OFFSET
1,3
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
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
FORMULA
Conjecture: There are about sqrt(x)*log(x) terms up to x. - Charles R Greathouse IV, Jul 11 2024
EXAMPLE
18 = 3*6 = t(2)*t(3) is a product of two triangular numbers and therefore in the sequence.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon Perry, Jul 23 2003
EXTENSIONS
More terms from Max Alekseyev and Jon E. Schoenfield, Sep 04 2009
STATUS
approved