A014442 -id:A014442 - cp's OEIS frontend

A173561 Numbers k such that gpf(k^2+1)/k sets a new record of low value, where gpf(k) is the greatest prime dividing k (A006530).

1, 3, 7, 38, 47, 57, 157, 239, 829, 882, 993, 1772, 2673, 2917, 2943, 4747, 4952, 5257, 6118, 9466, 12943, 17557, 18543, 34208, 44179, 72662, 85353, 114669, 219602, 260359, 320078, 330182, 478707, 485298, 1083493, 1143007, 1477034, 1528649, 1615463, 1635786, 1984933

Offset: 1

M. J. Knight (melknightdr(AT)verizon.net), Feb 21 2010

nonn

This is an infinite sequence, since the solutions to the Pell equations for primes p = 4*k+1 will give ratios with limit 0. For example, the entry 7 satisfies 7^2 - 2*5^2 = -1 and the ratio is 5/7. However, not all entries are given by this technique.

			a(3) = 7 because 7^2+1 = 2*5^2 and 5/7 is smaller than all previous results.

Jonathan Bober, Dan Fretwell, Greg Martin, and Trevor D. Wooley, Smooth values of polynomials, Journal of the Australian Mathematical Society, Vol. 108, No. 2 (2020), pp. 245-261. arXiv:1710.01970 [math.NT] [alternate link]

Cf. A006530, A014442.

f[n_] := FactorInteger[n^2 + 1][[-1, 1]]/n; s = {}; fm = 3; Do[f1 = f[n]; If[f1 < fm, fm = f1; AppendTo[s, n]], {n, 1, 2*10^4}]; s (* Amiram Eldar, Mar 03 2021 *)

More terms from Amiram Eldar, Mar 03 2021

A219586 Greatest prime factor of Product_{x=1..n} (x^2 + 1).

Original entry on oeis.org

2, 5, 5, 17, 17, 37, 37, 37, 41, 101, 101, 101, 101, 197, 197, 257, 257, 257, 257, 401, 401, 401, 401, 577, 577, 677, 677, 677, 677, 677, 677, 677, 677, 677, 677, 1297, 1297, 1297, 1297, 1601, 1601, 1601, 1601, 1601, 1601, 1601, 1601, 1601, 1601, 1601, 1601

Offset: 1

Michel Marcus, Nov 23 2012

nonn

C. Hooley, On the greatest prime factor of a quadratic polynomial, Acta Mathematica July 1967, Volume 117, Issue 1, pp 281-299.

Cf. A002496, A002522, A005574, A014442.

a:= proc(n) option remember; `if`(n=0, 0,
      max(a(n-1), numtheory[factorset](n^2+1)[]))
    end:
seq(a(n), n=1..55);  # Alois P. Heinz, Jan 03 2021

a[n_] := a[n] = If[n == 1, 2, Max[a[n-1], FactorInteger[n^2+1][[-1, 1]]]];
Table[a[n], {n, 1, 55}] (* Jean-François Alcover, May 14 2022, after Alois P. Heinz *)

a(m) = {for (n=1, m, f = factor(prod(x=1, n, x^2+1)); print1(f[length(f~), 1], ", "););}

A253596 Numbers k such that A002313(m) is the greatest prime divisor of k^2 + 1 and A002313(m+1) is the greatest prime divisor of (k+1)^2 + 1 for some m.

Original entry on oeis.org

1, 7, 31, 293, 1936, 2244, 4158, 5744, 11573, 25242, 285202, 339354

Offset: 1

Michel Lagneau, Jan 05 2015

A002313 contains the primes congruent to 1 or 2 (mod 4).
The corresponding indices m in A002313 are 1, 2, 6, 13, 69, 65, 322, 199, 130, 46, 1471, 866, ...
The corresponding primes A002313(m) are 2, 5, 37, 101, 809, 761, 4877, 2777, 1709, 509, 26821, 14957, ...

			31 is in the sequence because 31^2 + 1 = 2*13*37 and 32^2 + 1 = 5*5*41 with the property that 37 = A002313(6) and 41 = A002313(7).

