triangle.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 3, 4, 3, 6, 5, 6, 5, 6, 7, 8, 9, 10, 3, 8, 7, 10, 9, 10, 7, 10, 11, 8, 11, 12, 9, 10, 11, 14, 11, 14, 11, 12, 13, 12, 13, 12, 15, 12, 7, 18, 19, 16, 11, 14, 11, 14, 11, 18, 11, 18, 15, 18, 19, 22, 7, 16, 21, 20, 17, 22, 15, 18, 21, 20, 25, 20

Offset: 0

Michel Lagneau, Mar 15 2025

nonn

			a(4)=4 because P(4,z)= 4 + 5*z -5*z^3 -4*z^4  with 4 roots z1, z2, z2, z4 on the unit circle : z1 = -1, z2 = +1, z3 = -.625000 -.7806247*i, z4 = -.625000 +.7806247*i.
a(6)=6 because P(6,z)= 6 + 14*z +14*z^2 -14*z^4-14*z^5-6z^6  with 6 roots on the unit circle:
x1 = -1
x2 = +1
x2 = -.6666666667 - .7453559925*i
x3 = -.6666666667 + .7453559925*i
x5 = -.500000000 - .8660254038*i
x6 = -.500000000 + .8660254038*i

Cf. A007318, A214292, A382019 (on and inside the circle).

A382108:=proc(n) local m,y,it:
y:=[fsolve(add((binomial(n+1,k+1)-binomial(n+1,k))*x^k,k=0..n),x,complex)]:it:=0:
 for m from 1 to nops(y) do:
    if ((Re(y[m]))^2+(Im(y[m]))^2)=1
     then it:=it+1:
     else
    fi:
   od:
    A382108(n):=it:end proc:seq(A382108(n),n=1..80);

A382019 Number of zeros (counted with multiplicity) inside and on the unit circle of the polynomial P(n,z) = Sum_{k=0..n} T(n,k)*z^k where T(n,k) = A214292(n,k) is the first differences of rows in Pascal's triangle.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 5, 6, 7, 8, 9, 10, 9, 10, 11, 12, 13, 14, 13, 14, 15, 16, 17, 18, 17, 18, 19, 20, 21, 22, 21, 22, 23, 24, 25, 26, 25, 26, 27, 28, 29, 30, 29, 30, 31, 32, 33, 34, 33, 34, 35, 36, 37, 38, 37, 38, 39, 40, 41, 42, 41, 42, 43, 44, 45, 46, 45, 46, 47

Offset: 0

Michel Lagneau, Mar 12 2025

nonn

The polynomial is P(n,z) = z^(n+1) - ((z-1)*(z+1)^(n+1) +1)/z.

A root z (real or complex) is in or on the unit circle when its magnitude abs(z) <= 1.

			a(4)=4 because P(4,z)= 4 + 5*z -5*z^3 -4*z^4  with 4 roots z1, z2, z2, z4 on the unit circle : z1 = -1, z2 = +1, z3 = -.625000 -.7806247*i, z4 = -.625000 +.7806247*i.
a(6)=6 because P(6,z)= 6 + 14*z +14*z^2 -14*z^4-14*z^5-6z^6  with 6 roots on the unit circle:
  z1 = -1,
  z2 = +1,
  z3 = -.6666666667 - .7453559925*i,
  z4 = -.6666666667 + .7453559925*i,
  z5 = -.500000000 - .8660254038*i,
  z6 = -.500000000 + .8660254038*i.

Cf. A007318, A214292.

A382019:=proc(n) local m,y,it:
y:=[fsolve(add((binomial(n+1,k+1)-binomial(n+1,k))*x^k,k=0..n),x,complex)]:it:=0:
 for m from 1 to nops(y) do:
          if ((Re(y[m]))^2+(Im(y[m]))^2)<=1
          then
         it:=it+1:else fi:
   od: A382019(n):=it:end proc:
seq(A382019(n),n=1..70);

A381604 Least number k such that A381596(k) = n.

Original entry on oeis.org

1, 2, 5, 10, 17, 116, 501, 512

Offset: 0

Michel Lagneau, Mar 01 2025

Least number k such that the number of real zeros of the polynomial P(k,z) = Sum_{i=1..k} A001223(i)*z^(i-1) is equal to n, where A001223(i) = differences between consecutive primes.

			a(1) = 2 because P(2,z) = Sum_{i=1..2} A001223(i)*z^(i-1) = 1 + 2*z = 0 for z = -1/2.
a(2) = 5 because P(5,z) = Sum_{i=1..5} A001223(i)*z^(i-1) = 1 + 2*z + 2*z^2 + 4*z^3 + 2*z^4 = 0 for z = -1.6499348..., -0.5606729...

Cf. A001223, A381596.

with(numtheory):
for n from 0 to 20 do:
ii:=0:
  for k from 1 to 10^3 while(ii=0) do :
    P:=add((ithprime(i+1)-ithprime(i))*x^(i-1),i=1..k):
    y:=fsolve(P,x,real):z:=evalf({%}):y:=nops(z):
     if y=n
     then
      ii:=1:printf (`%d %d \n`,n,k):
       else
     fi:
  od:
od:

A381596 a(n) = number of real zeros (counted with multiplicity) of the polynomial P(n,z) = Sum_{i=1..n} A001223(i)*z^(i-1) where A001223(i) = differences between consecutive primes.

