A085780 Numbers that are a product of 2 triangular numbers.

0, 1, 3, 6, 9, 10, 15, 18, 21, 28, 30, 36, 45, 55, 60, 63, 66, 78, 84, 90, 91, 100, 105, 108, 120, 126, 135, 136, 150, 153, 165, 168, 171, 190, 198, 210, 216, 225, 231, 234, 253, 270, 273, 276, 280, 300, 315, 325, 330, 351, 360, 378, 396, 406, 408, 420, 435, 441

Offset: 1

Author

Jon Perry, Jul 23 2003

Keywords

• nonn

Comments

Is there a fast algorithm for detecting these numbers? - Charles R Greathouse IV, Jan 26 2013

The number of rectangles with positive width 1<=w<=i and positive height 1<=h<=j contained in an i*j rectangle is t(i)*t(j), where t(k)=A000217(k), see A096948. - Dimitri Boscainos, Aug 27 2015

Examples

			18 = 3*6 = t(2)*t(3) is a product of two triangular numbers and therefore in the sequence.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A085782, A068143, A000537 (subsequence), A006011 (subsequence), A033487 (subsequence), A188630 (subsequence).

Cf. A072389 (this times 4).

Programs

Maple

isA085780 := proc(n)
     local d;
     for d in numtheory[divisors](n) do
        if d^2 > n then
            return false;
        end if;
        if isA000217(d) then
            if isA000217(n/d) then
                return true;
            end if;
        end if;
    end do:
    return false;
end proc:
for n from 1 to 1000 do
    if isA085780(n) then
        printf("%d,",n) ;
    end if ;
end do: # R. J. Mathar, Nov 29 2015

Mathematica

t1 = Table[n (n+1)/2, {n, 0, 100}];Select[Union[Flatten[Outer[Times, t1, t1]]], # <= t1[[-1]] &] (* T. D. Noe, Jun 04 2012 *)

PARI

A003056(n)=(sqrtint(8*n+1)-1)\2
list(lim)=my(v=List([0]),t); for(a=1, A003056(lim\1), t=a*(a+1)/2; for(b=a, A003056(lim\t), listput(v,t*b*(b+1)/2))); vecsort(Vec(v),,8) \\ Charles R Greathouse IV, Jan 26 2013

Python

from itertools import count, islice
from sympy import divisors, integer_nthroot
def A085780_gen(startvalue=0): # generator of terms
    if startvalue <= 0:
        yield 0
    for n in count(max(startvalue,1)):
        for d in divisors(m:=n<<2):
            if d**2 > m:
                break
            if integer_nthroot((d<<2)+1,2)[1] and integer_nthroot((m//d<<2)+1,2)[1]:
                yield n
                break
A085780_list = list(islice(A085780_gen(),10)) # Chai Wah Wu, Aug 28 2022

Formula

Conjecture: There are about sqrt(x)*log(x) terms up to x. - Charles R Greathouse IV, Jul 11 2024

Extensions

More terms from Max Alekseyev and Jon E. Schoenfield, Sep 04 2009