A144255 -id:A144255 - 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

A234693 Primes of the form n^2 + 1 such that (n - 1)^2 + 1 and (n + 1)^2 + 1 are semiprimes.

Original entry on oeis.org

17, 101, 28901, 324901, 608401, 902501, 2016401, 5664401, 7452901, 14822501, 16974401, 18490001, 34222501, 40449601, 41731601, 46240001, 48580901, 50410001, 52417601, 76038401, 92736901, 103022501, 111936401, 121220101, 124768901, 139948901, 151290001

Offset: 1

Michel Lagneau, Dec 29 2013

nonn

The corresponding n are 4, 10, 170, 570, 780, 950, 1420, 2380...

Property: n^2 + 1 = p + q - 1 and for a(n) > 17, a(n) == 1 mod 100.

			101 = 10^2 + 1 is in the sequence because 9^2 + 1 = 2*41 and 11^2 + 1 = 2*61.

Donovan Johnson and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from Johnson)

Cf. A002496, A144255, A085722.

with(numtheory):for n from 1 to 10^5 do:n1:=n^2+1:n2:=(n+1)^2+1:n3:=(n+2)^2+1: if type(n2,prime)=true and bigomega(n1)=2 and bigomega(n3)=2 then printf(`%d, `,n2):else fi:od:

forstep(n=4,1e5,2,if(isprime(n^2+1) && isprime(n^2/2-n+1) && isprime(n^2/2+n+1), print1(n^2+1", "))) \\ Charles R Greathouse IV, Dec 29 2013

A261546 Numbers k such that the five numbers k^2+1, (k+1)^2+1, ..., (k+4)^2+1 are all semiprime.

Original entry on oeis.org

48, 58, 1688, 2948, 28338, 36998, 38648, 96248, 100308, 133458, 136798, 187538, 207088, 224508, 253808, 309738, 375348, 545048, 598348, 607688, 659548, 672398, 793958, 1055648, 1055688, 1140008, 1270408, 1317808, 1388398, 1399098, 1529488, 1597008, 1655338

Offset: 1

Michel Lagneau, Aug 24 2015

a(n) == 8 (mod 10).

a(15017) > 10^10. - Hiroaki Yamanouchi, Oct 03 2015

			48 is in the sequence because of these five semiprimes:
48^2+1 = 2305 = 5*461;
49^2+1 = 2402 = 2*1201;
50^2+1 = 2501 = 41*61;
51^2+1 = 2602 = 2*1301;
52^2+1 = 2705 = 5*541.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..15016

Subsequence of A085722.

Cf. A001358, A144255.

IsSemiprime:=func< n | &+[k[2]: k in Factorization(n)] eq 2 >; [ n: n in [1..3*10^5] | IsSemiprime(n^2+1) and IsSemiprime(n^2+2*n+2)and IsSemiprime(n^2+4*n+5)and IsSemiprime(n^2+6*n+10)and IsSemiprime(n^2+8*n+17)]; // Vincenzo Librandi, Aug 24 2015

with(numtheory):
  n:=5:
  for k from 1 to 10^6 do:
    jj:=0:
    for m from 0 to n-1 do:
       x:=(k+m)^2+1:d0:=bigomega(x):
       if d0=2
       then
       jj:=jj+1:
       else
       fi:
     od:
        if jj=n
        then
        printf(`%d, `,k):
        else
        fi:
    od:

PrimeFactorExponentsAdded[n_]:=Plus@@Flatten[Table[#[[2]], {1}]&/@FactorInteger[n]]; Select[Range[2 10^5], PrimeFactorExponentsAdded[#^2+1] == PrimeFactorExponentsAdded[#^2 + 2 # + 2]== PrimeFactorExponentsAdded[#^2 + 4 # + 5]== PrimeFactorExponentsAdded[#^2 + 6 # + 10]== PrimeFactorExponentsAdded[#^2 + 8 # + 17] == 2 &] (* Vincenzo Librandi, Aug 24 2015 *)

has(n) = bigomega(n^2+1)==2;
isok(n) = has(n) && has(n+1) && has(n+2) && has(n+3) && has(n+4); \\ Michel Marcus, Aug 24 2015

a261546() = {
  nterm = 0;
  for (i = 0, 10^9,
    if (isprime(20*i*i + 32*i + 13) &&
      isprime(50*i*i + 90*i + 41) &&
      isprime(50*i*i + 110*i + 61) &&
      isprime(20*i*i + 48*i + 29) &&
      bigomega(100*i*i + 200*i + 101) == 2,
      nterm += 1;
      print(nterm, " ", 10 * i + 8);
    );
  );
} \\ - Hiroaki Yamanouchi, Oct 03 2015

