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)))}
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
A226752
Possible total sums of three 3-digit primes that together use all nonzero digits 1-9.
Original entry on oeis.org
999, 1089, 1107, 1197, 1269, 1287, 1323, 1341, 1359, 1377, 1413, 1431, 1449, 1467, 1521, 1539, 1557, 1593, 1611, 1629, 1647, 1683, 1701, 1737, 1773, 1791, 1809, 1827, 1863, 1881, 1899, 1917, 1953, 1971, 1989, 2007, 2043, 2061, 2133, 2151, 2223, 2241, 2331, 2421
Offset: 1
149 + 263 + 587 = 999, and 149, 263, and 587 are all primes, so 999 is a (the smallest) term of the sequence. 653 + 827 + 941 = 2421, and 653, 827, and 941 are all primes, so 2421 is a (the largest) term of the sequence.
- David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 149 (entry for 999).
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Union[Transpose[Join[#,{Total[#]}]&/@(FromDigits/@Partition[#,3]&/@ Select[Permutations[Range[9]],And@@PrimeQ[FromDigits/@ Partition[ #,3]]&])][[4]]]
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from sympy import isprime
from itertools import permutations
aset = set()
for p in permutations("123456789"):
p = [int("".join(p[i*3:(i+1)*3])) for i in range(3)]
if all(isprime(pi) for pi in p): aset.add(sum(p))
print(sorted(aset)) # Michael S. Branicky, Jun 28 2021
Showing 1-4 of 4 results.
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