A134407 -id:A134407 - cp's OEIS frontend

A134406 Composite numbers of the form k^2 + 1.

10, 26, 50, 65, 82, 122, 145, 170, 226, 290, 325, 362, 442, 485, 530, 626, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1370, 1445, 1522, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501, 2602, 2705, 2810, 3026, 3250, 3365, 3482, 3601, 3722, 3845

Offset: 1

Jani Melik, Jan 18 2008

nonn

Square roots of these numbers are quadratic irrationals and corresponding chain fraction representations are periodic: sqrt(10) = [3;{2,3}], sqrt(26) = [5;{2,5}], sqrt(50) = [7;{2,7}], ..., where {} is denoted a period (we write {6} == {2,3}).

			10 is a term because 10 = 3^2 + 1 is composite,
26 is a term because 26 = 5^2 + 1 is composite,
50 is a term because 50 = 7^2 + 1 is composite.

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

Cf. A002496, A005574, A134407.

Supersequence of A144255.

ts_fn1:=proc(n) local i,tren,ans; ans:=[ ]: for i from 1 to n do tren := i^(2)+1: if (isprime(tren) = false) then ans:=[ op(ans), tren ]: fi od: RETURN(ans) end: ts_fn1(200);

Select[Range[70]^2+1,!PrimeQ[#]&] (* Harvey P. Dale, Aug 12 2012 *)

for(n=3,99, if(!isprime(t=n^2+1), print1(t", "))) \\ Charles R Greathouse IV, Aug 29 2016

from sympy import isprime
from itertools import count, takewhile
def aupto(limit):
    form = takewhile(lambda x: x <= limit, (k**2+1 for k in count(1)))
    return [number for number in form if not isprime(number)]
print(aupto(3845)) # Michael S. Branicky, Oct 26 2021

a(n) = 1 + A134407(n)^2. - R. J. Mathar, Oct 13 2019

A180507 Numbers k such that k^2 + 1 = p*q, p and q prime with p == q (mod k).

Original entry on oeis.org

3, 8, 12, 144, 1020, 8040, 13860, 34840, 729180, 1728240, 3232060, 17576520, 39279240, 85184880, 117649980, 778689840, 884737920, 1225045140, 1771563420, 3723878100, 3869896140, 4574299320, 7762395960, 12487172640, 14348911860, 14886940920, 21484957560, 24137574780

Offset: 1

Michel Lagneau, Jan 20 2011

nonn

q - p = k with k = 3, 8, 144.

The next terms with q - p = k are F(432) = 85738...5984 and F(570) where F(n) is the n-th Fibonacci number. All such terms are in A001906; the next such term, if one exists, has more than 25000 decimal digits. - Charles R Greathouse IV, Jan 21 2011

			a(3) = 12 because 12^2 + 1 = 5*29 and 29 - 5 = 2*12;
a(8) = 34840 because 34840^2 + 1 = 4289 * 283009 and 283009 - 4289 = 278720 = 8*34840.

Jinyuan Wang, Table of n, a(n) for n = 1..10000

Subset of A085722.

Cf. A001906, A002496, A005574, A027862, A134406, A134407.

with(numtheory):for k from 1 to 40000 do: x:=k^2+1:y:=factorset(x):yy:=bigomega(x):if
  yy=2 and irem(y[2],k) =y[1] then printf(`%d, `,k):else fi:od:

w(m, r) = Vec(x*(1-x)/(1-(m^2+2)*x+x^2) + O(x^r));
isok(s, t) = isprime(s) && isprime(s+t);
lista(nn) = {my(g, k, m=1, r, u=w(1, nn), v=List([])); for(i=2, r=#u, g=k=(u[i]+sqrtint(5*u[i]^2-4))/2; if(isok(u[i], k), listput(v, k))); while(r>2, u=w(m++, r); for(i=2, #u, k=(m*u[i]+sqrtint((m^2+4)*u[i]^2-4))/2; if(kJinyuan Wang, Mar 29 2020

More terms from Charles R Greathouse IV, Jan 24 2011

Missing terms inserted and more terms from Jinyuan Wang, Mar 30 2020

A187401 Numbers k such that k^2 + 1 = p*q, p and q primes and |p-q| is square.

Original entry on oeis.org

30, 100, 144, 274, 484, 516, 526, 756, 1046, 1250, 1714, 1806, 1834, 2284, 2440, 2610, 2940, 3524, 3824, 4190, 5084, 5746, 6766, 7486, 9746, 9920, 10310, 13024, 13210, 15396, 16916, 17546, 18726, 19256, 20000, 21194, 23214, 24964, 30370, 30394, 31126, 31496, 35180, 36680, 37816

Offset: 1

Michel Lagneau, Mar 09 2011

nonn

Note that if k^2+1 = p*q, then p+q cannot be a square. Proof by contradiction. There are two cases: p an odd prime and p=2. Case 1: suppose p and q are odd primes and q = y^2-p. Note that y must be an even number in order for q to be odd. Then p(y^2-p) = x^2+1 for some even x. Rearranging terms, we obtain p*y^2-1 = p^2+x^2. Looking at this equation modulo 4, we obtain -1 = 1, a contradiction. Case 2: Let p=2. Then we obtain 2y^2-x^2 = 5, which has no solutions in integers. - T. D. Noe, Mar 10 2011

			20000 is in the sequence because 20000^2+1 = 19801 * 20201 and 20201 - 19801 = 20^2.