Cf. A002313, A014442.

with(numtheory): nn:=500000:print(1):
for n from 1 to nn do:
   p:=n^2+1:x:=factorset(p):n0:=nops(x):p1:=x[n0]:
   q:=(n+1)^2+1:y:=factorset(q):n1:=nops(y):p2:=y[n1]:ii:=0:
     for j from 2 by 2 to 1000 while(ii=0) do:
      pp:=p1+j:
      if type(pp,prime)=true and irem(pp,4)=1
      then
      p3:=pp:ii:=1:
      else
      fi:
    od:
    if p3=p2
    then
    print(n):
     else
     fi:
    od:

lst={};Do[If[Mod[Prime[i],4]==1||Mod[Prime[i],4]==2,AppendTo[lst,Prime[i]]],{i,1,1000}];Do[Do[If[FactorInteger[n^2+1][[-1]][[1]]==Part[lst,j]&&FactorInteger[(n+1)^2+1][[-1]][[1]]==Part[lst,j+1],Print[n]],{n,1,20000}],{j,1,999}]

A258840 a(n) is the least integer k such that there are n values of i <= k for which gpf(i^2 + 1) = gpf(k^2 + 1), where gpf(x) is the greatest prime factor of x.

Original entry on oeis.org

1, 3, 7, 38, 47, 157, 302, 327, 515, 616, 697, 798, 818, 1303, 2818, 3141, 3323, 5648, 6962, 9193, 9872, 13213, 13747, 15445, 16271, 17149, 18263, 20491, 20727, 24389, 26915, 29078, 31867, 37848, 38007, 40182, 41508, 43328, 46349, 55025, 62258, 63133, 66893

Offset: 1

Michel Lagneau, Jun 12 2015

nonn

A014442(n) gives the largest prime factor of n^2 + 1.
The primes of the sequence are 3, 7, 47, 157, 1303, 3323, 46349, ...
The corresponding sequence Gpf(a(n)^2+1) is 2, 5, 5, 17, 17, 29, 37, 37, 101, 101, 101, 101, 101, 101, 101, 101, 101, 97, 97, 97, 97, 401, 349, 389, 557, 557, 557, 557, 557, 421, 421, 421, 557, ... and it is interesting to observe the frequency of repetitions for the numbers 5, 17, 37, 97, 101, 557, ...

			a(3) = 7 because gpf(7^2 + 1) = gpf(3^2 + 1) = gpf(2^2 + 1) = 5 => 3 occurrences.
a(4) = 38 because gpf(38^2 + 1) = gpf(21^2 + 1) = gpf(13^2 + 1) = gpf(4^2 + 1) = 17 => 4 occurrences.

Cf. A014442, A242012.

with(numtheory):nn:=70000:T:=array(1..nn):k:=0:kk:=1:
for m from 1 to nn do:
x:=factorset(m^2+1):n1:=nops(x):p:=x[n1]:k:=k+1:T[k]:=p:
od:
for n from 1 to 43 do:jj:=0:for k from kk to nn while(jj=0) do:
  q:=T[k]:ii:=0:jj:=0:
    for i from 1 to k do:
      if T[i]=q then ii:=ii+1:
      else
      fi:
    od:if ii=n then jj:=1:kk:=k:
    printf ( "%d %d \n",n,k):else fi:
  od:od:

gpf(n) = my(f=factor(n^2+1)); f[#f~,1];
nboc(k) = my(gpfk = gpf(k)); sum(i=1, k, gpf(i) == gpfk);
a(n) = my(k = 1); while (nbo(k) != n, k++); k; \\ Michel Marcus, Jun 12 2015

A339315 a(n) is the smallest number k such that k^2+1 divided by its largest prime factor is equal to F(2*n-1) for n > 0, or 0 if no such k exists, where F(n) is the Fibonacci sequence.

Original entry on oeis.org

1, 3, 8, 34, 55, 144, 610, 233, 12166, 2584, 4181, 68260, 46368, 75025, 3917414, 464656, 1346269, 16349962

Offset: 1

Michel Lagneau, Nov 30 2020

a(n) is the smallest number k such that A248516(k) = A001519(n) for n > 0, or 0 if no such k exists, where A001519(n) = F(2*n-1) (bisection of the Fibonacci sequence), with F(n) = A000045(n).
We observe that a(2 + 3m) = A001519(1 + 3m) = A000045(1 + 6m) for m = 2, 3, 4, 5. For n = 6, this property no longer works.
For k > 0, a(3k - 1) is odd, a(3k) and a(3k+1) are even.
We observe that a(n)^2 + 1 is the product of two prime Fibonacci numbers for n = 2, 3, 4, 6, 7.
The first 18 terms of the sequence are Fibonacci numbers, except a(9), a(12), a(15), a(16) and a(18).
The corresponding sequence b(n) = (a(n)^2+1)/ A001519(n) is 2, 5, 13, 89, 89, 233, 1597, 89, 92681, 1597, 1597, 162593, 28657, 28657, 29842993, 160373, 514229. We observe that a majority of terms of b(n) are prime Fibonacci numbers, except b(9), b(12), b(15) and b(16).

			a(4) = 34 because 34^2 + 1 = 13*89 = 1157, and 1157/89 = 13 = A248516(34) = A001519(4).
A curiosity: a(22) = 1134903170 = F(45) with F(45)^2 + 1 = F(43)*F(47) where F(43) and F(47) are prime Fibonacci numbers.

