A217393 Smallest k > 0 such that 1 + n^2 and 1 + (n+k)^2 have the same largest factor, or 0 if no such k exists.

0, 1, 4, 9, 3, 25, 0, 10, 23, 81, 39, 5, 8, 169, 83, 225, 24, 39, 143, 361, 17, 53, 7, 529, 263, 625, 19, 101, 363, 53, 12, 41, 43, 21, 543, 1225, 63, 9, 683, 1521, 29, 269, 25, 61, 923, 127, 221, 365, 1103, 22, 1199, 437, 175, 2809, 68, 3025, 182, 557, 1623, 157

Offset

1,3

Comments

Numbers k such that A014442(n) = A014442(n+k), otherwise 0.

A014442(n) is the largest prime factor of n^2 + 1.

a(n) = 0 when A014442(n) is the last possible largest prime, for instance a(1) = 0, a(7) = 0 whose corresponding largest primes are respectively 2 and 5. The general case for the numbers n such that a(n) = 0 is difficult.

Links

Table of n, a(n) for n=1..60.

Example

a(1) = 0 because A014442(1) = 2 is the unique largest prime of A014442(n);
a(2) = 1 because A014442(2) = 5 and A014442(2+1) = 5;
a(3) = 4 because A014442(3) = 5 and A014442(3+4) = 5;
a(4) = 9 because A014442(4) = 17 and A014442(4+17) = 17.
a(57) = 182 because A014442(57) = 13 and A014442(182+57) = 13.

Maple

with(numtheory):T:=array(1..300): for n from 1 to 300 do:x:=factorset(n^2+1):n1:=nops(x): T[n] := x[n1]:od:for a from 1 to 60 do:p:=T[a]:ii:=0:for b from a to 10000 do: z:=factorset(b^2+1): n2:=nops(z):if z[n2]=p and ii=0 then b0:=b:ii:=1:else if z[n2]=p and ii=1 then b1:=b:printf(`%d, `, b1-b0):ii:=2:else fi:fi:od:if ii=1 then printf(`%d, `, 0):else fi:od:

Crossrefs

Cf. A014442.

Sequence in context: A070436 A259450 A219731 * A285323 A365325 A321219

Adjacent sequences: A217390 A217391 A217392 * A217394 A217395 A217396

Keyword

nonn

Author

Michel Lagneau, Oct 02 2012

Status

approved