A089120 -id:A089120 - cp's OEIS frontend

A334080 Number of Pythagorean triples among the divisors of 60*n.

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 3, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 6, 4, 6, 2, 8, 2, 6, 4, 5, 4, 9, 2, 4, 6, 8, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 9, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 7, 6, 8, 2, 8, 4, 9, 2, 12, 2, 4, 6, 6, 4, 12, 2, 10, 5, 4, 2, 12, 4

Offset: 1

Michel Lagneau, Apr 14 2020

nonn

The odd numbers of the sequence are rare (see the table below).
The subsequence of odd terms begins with 1, 3, 3, 3, 5, 3, 5, 9, 3, 9, 7, 9, 5, 9, 9, 3, 11, 15, 5, 9, 5, 15, 9, 9, 9, 5, 19, 3, 15, 15, 9, ... (see the table at the link).
It is interesting to note that each set of divisors of A169823(n) contains m primitive Pythagorean triples for some n, m = 1, 2, ...
Examples:
- The set of divisors of A169823(1)= 60 contains only one primitive Pythagorean triple: (3, 4, 5).
- The set of divisors of A169823(136) = 8160 contains two primitive Pythagorean triples: (3, 4, 5) and (8, 15, 17).
- The set of divisors of A169823(910) = 54600 contains three primitive Pythagorean triples: (3, 4, 5), (5, 12, 13) and (7, 24, 25).
There is an interesting property: we observe that a(n) = A000005(n) except for n in the set {13, 26, 34, 39, 52, 65, 68, 70, 78, 91, 102, ...}. This set contains subset of numbers of the form 13*k, 34*k, 70*k, 203*k, 246*k, 259*k, ... for k = 1, 2, ...
We recognize the sequence A081752: {13, 34, 70, 203, 246, 259, 671, ...} (ordered product of the sides of primitive Pythagorean triangles divided by 60).
The following table shows the numbers of odd terms < 10^k for k = 2, 3, 4, 5, 6 and 7. For instance, among the 16 multiples of 60 less than 10^3, the divisors of the five numbers 60, 240, 540, 780 and 960 contain 1, 3, 3, 3 and 5 Pythagorean triples respectively, and that represents 31.25% of odd numbers.
+---------------+-----------------+---------------------+----------+
|   Intervals   |    Number of    | Number of odd terms |          |
|  D(k) < 10^k  | multiples of 60 |      in D(k)        |     %    |
| k = 2,3,...,7 |     in D(k)     |                     |          |
+---------------+-----------------+---------------------+----------+
|   < 10^2      |          1      |          1          | 100%     |
|   < 10^3      |         16      |          5          |  31.250% |
|   < 10^4      |        166      |         18          |  10.843% |
|   < 10^5      |       1666      |         72          |   4.321% |
|   < 10^6      |      16666      |        256          |   1.536% |
|   < 10^7      |     166666      |        879          |   0.527% |
|---------------+-----------------+---------------------+----------+

			a(4) = 3 because the divisors of A169823(4) = 240 are {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240} with 3 Pythagorean triples: (3, 4, 5), (6, 8, 10) and (12, 16, 20). The first triple is primitive.

Michel Lagneau, Odd terms.

Cf. A000005, A009003, A014442, A081752, A089120, A144255, A169823.

with(numtheory):
for n from 60 by 60 to 5400 do :
   d:=divisors(n):n0:=nops(d):it:=0:
    for i from 1 to n0-1 do:
     for j from i+1 to n0-2 do :
      for m from i+2 to n0 do:
       if d[i]^2 + d[j]^2 = d[m]^2
        then
        it:=it+1:
        else
       fi:
      od:
     od:
    od:
    printf(`%d, `,it):
   od:

ishypo(n) = setsearch(Set(factor(n)[, 1]%4), 1); \\ A009003
a(n) = {n *= 60; my(d=divisors(n), nb=0); for (i=3, #d, if (ishypo(d[i]), for (j=2, i-1, for (k=3, j-1, if (d[j]^2 + d[k]^2 == d[i]^2, nb++););););); nb;} \\ Michel Marcus, Apr 26 2020

A217448 Least k > 0 such that 1 + n^2 and 1 + (n+k)^2 have the same smallest prime factor.