Robert Israel, Table of n, a(n) for n = 1..600

Cf. A134406, A134407, A002522, A005574.

with(numtheory):nn:=50000:for i from 1 to nn do: n:=i^2+1:x:=factorset(n):x1:=nops(x):x2:=bigomega(n):if  x1=2 and x2=2 then z:=x[2]-x[1] :w:=sqrt(z):if w= floor(w) then printf(`%d,  `, i):else fi:else fi :od:
# Alternative:
N:= 500: # to get a(1) to a(N)
count:= 0:
for k from 2 by 2 while count < N do
  f:= ifactors(k^2+1)[2];
  if nops(f) = 2 and {f[1,2],f[2,2]}={1} and issqr(abs(f[1,1]-f[2,1])) then
    count:= count+1;
    A[count]:= k;
  fi
od:
seq(A[i],i=1..count); # Robert Israel, Jun 09 2014

okQ[k_] := Module[{ff = FactorInteger[k^2+1]}, Length[ff] == 2 && ff[[All, 2]] == {1, 1} && IntegerQ[Sqrt[ff[[2, 1]] - ff[[1, 1]]]]];
Select[Range[2, 40000, 2], okQ] (* Jean-François Alcover, Jun 25 2019 *)

A = []
for k in range(2, 2000, 2):
    K = k^2 + 1
    f = prime_divisors(K)
    if len(f) == 2:
        if mul(f) == K:
            if is_square(abs(f[0]-f[1])):
                A.append(k)
print(A) # Peter Luschny, Jun 10 2014

A238947 Numbers k such that k^2 + 1 = p*q, p < q primes and q-p is a power of 2.

Original entry on oeis.org

8, 100, 3524, 5084, 36680, 77980, 21474824, 134201344, 148647496, 300741464, 73851531256, 153122539756, 778318386944, 6669171349484, 16526971109344, 596403262068016, 9376599920124524, 26698166963373164, 140144514160214876, 1613032378604451500

Offset: 1

Michel Lagneau, Mar 07 2014

Note that if n^2+1 = p*q, then p+q cannot be a power of 2. Proof by contradiction: There are two cases: p an odd prime and p=2. Case 1: suppose p and q are odd primes and q = 2^m-p. Then p(2^m-p) = n^2+1 for some even n. Rearranging terms, we obtain p*2^m-1 = p^2+n^2. Looking at this equation modulo 4, we obtain -1 = 1, a contradiction. Case 2: Let p=2. Then we obtain 2^(m+1)-n^2 = 5, which has no solutions in integers.

			8^2+1 = 65 = 5*13 and 13-5 = 2^3;
100^2+1 = 10001 = 73*137 and 137-73 = 2^6;
3524^2+1 = 12418577 = 3049*4073 and 4073-3049 = 2^10.

Giovanni Resta, Some terms greater than a(20)

Cf. A134406, A134407, A002522.

Subsequence of A085722.

with(numtheory):for a from 1 to 200000 do:p:=ithprime(a):for i from 1 to 50 do:q:=p+2^i:if type(q,prime)=true then x:=sqrt(p*q-1):if x=floor(x) then print(x):else fi:fi:od:od:

Select[Range[10^5],!PrimeQ[#^2+1]&&Plus@@Last/@FactorInteger[#^2+1]==2&&PrimeNu[#^2+1]==2&&IntegerQ[Log[2,FactorInteger[#^2+1][[2]][[1]]-FactorInteger[#^2+1][[1]][[1]]]]&]

isok(n) = (bigomega(n^2+1) == 2) && (f = factor(n^2+1)) && ((f[2, 1] - f[1, 1])== 2^(valuation(f[2, 1] - f[1, 1], 2))); \\ Michel Marcus, Mar 07 2014

a(7)-a(20) from Giovanni Resta, Mar 07 2014

A247018 Numbers of the form 3*z^2 + z + 3 for some integer z.

