A225844 Least k>0 such that triangular(n) + k*(k+1) is a triangular number.

2, 1, 3, 5, 7, 2, 11, 13, 5, 17, 19, 3, 6, 25, 27, 9, 31, 33, 35, 4, 9, 41, 8, 45, 47, 10, 14, 53, 9, 5, 59, 61, 21, 18, 67, 69, 21, 73, 75, 14, 22, 6, 11, 13, 87, 15, 91, 26, 20, 34, 12, 101, 26, 105, 30, 7, 20, 33, 115, 117, 119, 34, 21, 125, 37, 129, 29, 133, 14, 137

Offset

0,1

Comments

For n>0, a(n) <= 2*n-1, because n*(n+1)/2 + (2*n-1)*2*n = (9*n^2 - 3*n)/2 = 3*n*(3*n-1)/2 = triangular(3*n-1).

The subsequence with terms less than 2*n-1 begins: 2, 5, 3, 6, 9, 4, 9, 8, 10, 14, 9, 5, 21, 18, 21, 14, 22, 6, 11, 13, 15, ...

The sequence of n's such that a(n) < 2*n-1 begins: 5, 8, 11, 12, 15, 19, 20, 22, 25, 26, ...

Links

Table of n, a(n) for n=0..69.

Mathematica

lktrno[n_]:=Module[{t=(n(n+1))/2, k=1}, While[!IntegerQ[(Sqrt[ 8(t+k(k+1))+1]-1)/2], k++]; k]; Array[lktrno, 70, 0] (* Harvey P. Dale, Aug 19 2014 *)

Prog

(Python)
def isTriangular(a):
    sr = 1L << (long.bit_length(long(a)) >> 1)
    a += a
    while a < sr*(sr+1):  sr>>=1
    b = sr>>1
    while b:
      s = sr+b
      if a >= s*(s+1):  sr = s
      b>>=1
    return (a==sr*(sr+1))
n = tn = 0
while 1:
for m in range(1, 1000000000):
    if isTriangular(tn + m*(m+1)): break
  print str(m)+', ',
  n += 1
  tn += n
(PARI) a(n)=for(k=1, 2*n, t=n*(n+1)/2+k*(k+1); x=sqrtint(2*t); if(t==x*(x+1)/2, return(k))) /* from Ralf Stephan */

Crossrefs

Cf. A000217, A002378, A082183.

Cf. A101157 (least k>0 such that triangular(n) + k^2 is a triangular number).

Sequence in context: A275705 A217036 A127201 * A006769 A075643 A076074

Adjacent sequences: A225841 A225842 A225843 * A225845 A225846 A225847

Keyword

nonn

Author

Alex Ratushnyak, May 17 2013

Status

approved