Original entry on oeis.org

2, 6, 2, 26, 2, 74, 2, 4, 2, 404, 2, 6, 2, 366, 2, 514, 2, 4, 2, 1564, 2, 6, 2, 1106, 2, 4010, 2, 4, 2, 34, 2, 6, 2, 10, 2, 2594, 2, 4, 2, 22334, 2, 6, 2, 16, 2, 58, 2, 4, 2, 64, 2, 6, 2, 29062, 2, 18710, 2, 4, 2, 10, 2, 6, 2, 42, 2, 17428, 2, 4, 2, 16, 2, 6

Offset: 1

Michel Lagneau, Oct 03 2012

Alternate title: Least k > 0 such that A089120(n) = A089120(n+k).
A089120(n): smallest prime factor of n^2 + 1.
Conjecture: a(n) exists for all n.

			a(10) = 404 because 10^2 + 1 = 101, (10+404)^2+1 = 101*1697 so A089120(10) = A089120(414) = 101;
a(170) = 404274 because 170^2 + 1 = 28901, (170+404274)^2+1 = 163574949137 = 28901* 5659837 so A089120(170) = A089120(40444) = 28901.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A089120, A217393, A014442.

with(numtheory):T:=array(1..100): for n from 1 to 100 do:x:=factorset(n^2+1):n1:=nops(x): T[n] := x[1]:od:for a from 1 to 80 do:p:=T[a]:ii:=0:for k from 1 to 50000 while(ii=0) do: z:=factorset((a+k)^2+1): n2:=nops(z):if z[1]=p then printf(`%d, `,k):ii:=1:else fi:od:od:

sspf[n_]:=Module[{c=FactorInteger[1+n^2][[1,1]],k=1},While[ FactorInteger[ 1+ (n+k)^2][[1,1]]!=c,k++];k]; Array[sspf,80] (* Harvey P. Dale, Oct 12 2012 *)

A020639(n) = if(1==n, n, factor(n)[1, 1]);
A217448(n) = { my(spf=A020639(1+(n^2)), x); for(k=1,oo,x=1+((n+k)^2); if(!(x%spf) && A020639(x)==spf,return(k))); }; \\ Antti Karttunen, May 24 2021

A247339 a(n) is the least number k such that the greatest prime divisor of k^2+1 is the smallest prime divisor of n^2+1.

Original entry on oeis.org

1, 2, 1, 4, 1, 6, 1, 2, 1, 10, 1, 2, 1, 14, 1, 16, 1, 2, 1, 20, 1, 2, 1, 24, 1, 26, 1, 2, 1, 4, 1, 2, 1, 5, 1, 36, 1, 2, 1, 40, 1, 2, 1, 5, 1, 12, 1, 2, 1, 9, 1, 2, 1, 54, 1, 56, 1, 2, 1, 5, 1, 2, 1, 4, 1, 66, 1, 2, 1, 5, 1, 2, 1, 74, 1, 23, 1, 2, 1, 6, 1, 2

Offset: 1

Michel Lagneau, Sep 14 2014

nonn

a(n)=n if n^2+1 is prime and a(n)=1 if n is odd.

Conjecture: for all integer n, there exists at least an integer m <= n such that the smallest prime factor of n^2+1 is also the greatest prime factor of m^2+1. - Michel Lagneau, Sep 27 2015

			a(34)=5 because the greatest prime divisor of 5^2+1 = 2*13 is the smallest prime divisor of 34^2+1 =13*89.

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

Cf. A002522, A089120, A014442.

with(numtheory):nn:=2000:T:=array(1..nn):U:=array(1..nn):
  for i from 1 to nn do:
    x:=factorset(i^2+1):T[i]:=x[1]:U[i]:=i:
  od:
    for n from 1 to 100 do:
     ii:=0:
      for k from 1 to 50000 while(ii=0) do:
       y:=factorset(k^2+1):n0:=nops(y):q:=y[n0]:
        if q=T[n]
         then
         ii:=1: printf(`%d, `,k):
         else
        fi:
     od:
   od:

Table[k = 1; While[FactorInteger[k^2 + 1][[-1, 1]] != FactorInteger[n^2 + 1][[1, 1]], k++]; k, {n, 82}] (* Michael De Vlieger, Sep 27 2015 *)

a(n) = {f = factor(n^2+1)[1,1]; k = 1; while (! ((g=factor(k^2+1)) && (g[#g~,1] == f)), k++); k;} \\ Michel Marcus, Sep 14 2014

A248511 Difference between k and the least prime factor of k^2+1 where k is the n-th number with k^2+1 composite.

