A188621 -id:A188621 - cp's OEIS frontend

A069752 Smallest k>n such that the triangular number n(n+1)/2 divides the triangular number k(k+1)/2.

2, 3, 8, 15, 9, 14, 48, 63, 35, 44, 32, 39, 77, 20, 80, 255, 135, 152, 75, 35, 77, 230, 183, 200, 299, 324, 188, 203, 144, 155, 960, 351, 153, 84, 224, 296, 665, 246, 104, 615, 245, 258, 472, 99, 414, 1034, 704, 735, 1175, 374, 272, 636, 1377, 539, 175, 399, 551

Offset: 1

Benoit Cloitre, May 01 2002

Note that k <= n^2-1, with equality occurring only if n and n+1 are a prime and a power of 2 (in either order); that is, when n is a Mersenne prime or n+1 is a Fermat prime. - T. D. Noe, Apr 08 2011

Zak Seidov, Table of n, a(n) for n = 1..10000

Cf. A000217, A068084, A188621.

Clear[k]; Join[{2}, Table[Reduce[k*(k+1) == 0, k, Modulus -> n*(n+1)][[3, 2]], {n, 2, 100}]] (* T. D. Noe, Apr 08 2011 *)

for(s=1,1000,s1=s*(s+1);n=s+1; while(n*(n+1)%s1>0,n++); print1(n,","); ) \\ Zak Seidov, Apr 08 2011

A267077 Least m>0 for which mn^2 + 1 is a square and mtriangular(n) + 1 is a triangular number (A000217). Or -1 if no such m exists.

Original entry on oeis.org

1, 35, 30, 18135, 189, 27, 321300, 23760, 1188585957, 1656083, 26, 244894427400, 82093908624206325, 1858717755529547, 86478, 21491811639746039592, 26135932603458945934958445, 353382195058506640426335, 26780050, 7859354769338288038121982384, 274554988002

Offset: 0

Alex Ratushnyak, Jan 10 2016

nonn

			26*10^2+1 = 2601 is a square, and 26*10*11/2+1 = 1431 = triangular(53), and 26 is the smallest such multiplier, therefore a(10) = 26.

Don Reble, Table of n, a(n) for n = 0..300

Cf. A000217, A000290, A035096, A061782, A067872, A188621.

from math import sqrt
def A267077(n):
    if n == 0:
        return 1
    u,v,t,w = max(8,2*n),max(4,n)**2-9,4*n*(n+1),n**2
    while True:
        m,r = divmod(v,t)
        if not r and int(sqrt(m*w+1))**2 == m*w+1:
            return m
        v += u+1
        u += 2 # Chai Wah Wu, Jan 15 2016

#!/usr/bin/python3
# This sequence is easy if you use a Pell-equation solver such as labmath.py
# Solve the A267077 Pell equation:
# nx^2 - (4n+4)y^2 = 5n-4; but also y^2 == 1 mod n^2
# Let u = nx, then # u^2 - n*(4n+4)y^2 = n*(5n-4)
#   and (y > n) and (u == 0 mod n) and (y^2 == 1 mod n^2)
# (y > n makes m > 0)
# Report m = (y^2 - 1) / n^2
import labmath
print(0, 1)
print(1, 35) # When n<2, the Pell equation is elliptical.
for nn in range(2,1001):
    nsq = nn * nn
    ps = labmath.pell(nn*(4*nn+4), nn*(5*nn-4))
    uu,yy = next(ps[0])
    while (yy <= nn) or ((uu % nn) != 0) or ((yy*yy) % nsq != 1):
        uu,yy = next(ps[0])
    print(nn, (yy*yy - 1) // nsq)
# From Don Reble, Apr 15 2022, added by N. J. A. Sloane, Apr 15 2022.

a(12)-a(15) from Chai Wah Wu, Jan 16 2016

a(16) and beyond from Don Reble, Apr 15 2022

cp's OEIS Frontend

A069752 Smallest k>n such that the triangular number n(n+1)/2 divides the triangular number k(k+1)/2.

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A267077 Least m>0 for which mn^2 + 1 is a square and mtriangular(n) + 1 is a triangular number (A000217). Or -1 if no such m exists.

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cp's OEIS Frontend

A069752 Smallest k>n such that the triangular number n*(n+1)/2 divides the triangular number k*(k+1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

A267077 Least m>0 for which m*n^2 + 1 is a square and m*triangular(n) + 1 is a triangular number (A000217). Or -1 if no such m exists.

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Author

Keywords

Examples

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Python

Python

Extensions

A069752 Smallest k>n such that the triangular number n(n+1)/2 divides the triangular number k(k+1)/2.

A267077 Least m>0 for which mn^2 + 1 is a square and mtriangular(n) + 1 is a triangular number (A000217). Or -1 if no such m exists.