cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A226742 Triangular numbers obtained as the concatenation of 2*k and k.

Original entry on oeis.org

21, 105, 2211, 9045, 222111, 306153, 742371, 890445, 1050525, 22221111, 88904445, 107905395, 173808690, 2222211111, 8889044445, 12141260706, 15754278771, 222222111111, 888890444445, 22222221111111, 36734701836735, 65306123265306
Offset: 1

Views

Author

Antonio Roldán, Jun 18 2013

Keywords

Comments

Includes (2*10^k+1)*(10^k-1)/9 and (2*10^k+1)*(4*10^k+5)/9 for k >= 1. - Robert Israel, Feb 06 2025

Examples

			If k=111, 2k=222, 2k//k = 222111 = 666*667/2, a triangular number.
		

Crossrefs

Programs

  • Maple
    g:= proc(d) local a, b, n, Res, x, y;
          Res:= NULL:
          for a in numtheory:-divisors(2*(2*10^d+1)) do
            b:= 2*(2*10^d+1)/a;
            if igcd(a, b)>1 then next fi;
            n:= chrem([0, -1], [a, b]);
            x:= n*(n+1)/2;
            y:= x/(2*10^d+1);
            if y < 10^(d-1) or y >= 10^d  then next fi;
            Res:= Res, (2*10^d+1)*y
          od;
          op(sort([Res]))
    end proc:
    map(g, [$1..10]); # Robert Israel, Feb 06 2025
  • Mathematica
    TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; t = {}; Do[s = FromDigits[Join[IntegerDigits[2*n], 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=1,10^5,a=concatint(2*n,n);if(istriang(a),print(a)))}

A226772 Triangular numbers obtained as the concatenation of n and 2n.

Original entry on oeis.org

36, 1326, 2346, 3570, 125250, 223446, 12502500, 22234446, 1250025000, 2066441328, 2222344446, 2383847676, 3673573470, 125000250000, 222223444446, 5794481158896, 12500002500000, 12857132571426, 22222234444446, 49293309858660, 804878916097578, 933618918672378, 971908519438170
Offset: 1

Views

Author

Antonio Roldán, Jun 18 2013

Keywords

Comments

Includes 125*10^(2*k+1)+25*10^k and (10^k+2)*(1+(10^k-1)*2/9) for k >= 1. - Robert Israel, Nov 09 2020

Examples

			If n=23, 2n=46, n//2n = 2346 = 68*69/2, a triangular number.
		

Crossrefs

Programs

  • Maple
    F:= proc(d) local D,R,M,m,w,x,x1,x2;
       R:= NULL;
       M:= 10^d/2+1;
       D:= numtheory:-divisors(M);
       for m in D do if igcd(m,M/m)=1 then
         for w in [chrem([-1,1],[8*m,M/m]), chrem([1,-1],[8*m,M/m])] do
         x:= (w^2-1)/8;
         x1:= x mod 10^d;
         x2:= floor(x/10^d);
         if x1 = 2*x2 and x1 >= 10^(d-1) then R:= R, x fi
       od fi od;
       op(sort([R]))
    end proc:
    36, seq(F(d),d=2..10); # Robert Israel, Nov 09 2020
  • Mathematica
    TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; t = {}; Do[s = FromDigits[Join[IntegerDigits[n], IntegerDigits[2*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=1,10^5,a=concatint(n,2*n);if(istriang(a),print(a)))}

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

Original entry on oeis.org

10, 21, 26519722651971, 33388573338856, 69954026995401, 80863378086336
Offset: 1

Views

Author

Antonio Roldán, Jun 18 2013

Keywords

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

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
Showing 1-3 of 3 results.