Original entry on oeis.org

1, 3, 5, 3, 7, 9, 7, 11, 13, 15, 13, 17, 19, 17, 21, 23, 25, 23, 27, 13, 29, 27, 31, 21, 33, 35, 33, 37, 39, 37, 41, 31, 43, 17, 45, 43, 47, 9, 49, 47, 51, 53, 55, 53, 57, 47, 59, 57, 61, 47, 63, 65, 63, 67, 57, 69, 67, 71, 73, 23, 75, 73, 77, 43, 79, 77, 81

Offset: 1

Michel Lagneau, Oct 07 2014

nonn

a(n) = A134407(n) - least prime divisor of A134406(n).

			a(1) = 1 because the first composite is 3^2+1 = 2*5 and 3-2 = 1.

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

Cf. A002496, A020639, A089120, A134406, A134407.

with(numtheory):
   for n from 1 to 200 do:
    p:=n^2+1:x:=factorset(p):d:=n-x[1]:
    if type(p,prime)=false
    then
    printf(`%d, `,d):
    else
    fi:
   od:

A248532 Numbers n such that the smallest prime divisor of n^2+1 is 53.

Original entry on oeis.org

76, 136, 454, 500, 560, 666, 924, 984, 1196, 1454, 1514, 1666, 1726, 2090, 2196, 2256, 2620, 2726, 2786, 3044, 3104, 3150, 3210, 3256, 3316, 3680, 3786, 4104, 4210, 4270, 4316, 4634, 4694, 4800, 4846, 5224, 5330, 5694, 5800, 5860, 5906, 5966, 6224, 6330, 6390

Offset: 1

Michel Lagneau, Oct 08 2014

Or numbers n such that the smallest prime divisor of n^2+1 is A002313(8).

a(n)== 30 or 76 (mod 106).

			76 is in the sequence because 76^2+1= 53*109.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A089120, A002313, A209874, A248527, A248528, A248529, A248530, A248531.

[n: n in [2..7000] | PrimeDivisors(n^2+1)[1] eq 53]; // Bruno Berselli, Oct 08 2014

lst={};Do[If[FactorInteger[n^2+1][[1, 1]]==53, AppendTo[lst, n]], {n, 2, 2000}]; lst
Select[Range[7000],FactorInteger[#^2+1][[1,1]]==53&] (* Harvey P. Dale, Aug 04 2016 *)
p = 53; ps = Select[Range[p - 1], Mod[#, 4] != 3 && PrimeQ[#] &]; Select[Range[7000], Divisible[(nn = #^2 + 1), p] && ! Or @@ Divisible[nn, ps] &] (* Amiram Eldar, Aug 16 2019 *)

A248550 Numbers n such that the smallest prime divisor of n^2+1 is 73.

Original entry on oeis.org

100, 246, 484, 630, 776, 830, 976, 1506, 1706, 1944, 2144, 2236, 2290, 2874, 3020, 3604, 3696, 3750, 3896, 4134, 4426, 4626, 4864, 5064, 5210, 5356, 5594, 5740, 5794, 5940, 6086, 6324, 6470, 6616, 6670, 6816, 7200, 7254, 7346, 7400, 7546, 7930, 7984, 8076

Offset: 1

Michel Lagneau, Oct 08 2014

Or numbers n such that the smallest prime divisor of n^2+1 is A002313(10).
a(n) == 46 or 100 (mod 146).
No need to completely factorize n^2+1. - David A. Corneth, Apr 29 2017

			100 is in the sequence because 100^2+1= 73*137.
246 is in the sequence because 246^2+1 isn't divisible by any prime less than 73 and is divisible by 73. - _David A. Corneth_, Apr 29 2017

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A089120, A002313, A209874, A248527-A248531, A248549-A248553.

