A186710 -id:A186710 - cp's OEIS frontend

A186713 For the starting base k = A118119(n), a(n) is the largest value q such that gcd(k^n+1, (k+1)^n+1, ..., (k+q)^n+1) > 1.

1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2

Offset: 2

Michel Lagneau, Feb 26 2011

nonn

			a(2) = 1 because 2^2+1 = 5 and 3^2+1 = 2*5 => gcd(..) = 5 and q = 1;
a(53) = 3 because
5^53 + 1 = 2 * 3 * 107 * 28838378869 * 599659003321309822423087;
6^53 + 1 = 7 * 107 * 97351567 * 33685364386033 * 71080464397105403;
7^53 + 1 = 2^3 * 107 * 345449549 * 35416476134069*58902316970027001503;
8^53 + 1 = 3^2 * 107 * 6043 * 28059810762433 * 4475130366518102084427698737 => gcd(..) = 107 and q=3.

Cf. A186710, A118119.

A186713 := proc(n) local k ,g,q; k := A118119(n) ; for q from 1 do g := igcd(seq((k+i)^n+1,i=0..q)) ; if g=1 then return q-1 ; end if; end do: end proc: # R. J. Mathar, Mar 07 2011

a(55), a(56) corrected by R. J. Mathar, Mar 07 2011

A186814 a(n) = smallest number m such that A002144(n) divides gcd(A002314(n)^2+1,(A002314(n)+m)^2+1).

Original entry on oeis.org

1, 3, 9, 5, 25, 23, 7, 39, 19, 21, 53, 81, 43, 83, 63, 61, 101, 13, 143, 31, 169, 15, 55, 113, 225, 105, 157, 175, 17, 263, 89, 41, 77, 269, 165, 159, 271, 361, 123, 363, 75, 315, 239, 365, 93, 51, 437, 321, 397, 529, 439, 351, 543, 229, 333, 355, 449, 557, 625, 431, 517, 27, 583

Offset: 1

Michel Lagneau, Feb 27 2011

nonn

Sequence A002314 gives the minimal integer square root of -1 modulo p(n),where p(n) = n-th prime of form 4k+1.

			for n=1, k = A002314(1) = 2 => a(1) = 1, because 2^2+1 = 5 and (2+1)^2+1 = 2*5 ;
for n=2, k = A002314(2) = 5 => a(2) = 3, because 5^2+1 = 2*13 and (5+3)^2+1 = 5*13 ;
for n=3, k = A002314(3) = 4 => a(3) = 9, because 4^2+1 = 17 and (4+9)^2+1 = 2*5*17;
for n=4, k = A002314(4)= 12 => a(4)= 5, because 12^2+1
= 5*29 and (12+5)^2+1 = 2*5*29, and 29 divides
GCD(5*29, 2*5*29)=145.

Cf. A002314, A002144, A186710, A186713.

with(numtheory):T:=array(1..90):j:=1:for i from 1 to 250 do:x:=4*i+1:if type(x,prime)=true
  then T[j]:=x:j:=j+1:else fi:od:for p from 1 to j do:u:=T[p]:id:=0: for m from
  1 to 1000 while(id=0) do: z:=m^2+1:for d from 1 to u while(id=0) do: z1:=(m+d)^2+1:zz:=
  gcd(z,z1):if irem(zz,u)=0 then id:=1:printf(`%d, `,d):else fi:od:od:od:

cp's OEIS Frontend

A186713 For the starting base k = A118119(n), a(n) is the largest value q such that gcd(k^n+1, (k+1)^n+1, ..., (k+q)^n+1) > 1.

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