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
If k=111, 2k=222, 2k//k = 222111 = 666*667/2, a triangular number.
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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
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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 *)
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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
If n=23, 2n=46, n//2n = 2346 = 68*69/2, a triangular number.
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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
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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 *)
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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)))}
A226788
Triangular numbers obtained as the concatenation of n and n+1.
Original entry on oeis.org
45, 78, 4950, 5253, 295296, 369370, 415416, 499500, 502503, 594595, 652653, 760761, 22542255, 49995000, 50025003, 88278828, 1033010331, 1487714878, 4999950000, 5000250003, 490150490151, 499999500000, 500002500003, 509949509950, 33471093347110, 49999995000000, 50000025000003
Offset: 1
If n=295, n//n+1 = 295296 = 768*769/2, a triangular number.
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TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; t = {}; Do[s = FromDigits[Join[IntegerDigits[n], IntegerDigits[n+1]]]; If[TriangularQ[s], AppendTo[t, s]], {n, 100000}]; t (* T. D. Noe, Jun 18 2013 *)
Select[FromDigits[Join[Flatten[IntegerDigits[#]]]]&/@Partition[ Range[ 5000010],2,1], OddQ[Sqrt[8#+1]]&] (* Harvey P. Dale, Jun 11 2015 *)
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concatint(a,b)=eval(concat(Str(a),Str(b)))
istriang(x)=issquare(8*x+1)
{for(n=1,10^7,a=concatint(n,n+1);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
26519722651971 is the concatenation of 2651972 and 2651971 and a triangular number, because 26519722651971 = 7282818*7282819/2.
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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 *)
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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)))}
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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
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# 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
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