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

Author

Alex Ratushnyak, May 17 2013

Keywords

• nonn

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

• Alois P. Heinz, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000217, A002378, A082183, A088572.

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

Programs

Maple

a:= proc(n) option remember; local w, k; w:= n*(n+1)/2;
      for k while not issqr(8*(w+k*(k+1))+1) do od; k
    end:
seq(a(n), n=0..69);  # Alois P. Heinz, Nov 13 2024

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 *)

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 */

Python

def isTriangular(a):
    sr = 1 << (a.bit_length() >> 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(m, end=', ')
  n += 1
  tn += n