issemi(n)=forprime(p=2,97, if(n%p==0, return(isprime(n/p)))); bigomega(n)==2
list(lim)=my(v=List()); forstep(k=48,lim,[10,30,10], if(issemi(k^2+1) && issemi((k+1)^2+1) && issemi((k+3)^2+1) && issemi((k+4)^2+1) && issemi((k+2)^2+1), listput(v,k))); Vec(v) \\ Charles R Greathouse IV, Jul 06 2017

a(18)-a(33) from Hiroaki Yamanouchi, Oct 03 2015

A333635 Numbers m such that m^2 + 1 has at most 2 prime factors.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 15, 16, 19, 20, 22, 24, 25, 26, 28, 29, 30, 34, 35, 36, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 56, 58, 59, 60, 61, 62, 64, 65, 66, 69, 71, 74, 76, 78, 79, 80, 84, 85, 86, 88, 90, 92, 94, 95, 96, 100

Offset: 1

Bernard Schott, Mar 30 2020

nonn

Equivalently, numbers m such that m^2 + 1 is prime or semiprime.
Henryk Iwaniec proved in 1978 that this sequence is infinite (see link). By contrast, it is not known whether there are infinitely many primes of the form m^2 + 1 (or infinitely many semiprimes of that form).
The integers that have at most 2 prime factors counted with multiplicity are called almost-primes of order 2 and they are in A037143. Here, as m^2 + 1 is not a square for m > 0, all the semiprimes of this form have two distinct prime factors (A144255), and with the primes of the form m^2 + 1 (A002496), they constitute A248742.

			10^2 + 1 = 101, which is prime, so 10 is in the sequence.
11^2 + 1 = 122 = 2 * 61, so 11 is in the sequence.
12^2 + 1 = 145 = 5 * 29, so 12 is in the sequence.
13^2 + 1 = 170 = 2 * 5 * 17, so 13 is not in the sequence.

R. K. Guy, Unsolved Problems in Number Theory, A1.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Henryk Iwaniec, Almost-primes represented by quadratic polynomials, Inventiones Mathematicae 47 (2) (1978), pp. 171-188.

Union of A005574 and A085722.
Cf. A002496 (m^2 + 1 is prime), A005574 (corresponding m).
Cf. A144255 (m^2 + 1 is semiprime), A085722 (corresponding m).
Cf. A248742 (m^2 + 1 is prime or semiprime), this sequence (corresponding m).
Cf. A037143 (numbers with at most 2 prime factors counted with multiplicity).

Select[Range[100], PrimeQ[(k = #^2 + 1)] || PrimeOmega[k] == 2 &]  (* Amiram Eldar, Mar 30 2020 *)
Select[Range[100],PrimeOmega[#^2+1]<3&] (* Harvey P. Dale, Aug 08 2025 *)

A360739 Semiprimes of the form k^2 + 2.

Original entry on oeis.org

6, 38, 51, 123, 146, 291, 326, 731, 843, 1227, 1371, 1766, 1851, 2306, 2603, 2811, 2918, 3027, 3602, 4227, 4358, 4763, 5186, 5331, 5627, 6243, 6891, 7058, 7571, 8102, 8651, 9411, 13227, 14163, 15627, 17426, 17691, 18227, 18771, 19883, 20738, 22502, 23411, 24027

Offset: 1

Elmo R. Oliveira, Feb 18 2023

A242330 gives the corresponding values of k.

			123 is a term because 11^2 + 2 = 123 = 3*41.

Cf. A001358, A059100, A085722, A144255, A242330.

Select[Range[0, 200]^2 + 2, PrimeOmega[#] == 2 &] (* Amiram Eldar, Feb 18 2023 *)

a(n) = A242330(n)^2 + 2.

A360740 Semiprimes of the form k^2 + 3.

Original entry on oeis.org

4, 39, 259, 327, 403, 579, 679, 1027, 1159, 1299, 1603, 1939, 2119, 2307, 3139, 3603, 4359, 4627, 6087, 6403, 7747, 9607, 10003, 10407, 10819, 11667, 13459, 13927, 14403, 16387, 18499, 21907, 23107, 26899, 28903, 30279, 30979, 33127, 35347, 36103, 36867, 38419

Offset: 1

Elmo R. Oliveira, Feb 18 2023

A242331 gives the corresponding values of k.

			259 is a term because 16^2 + 3 = 259 = 7*37.

Cf. A001358, A085722, A117950, A144255, A242330, A242331.

Select[Range[0, 200]^2 + 3, PrimeOmega[#] == 2 &] (* Amiram Eldar, Feb 18 2023 *)

a(n) = A242331(n)^2 + 3.

A360741 Semiprimes of the form k^2 + 4.

Original entry on oeis.org

4, 85, 365, 445, 533, 629, 965, 1685, 1853, 2605, 2813, 3029, 3973, 4765, 5045, 5629, 5933, 6245, 6893, 8285, 8653, 11029, 11453, 11885, 12773, 14165, 15133, 16645, 17165, 17693, 20453, 21029, 22205, 22805, 23413, 24653, 27229, 29245, 29933, 30629, 32765, 34229

Offset: 1

Elmo R. Oliveira, Feb 18 2023

A242332 gives the corresponding values of k.