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1

Offset: 1

Michel Lagneau, Mar 01 2025

nonn

			a(4) = 1 because P(4,z) = Sum_{i=1..4} A001223(i)*z^(i-1) = 1 + 2*z + 2*z^2 + 4*z^3 = (2*z + 1)*(2*z^2 + 1) = 0 for z = -1/2.
a(5) = 2 because P(5,z) = Sum_{i=1..5} A001223(i)*z^(i-1) = 1 + 2*z + 2*z^2 + 4*z^3 + 2*z^4 = 0 for z = -1.6499348..., -0.5606729...

Cf. A000040, A001223, A381604.

with(numtheory):
for n from 1 to 100 do :
  P:=add((ithprime(i+1)-ithprime(i))*x^(i-1),i=1..n):
   y:=fsolve(P,x,real):
   z:=evalf({%}):k:=nops(z):
   printf(`%d, `,k):
od:

a(n) = my(v=primes(n+1)); #polrootsreal(sum(i=1, n, (v[i+1]-v[i])*z^(i-1))); \\ Michel Marcus, Mar 01 2025

A381097 Consider the polynomial P(m,z) = Sum_{i=1..k} d(i)*z^(i-1) where d(1), d(2), ..., d(k) are the k divisors of m. The sequence lists the numbers m such that P(m,z) is irreducible.

Original entry on oeis.org

2, 3, 4, 5, 7, 9, 11, 12, 13, 16, 17, 19, 23, 24, 25, 29, 30, 31, 36, 37, 40, 41, 43, 45, 47, 48, 49, 53, 56, 59, 60, 61, 63, 64, 67, 70, 71, 72, 73, 79, 80, 81, 83, 84, 89, 90, 96, 97, 101, 103, 105, 107, 108, 109, 112, 113, 120, 121, 126, 127, 131, 132, 135

Offset: 1

Michel Lagneau, Feb 14 2025

nonn

The squares>1 and the prime numbers are in the sequence.

			The prime numbers q are in the sequence because P(q,z) = qz + 1.
6 is not in the sequence because P(6,z)=(2z+1)*(3z^2+1).
The following table gives the irreducible polynomials.
+-----------------------------------------------------------+
|  m |                 P(m,z)                               |
+-----------------------------------------------------------+
|  4 | 1 + 2z + 4z^2                                        |
+-----------------------------------------------------------+
|  9 | 1 + 3z + 9z^2                                        |
+-----------------------------------------------------------+
| 12 | 1 + 2z + 3z^2 + 4z^3 + 6z^4 + 12z^5                  |
+-----------------------------------------------------------+
| 16 | 1 + 2z + 4z^2 + 8z^3 + 16z^4                         |
+----------------------------+------------------------------+
| 24 | 1 + 2z + 3z^2 + 4z^3 + 6z^4 + 8z^5 + 12z^6 + 24z^7   |
+-----------------------------------------------------------+
| 25 | 1 + 5z + 25z^2                                       |
+-----------------------------------------------------------+
| 30 | 1 + 2z + 3z^2 + 5z^3 + 6z^4 + 10z^5 + 15z^6 + 30z^7  |
+-----------------------------------------------------------+

Cf. A291127.

with(numtheory):
for n from 1 to 135 do :
  it:=0:d:=divisors(n):P:=add(op(i,d)*x^(i-1),i=1..nops(d)):
   y:=fsolve(P,x,complex):z:=evalf({%}):k:=nops(z):
    if irreduc(P) then printf(`%d, `,n):else fi:
od:

isok(n) = my(d=divisors(n)); polisirreducible(sum(i=1, #d, d[i]*z^(i-1))); \\ Michel Marcus, Feb 14 2025

cp's OEIS Frontend

User: Michel Lagneau

A386516 Least k such that k^2+1 contains exactly n distinct prime factors of the form m^2+1 or 0 if no such k exists.

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A383644 a(n) is the number of zeros in the left half-plane of the Maclaurin polynomial of degree n for exp(z).

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A383747 Consider the polynomial P(m,z) = Sum_{k=1..r} d(k)*z^(k-1) where d(1) < d(2) < ... < d(r) are the r divisors of m. The sequence lists the numbers m such that P(m,z) contains at least three zeros of the form -1/q, i/q, -i/q, for some integer q, i = sqrt(-1).

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A383748 a(n) = q is the smallest integer, such that the numbers -1/q, i/q, -i/q with i = sqrt(-1), are three zeros of the polynomial P(A783747(n),z) = Sum_{k=1..r} d(k)*z^(k-1) where d(1) < d(2), ..., < d(r) are the r divisors of A383747(n).

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A383109 Consider the isosceles triangle whose vertices are the Gaussian integers z1=0, z2 = x+i*y, z3=x-i*y. The sequence lists the pairs of positive integer foci (f_i, f_j), f_i < f_j of Steiner inellipse for some z2, z3.

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A382108 Number of zeros (counted with multiplicity) on the unit circle of the polynomial P(n,z) = Sum_{k=0..n} T(n,k)*z^k where T(n,k) = A214292(n,k) is the first differences of rows in Pascal's triangle.

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A382019 Number of zeros (counted with multiplicity) inside and on the unit circle of the polynomial P(n,z) = Sum_{k=0..n} T(n,k)*z^k where T(n,k) = A214292(n,k) is the first differences of rows in Pascal's triangle.

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A381604 Least number k such that A381596(k) = n.

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A381596 a(n) = number of real zeros (counted with multiplicity) of the polynomial P(n,z) = Sum_{i=1..n} A001223(i)*z^(i-1) where A001223(i) = differences between consecutive primes.

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A383109 Consider the isosceles triangle whose vertices are the Gaussian integers z1=0, z2 = x+iy, z3=x-iy. The sequence lists the pairs of positive integer foci (f_i, f_j), f_i < f_j of Steiner inellipse for some z2, z3.