lst={};Do[If[FactorInteger[n^2+1][[1, 1]]==73, AppendTo[lst, n]], {n, 2, 10000}]; lst
Select[Range[8100],FactorInteger[#^2+1][[1,1]]==73&] (* Harvey P. Dale, Apr 29 2017 *)
p = 73; ps = Select[Range[p - 1], Mod[#, 4] != 3 && PrimeQ[#] &]; Select[Range[8000], Divisible[(nn = #^2 + 1), p] && ! Or @@ Divisible[nn, ps] &] (* Amiram Eldar, Aug 16 2019 *)

is(n) = {my(m=n%146, p=2, n21 = n^2+1, v=[5, 13, 17, 29, 37, 41, 53, 61]);
return(abs(73-m)==27&&sum(i=1, #v, p=nextprime(p+1); valuation(n21,v[i]))==0)}
upto(n) = {my(l=List(), i=54, m=46); while(mDavid A. Corneth, Apr 29 2017

A248551 Numbers n such that the smallest prime divisor of n^2+1 is 89.

Original entry on oeis.org

144, 390, 856, 1746, 1814, 1924, 2170, 2526, 2636, 2704, 2814, 3170, 3416, 3594, 3704, 3950, 4060, 4306, 4840, 4950, 5306, 5484, 6374, 6620, 6730, 7086, 7154, 7264, 7866, 7976, 8044, 8154, 8400, 8756, 8866, 9044, 9400, 9646, 9824, 10180, 10290, 11070, 11426

Offset: 1

Michel Lagneau, Oct 08 2014

Or numbers n such that the smallest prime divisor of n^2+1 is A002313(11).

a(n)== 34 or 144 (mod 178).

			144 is in the sequence because 144^2+1= 89*233.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A089120,A002313,A209874,A248527-A248531,A248549-A248553.

lst={};Do[If[FactorInteger[n^2+1][[1, 1]]==89, AppendTo[lst, n]], {n, 2, 10000}]; lst
p = 89; ps = Select[Range[p - 1], Mod[#, 4] != 3 && PrimeQ[#] &]; Select[Range[12000], Divisible[(nn = #^2 + 1), p] && ! Or @@ Divisible[nn, ps] &] (* Amiram Eldar, Aug 16 2019 *)

A248552 Numbers n such that the smallest prime divisor of n^2+1 is 97.

Original entry on oeis.org

366, 410, 604, 754, 1336, 1530, 1574, 2156, 2500, 2544, 2694, 3126, 3276, 3470, 3514, 3664, 4096, 4290, 4440, 5066, 5454, 5604, 6186, 6230, 6380, 6424, 6574, 7156, 8126, 8170, 8320, 9140, 9334, 9484, 9916, 10066, 10110, 10260, 10454, 11036, 11230, 11424, 11856

Offset: 1

Michel Lagneau, Oct 08 2014

Or numbers n such that the smallest prime divisor of n^2+1 is A002313(12).

a(n)== 22 or 172 (mod 194).

			366 is in the sequence because 366^2+1= 97*1381.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A089120, A002313, A209874, A248527-A248531, A248549-A248553.

lst={};Do[If[FactorInteger[n^2+1][[1, 1]]==97, AppendTo[lst, n]], {n, 2, 10000}]; lst
Select[Range[12000],FactorInteger[#^2+1][[1,1]]==97&] (* Harvey P. Dale, Aug 11 2017 *)
p = 97; ps = Select[Range[p - 1], Mod[#, 4] != 3 && PrimeQ[#] &]; Select[Range[12000], Divisible[(nn = #^2 + 1), p] && ! Or @@ Divisible[nn, ps] &] (* Amiram Eldar, Aug 16 2019 *)

A249122 a(n) = floor(n / lpf(n^2 + 1)) where lpf(n^2 + 1) is the smallest prime divisor of n^2 + 1.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 3, 1, 4, 0, 5, 2, 6, 0, 7, 0, 8, 3, 9, 0, 10, 4, 11, 0, 12, 0, 13, 5, 14, 1, 15, 6, 16, 2, 17, 0, 18, 7, 19, 0, 20, 8, 21, 3, 22, 1, 23, 9, 24, 1, 25, 10, 26, 0, 27, 0, 28, 11, 29, 4, 30, 12, 31, 3, 32, 0, 33, 13, 34, 5, 35, 14, 36, 0, 37, 1

Offset: 1

Michel Lagneau, Oct 21 2014

nonn

a(n) = floor(n / A089120(n)).

a(A002496(n)) = 0 and a(A247340(n)) = 1 where A002496 are the primes of form m^2 + 1 and A247340(n) = {3, 8, 30, 46, 50, 76, ...} are the numbers m such that m^2 + 1 = p*q, p and q primes => p | a^2+1 and q | b^2+1 for some a,b < m.

			a(8) = 1 because 30^2 + 1 = 17*53 and floor(30/17) = 1.
Or a(8) = a(A247340(2)) = 1.