Cf. A000045, A001519, A001906, A002522, A005478, A014442, A245236, A245306, A248516, A281618.

with(numtheory):with(combinat,fibonacci):
nn:=100:n0:=20:
for n from 1 to n0 do:
  ii:=0:
  for m from 1 to 10^10 while(ii=0) do:
   x:=m^2+1:y:=factorset(x):n1:=nops(y):
   z:=x/y[n1]:
    if z = fibonacci(2*n-1)
     then
     ii:=1:printf(`%d %d \n`,n,m):
     else
    fi:
  od:
od:

a(n) = {my(k=1, f=fibonacci(2*n-1)); while ((k^2+1)/vecmax(factor(k^2+1)[,1]) != f, k++); k;} \\ Michel Marcus, Nov 30 2020

A348594 Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

Original entry on oeis.org

8, 50, 1250, 1800, 2450, 9800, 14450, 20000, 24200, 101250, 105800, 135200, 162450, 168200, 204800, 304200, 336200, 451250, 480200, 490050, 530450, 696200, 924800, 966050, 1008200, 1125000, 1155200, 1428050, 1805000, 2332800, 2420000, 2576450, 2761250, 2832200

Offset: 1

Michel Lagneau, Jan 26 2022

nonn

Subsequence of A085722.
The corresponding pairs (p, q) of the sequence are (5, 13), (41, 61), (1201, 1301), (1741, 1861), (2381, 2521), (9661, 9941), (14281, 14621), (19801, 20201), (23981, 24421), (100801, 101701), ...
Property:
a(n) = 2* A109306(n)^2 and a(n) == 0 (mod 50) for n > 1. Proof:
From the relations:
(1)        m^2 + 1 = p*q
(2)        (p + q)/2 = m + 1
We obtain:
(3)        p = m + 1 - sqrt(8*m)/2
(4)        q = m + 1 + sqrt(8*m)/2
with m = 2*k^2 we obtain:
(5)        p = k^2 + (k-1)^2
(6)        q = k^2 + (k+1)^2
For n > 1, A109306(n) == 0 (mod 5) => 2*A109306(n)^2 == 0 (mod 50).

			50 = 2*5^2 is in the sequence because 50^2 + 1 = 41*61 with 50 = (41 + 61)/2 - 1.

Cf. A014442, A085722, A089120, A109306, A144255.

with(numtheory):nn:=250:printf(`%d, `,8):
for k from 0 to nn do:
n:=50*k^2:d:=factorset(n^2+1):
  if bigomega(n^2+1)=2 and (d[1]+d[2])/2 - 1 = n
   then
    printf(`%d, `,n):
    else
  fi:
od:

q[n_] := Module[{f = FactorInteger[n^2 + 1]}, f[[;; , 2]] == {1, 1} && f[[1, 1]] + f[[2, 1]] == 2*n + 2]; Select[Range[3*10^5], q] (* Amiram Eldar, Jan 26 2022 *)

isok(m) = my(x); if (bigomega(x=m^2+1)==2, my(f=factor(x)); (f[1,1]+f[2,1] == 2*(m+1))); \\ Michel Marcus, Jan 26 2022

cp's OEIS Frontend

A173561 Numbers k such that gpf(k^2+1)/k sets a new record of low value, where gpf(k) is the greatest prime dividing k (A006530).

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A219586 Greatest prime factor of Product_{x=1..n} (x^2 + 1).

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A253596 Numbers k such that A002313(m) is the greatest prime divisor of k^2 + 1 and A002313(m+1) is the greatest prime divisor of (k+1)^2 + 1 for some m.

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A258840 a(n) is the least integer k such that there are n values of i <= k for which gpf(i^2 + 1) = gpf(k^2 + 1), where gpf(x) is the greatest prime factor of x.

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A339315 a(n) is the smallest number k such that k^2+1 divided by its largest prime factor is equal to F(2*n-1) for n > 0, or 0 if no such k exists, where F(n) is the Fibonacci sequence.

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PARI

A348594 Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

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