Original entry on oeis.org

3, 5, 7, 13, 17, 27, 33, 47, 55, 73, 83, 105, 117, 143, 157, 187, 203, 237, 255, 293, 313, 355, 377, 423, 447, 497, 523, 577, 605, 663, 693, 755, 787, 853, 887, 957, 993, 1067, 1105, 1183, 1223, 1305, 1347, 1433, 1477, 1567, 1613, 1707, 1755, 1853, 1903

Offset: 1

Matt C. Anderson, Sep 09 2014

Note that z is allowed to be negative. - N. J. A. Sloane, Jul 09 2021
Subsequence of A134407.
Numbers k such that 12*k - 35 is a square. - Robert Israel, Sep 18 2014

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A002522, A001105.

select(t -> issqr(12*t-35), [$1..1000]); # Robert Israel, Sep 18 2014

Union[Flatten[Table[3z^2+{z,-z}+3,{z,0,40}]]] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{3,5,7,13,17},60] (* Harvey P. Dale, Jul 10 2021 *)

Vec(x*(3 + 2*x - 4*x^2 + 2*x^3 + 3*x^4) / ((1 - x)^3*(1 + x)^2) + O(x^60)) \\ Colin Barker, Feb 01 2018

From Colin Barker, Feb 01 2018: (Start)
G.f.: x*(3 + 2*x - 4*x^2 + 2*x^3 + 3*x^4) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5. (End)

At some point in the history of this entry the definition was changed from the correct definition to the erroneous "a(n) = 3*n^2 + n + 3". I have restored the original definition, and I deleted some incorrect programs. Thanks to Harvey P. Dale for pointing out that something was wrong. - N. J. A. Sloane, Jul 09 2021.

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:

A180508 Numbers k such that k^2 + 1 = p*q, p and q primes and | p-q | <= k.

Original entry on oeis.org

3, 8, 46, 50, 76, 100, 144, 266, 274, 334, 504, 516, 526, 566, 670, 726, 756, 810, 836, 844, 1064, 1086, 1164, 1250, 1300, 1714, 1740, 1800, 1826, 1834, 1946, 1950, 2014, 2194, 2220, 2440, 2450, 2466, 2494, 2560, 2610

Offset: 1

Michel Lagneau, Jan 20 2011

nonn

|p - q| = k for k = 3, 8, 144.

			46 is in the sequence because 46^2 + 1 = 29*73, and 73-29 = 44 < 46.

Cf. A134406 A027862 A134407 A002522 A005574.

with(numtheory):for n from 1 to 4000 do: x:=n^2+1:y:=factorset(x):yy:=bigomega(x):if
  yy=2 and (y[2]-y[1] < n or y[2]-y[1] = n) then printf(`%d, `,n):else fi:od:

A248015 Positive numbers n such that n^2 + 1 is composite and there are no positive integers c and z such that n = c*z^2 + z + c.

Original entry on oeis.org

8, 18, 28, 30, 34, 44, 46, 48, 50, 58, 60, 64, 68, 70, 76, 78, 86, 88, 96, 98, 100, 104, 108, 114, 118, 128, 136, 144, 148, 158, 164, 166, 168, 178, 186, 188, 190, 194, 196, 198, 200

Offset: 1

Matt C. Anderson, Sep 29 2014

nonn

Subset of A134407.
If f(x) = x^2 + 1 and g(c,y) = c*y^2 + y + c then the algebraic substitution of g for x gives a factorization: f(g(c,y)) = (y^2 + 1)*(c^2*y^2 + c^2 + 2*c*y + 1).  Since both factors of f(g(c,y)) are integers greater than one, f(g(c,y)) is a composite number.
The numbers are necessarily even terms from A134407 since for odd n = 2c + 1 one has the "forbidden" decomposition with z = 1. - M. F. Hasler, Oct 04 2014

Eric Weisstein's World of Mathematics, Landau's Problems

Cf. A134407.

maxn:=200:
mb:=proc(n::integer)::integer;
  if isprime(n^2+1)=false then return n else return -1 fi;
end proc:
A134407 := Vector(maxn):
for a from 1 to maxn do A134407[a]:= mb(a): end do:
A134407s:=convert(A134407,'set') minus {-1}:
A134407l:=convert(A134407s,'list'):
for c from 1 to 200 do
  for z from 1 to 20 do
    A134407s := A134407s minus {c*z^2 + z + c}:
  end do:
end do:
A134407s;

is(n)={!bittest(n,0)&&!isprime(n^2+1)&&!for(z=2,sqrtint(n),(n-z)%(z^2+1)||return)} \\ M. F. Hasler, Oct 04 2014

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A134406 Composite numbers of the form k^2 + 1.

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A180507 Numbers k such that k^2 + 1 = p*q, p and q prime with p == q (mod k).

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A187401 Numbers k such that k^2 + 1 = p*q, p and q primes and |p-q| is square.

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A238947 Numbers k such that k^2 + 1 = p*q, p < q primes and q-p is a power of 2.

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A247018 Numbers of the form 3*z^2 + z + 3 for some integer z.

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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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A180508 Numbers k such that k^2 + 1 = p*q, p and q primes and | p-q | <= k.

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