A226736 Triangular numbers such that the binary representation is of the form either Tbt or Tt, where T and t are binary representations of triangular numbers with equal binary length, and b is a binary digit: b = 1 or 0.

3, 6, 15, 28, 105, 351, 496, 1830, 7140, 30876, 129795, 508536, 2029105, 8260080, 32817151, 33550336, 133179360, 533811475, 2139135936, 8564460003, 34292793216, 137220150385, 549219598080, 2197036652910, 8791800675840, 35164716131980, 140703139101696, 562926884985640

Offset: 1

Author

Alex Ratushnyak, Jun 16 2013

Keywords

• nonn
• base
• less

Comments

The binary representation of the triangular number t (least significant bits of a(n)) is allowed to have leading zeros.

The sequence of indices of a(n) begins: 2, 3, 5, 7, 14, 26, 31, 60, 119, 248, 509, 1008, 2014, 4064, 8101, 8191, 16320, 32674, 65408, 130877, 261888, 523870, 1048064, 2096204, 4193280, 8386264, 16775168, 33553744, ...

Examples

			105 = (1101001)_2 = (110)_2//(1)_2//(001)_2 is in the sequence, where 105, (110)_2 = 6 and (001)_2=1 are each triangular numbers.
		

Crossrefs

Cf. A000217, A070939, A226836.

Programs

C

#include 
#include 
typedef unsigned long long U64;
U64 isTriangular(U64 a) {  // ! Must be a < (1<<63)
    U64 s = sqrt(a*2);
    return (s*(s+1) == a*2);
}
int main() {
  U64 i, j, n, tn, t, T, prev=0;
  for (n = tn = 3; tn > prev; prev = tn, tn += n, ++n) {
    for (i = 64, j = tn; j < (1ULL<<63); j += j)
      --i;  // binary length of tn
    j = i >> 1;  //  TOt or Tt, binary length of t is j
    t = tn & ((1ULL<> (j+(i&1));
    if (isTriangular(t) && isTriangular(T))
        printf("%20llu %10llu %10llu\n", tn, T, t);
  }
  return 0;
}

Maple

A000217 := proc(n)
    n*(n+1) /2 ;
end proc:
isA000217 := proc(n)
    local t1;
    t1:=floor(sqrt(2*n));
    if n = t1*(t1+1)/2 then
        true
    else
        false;
    end if;
end proc:
for n from 2 do
    tmain := A000217(n) ;
    dgs := convert(tmain,base,2) ;
    ndgs := floor(nops(dgs)/2) ;
    tlo := [op(1..ndgs,dgs)] ;
    if type(nops(dgs),'even') then
        thi := [op(ndgs+1..2*ndgs,dgs)] ;
    else
        thi := [op(ndgs+2..2*ndgs+1,dgs)] ;
    end if;
    tlo := add(op(i,tlo)*2^(i-1),i=1..nops(tlo)) ;
    if isA000217(tlo) then
        thi := add(op(i,thi)*2^(i-1),i=1..nops(thi)) ;
        if isA000217(thi) then
            printf("%d,\n",tmain) ;
        end if;
    end if;
end do: # R. J. Mathar, Jun 18 2013