A226789 Triangular numbers obtained as the concatenation of n+1 and n.

10, 21, 26519722651971, 33388573338856, 69954026995401, 80863378086336

Offset: 1

Author

Antonio Roldán, Jun 18 2013

Keywords

• nonn
• base
• more

Comments

There are only six terms less than 10^20.

Examples

			26519722651971 is the concatenation of 2651972 and 2651971 and a triangular number, because 26519722651971 = 7282818*7282819/2.
		

Crossrefs

Cf. A003098, A068899, A226742, A226772, A226788.

Programs

Mathematica

TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; t = {}; Do[s = FromDigits[Join[IntegerDigits[n+1], IntegerDigits[n]]]; If[TriangularQ[s], AppendTo[t, s]], {n, 100000}]; t (* T. D. Noe, Jun 18 2013 *)

PARI

concatint(a,b)=eval(concat(Str(a),Str(b)))
istriang(x)=issquare(8*x+1)
{for(n=1,10^7,a=concatint(n+1,n);if(istriang(a),print(a)))}

Python

from math import isqrt
def istri(n): t = 8*n+1; return isqrt(t)**2 == t
def afind(klimit, kstart=0):
    strk = "0"
    for k in range(kstart, klimit+1):
        strkp1 = str(k+1)
        t = int(strkp1 + strk)
        if istri(t):
            print(t, end=", ")
        strk = strkp1
afind(81*10**5) # Michael S. Branicky, Oct 21 2021

Python

# alternate version
def isconcat(n):
    if n < 10: return False
    s = str(n)
    mid = (len(s)+1)//2
    lft, rgt = int(s[:mid]), int(s[mid:])
    return lft - 1 == rgt
def afind(tlimit, tstart=0):
    for t in range(tstart, tlimit+1):
        trit = t*(t+1)//2
        if isconcat(trit):
            print(trit, end=", ")
afind(13*10**6) # Michael S. Branicky, Oct 21 2021