Except for 4, all terms == 5 (mod 8). - Robert Israel, Feb 18 2023

			85 is a term because 9^2 + 4 = 85 = 5*17.

Cf. A001358, A085722, A087475, A144255, A242330, A242331, A242332.

select(t -> numtheory:-bigomega(t)=2, [seq(i^2+4,i=0..1000)]); # Robert Israel, Feb 18 2023

Select[Range[0, 200]^2 + 4, PrimeOmega[#] == 2 &] (* Amiram Eldar, Feb 18 2023 *)

a(n) = A242332(n)^2 + 4.

A361696 Semiprimes of the form k^2 + 5.

Original entry on oeis.org

6, 9, 14, 21, 69, 86, 201, 329, 446, 489, 581, 681, 734, 789, 905, 1094, 1769, 1941, 2606, 2921, 3254, 3369, 3849, 3974, 4101, 4629, 4766, 6729, 7061, 7401, 8105, 8654, 9609, 9806, 10409, 10821, 11669, 12326, 13929, 17429, 17961, 19049, 20741, 23109, 23721, 24341, 27561, 30281, 31334, 32405

Offset: 1

Elmo R. Oliveira, Mar 20 2023

			69 is a term because 8^2 + 5 = 69 = 3*23.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A144255, A242333, A360739, A360740, A360741.

Intersection of A117951 and A001358.

Select[Range[200]^2 + 5, PrimeOmega[#] == 2 &] (* Paolo Xausa, Aug 21 2025 *)

isok(k) = issquare(k-5) && (bigomega(k)==2); \\ Michel Marcus, Mar 27 2023

a(n) = A242333(n)^2 + 5.

A258780 a(n) is the least k such that k^2 + 1 is a semiprime p*q, p < q, and (q - p)/2^n is prime.

Original entry on oeis.org

8, 12, 140, 64, 2236, 196, 1300, 1600, 6256, 5084, 248756, 246196, 484400, 36680, 887884, 821836, 1559116, 104120, 126072244, 9586736, 4156840, 542759984, 1017981724, 2744780140, 405793096, 148647496, 1671024916

Offset: 2

Michel Lagneau, Jun 10 2015

The corresponding primes are 2, 3, 71, 7, 1069, 7, 5, 5, 59, 2, 368471, 180463, 12421, 2, 29, 125683, 226169, 5, 369704891, 197, 5, 263, 7444559, 239621423, 594271, 2, 474359, ...

All terms are even, in order for k^2+1 to be odd. Otherwise, with k^2+1 being even, p-q would be odd and hence not a multiple of 2^n. - Michel Marcus, Apr 13 2019

			a(2)=8 because 8^2+1 = 5*13 and (13-5)/2^2 = 2 is prime. The number 8 is the first term of the sequence 8, 22, 34, 46, 50, 58, ...
a(3)=12 because 12^2+1 = 5*29 and (29-5)/2^3 = 3 is prime. The number 12 is the first term of the sequence 12, 28, 44, 52, 76, 80, ...
a(4)=140 because 140^2+1 = 17*1153 and (1153-17)/2^4 = 71 is prime. The number 140 is the first term of the sequence 140, 296, 404, 604, ...

Cf. A085722, A144255.

lst={};Do[k=2;While[!(Plus@@Last/@FactorInteger[k^2+1]==2&&PrimeQ[(FactorInteger[k^2+1][[-1,1]]-FactorInteger[k^2+1][[1,1]])/2^n]),k=k+2];Print[n," ",k],{n,2,19}];lst

isok(k, n) = my(kk=k^2+1, f=factor(kk)[,1]~); (bigomega(kk) == 2) && (#f == 2) && (p=f[1]) && (q=f[2]) && (qq=(q-p)/2^n) && !frac(qq) && isprime(qq);
a(n) = my(k=2); while (!isok(k,n), k+=2); k; \\ Michel Marcus, Apr 13 2019

Name edited by Jon E. Schoenfield, Sep 12 2017
a(20)-a(22) from Daniel Suteu, Apr 13 2019
a(23)-a(28) from Daniel Suteu, Nov 09 2019

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

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

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A234693 Primes of the form n^2 + 1 such that (n - 1)^2 + 1 and (n + 1)^2 + 1 are semiprimes.

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A261546 Numbers k such that the five numbers k^2+1, (k+1)^2+1, ..., (k+4)^2+1 are all semiprime.

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A333635 Numbers m such that m^2 + 1 has at most 2 prime factors.

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A360739 Semiprimes of the form k^2 + 2.

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A360740 Semiprimes of the form k^2 + 3.

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A360741 Semiprimes of the form k^2 + 4.

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