A180313 A sequence a(n) such that a(n+1)^2 - a(n)^2 are perfect squares.

3, 5, 13, 85, 221, 1445, 3757, 24565, 63869, 417605, 1085773, 7099285, 18458141, 120687845, 313788397, 2051693365, 5334402749, 34878787205, 90684846733, 592939382485, 1541642394461, 10079969502245, 26207920705837, 171359481538165, 445534651999229, 2913111186148805

Offset: 1

Author

Valentin Tiriac (valtron2000(AT)gmail.com), Aug 26 2010

Keywords

• nonn

Comments

The lexically smallest sequence with a(n+1)^2-a(n)^2 representing perfect squares is A018928.

This version here is constructed via a(n+1) = a(n)* sqrt( 1+((p^2-1)/(2p))^2) where p = A020639(a(n)) is the smallest prime divisor of the previous term.

Examples

			After a(1)=3, p=3 (again) and a(2) = 3*sqrt(1+ (8/6)^2) = 5.
After a(4)=85, p=5 and a(5) = 85*sqrt(1+ (24/10)^2) = 85*sqrt(169/25) = 221.
		

Programs

Maple

A020639 := proc(n) min(op(numtheory[factorset](n))) ; end proc:
A180313 := proc(n) option remember; if n = 1 then 3; else aprev := procname(n-1) ; p := A020639(aprev) ; aprev* sqrt(1+((p^2-1)/2/p)^2) ; end if; end proc:
for n from 1 to 30 do printf("%d,",A180313(n)) ; end do: # R. J. Mathar, Sep 23 2010

Mathematica

spd[n_] := FactorInteger[n][[1, 1]];
a[n_] := a[n] = If[n == 1, 3, aprev = a[n-1];
    p = spd[aprev]; aprev*Sqrt[1+((p^2-1)/2/p)^2]];
Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Feb 28 2024, after R. J. Mathar *)

Perl

# use 5.12.0; use warnings; use Math::Prime::TiedArray; tie my @primes, 'Math::Prime::TiedArray';
sub SmallestPrimeDivisor ($) { my ($n) = @_; for my $p (@primes) { if ($n % $p == 0) { return $p; } } }
sub FindIncrement ($) { my ($n) = @_; my $p = SmallestPrimeDivisor $n; my $k = $n / $p; return $k * ($p ** 2 - 1) / 2; }
my $n = 3; say $n; for my $i (0 .. 23) { my $d = FindIncrement $n; $n = sqrt($d ** 2 + $n ** 2); say $n; }

Extensions

Nomenclature normalized by R. J. Mathar, Sep 23 2010

Corrected indexing error introduced with previous edit - R. J. Mathar, Oct 01 2010