Michel Lagneau, Table of n, a(n) for n = 1..20000

Cf. A002496, A089120, A134406, A247340.

with(numtheory):
   for n from 1 to 200 do:
    p:=n^2+1:x:=factorset(p):d:=floor(n/x[1]):
    printf(`%d, `, d):
   od:

Table[Floor[n/ FactorInteger[n^2+1][[ 1, 1]]], {n, 100}]

a(n) = n\factor(n^2+1)[1, 1]; \\ Michel Marcus, Oct 25 2014

A250028 a(n) is the number of positive integers k <= n such that lpf(k^2 + 1) = lpf(n^2 + 1), where lpf() is the least prime factor function.

Original entry on oeis.org

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

Offset: 1

Michel Lagneau, Nov 11 2014

nonn

The least prime factor of n^2 + 1 is A089120(n).
a(2*j+1) = j+1 because 2 is the least prime factor of the even numbers.
a(n) = 1 if n is a term in A005574 (numbers n such that n^2 + 1 is prime).
a(n) = 1 if lpf(n^2 + 1) appears for the first time (example: a(50) = 1 because lpf(50^2 + 1) = lpf(41*61) = 41).
Property: if p = lpf(n^2 + 1), then p divides (n-p)^2 + 1.

			a(3) = 2 because the least prime factor of 3^2 + 1 is 2 and 2 is the 2nd positive integer k for which lpf(k^2 + 1) is 2 (the 1st occurrence is 1^2 + 1 = 2).
a(12) = 3 because the least prime factor of 12^2 + 1 = 5*29 is 5, and 5 is the 3rd occurrence (the 1st and 2nd are 2^2 + 1 = 5 and 8^2 + 1 = 5*13, respectively).

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

Cf. A002496, A005574, A089120, A242012.

with(numtheory):nn:=200:T:=array(1..nn):k:=0:
for m from 1 to nn do:
x:=factorset(m^2+1):n1:=nops(x):p:=x[1]:k:=k+1:T[k]:=p:
od:
  for n from 1 to 150 do:
  q:=T[n]:ii:=0:
    for i from 1 to n do:
      if T[i]=q then ii:=ii+1:
      else
      fi:
    od:
    printf(`%d, `, ii):
  od:

With[{s = Array[FactorInteger[#^2 + 1][[1, 1]] &, {76}]}, Reap[Do[Sow@ Count[Take[s, i], k_ /; k == FactorInteger[i^2 + 1][[1, 1]]], {i, Length@ s}]][[-1, 1]]] (* Michael De Vlieger, Sep 12 2017 *)

a(n) = my(gn = vecmin(factor(n^2+1)[,1])); sum(k=1, n, vecmin(factor(k^2+1)[,1]) == gn); \\ Michel Marcus, Sep 11 2017

Edited by Jon E. Schoenfield, Sep 11 2017

cp's OEIS Frontend

A334080 Number of Pythagorean triples among the divisors of 60*n.

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A217448 Least k > 0 such that 1 + n^2 and 1 + (n+k)^2 have the same smallest prime factor.

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A247339 a(n) is the least number k such that the greatest prime divisor of k^2+1 is the smallest prime divisor of n^2+1.

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A248511 Difference between k and the least prime factor of k^2+1 where k is the n-th number with k^2+1 composite.

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Maple

A248532 Numbers n such that the smallest prime divisor of n^2+1 is 53.

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A248550 Numbers n such that the smallest prime divisor of n^2+1 is 73.

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A248551 Numbers n such that the smallest prime divisor of n^2+1 is 89.

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A248552 Numbers n such that the smallest prime divisor of n^2+1 is 97.

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