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)))}
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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