A067872 -id:A067872 - cp's OEIS frontend

A068310 n^2 - 1 divided by its largest square divisor.

3, 2, 15, 6, 35, 3, 7, 5, 11, 30, 143, 42, 195, 14, 255, 2, 323, 10, 399, 110, 483, 33, 23, 39, 3, 182, 87, 210, 899, 15, 1023, 17, 1155, 34, 1295, 38, 1443, 95, 1599, 105, 1763, 462, 215, 506, 235, 138, 47, 6, 51, 26, 2703, 78, 2915, 21, 3135, 203, 3363, 870, 3599

Offset: 2

Lekraj Beedassy, Feb 25 2002

In other words, squarefree part of n^2-1.

Least m for which x^2 - m*y^2 = 1 has a solution with x = n.

			a(6) = 35, as 6^2 - 1 = 35 itself is squarefree.
7^2-1 = 48 = A005563(6), whose largest square divisor is A008833(48) = 16, so a(7) = 48/16 = 3.

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000

Cf. A002350, A007913, A067872, A033314, A027746, A175607.

a068310 n = f 1 $ a027746_row (n^2 - 1) where
   f y [] = y
   f y [p] = y*p
   f y (p:ps'@(p':ps)) | p == p' = f y ps
                       | otherwise = f (y*p) ps'
-- Reinhard Zumkeller, Nov 26 2011

a[n_] := Times@@(#[[1]] ^ Mod[ #[[2]], 2]&/@FactorInteger[n^2-1])
Table[(n^2-1)/Max[Select[Divisors[n^2-1],IntegerQ[Sqrt[#]]&]],{n,2,60}] (* Harvey P. Dale, Dec 08 2019 *)

a(n) = core(n*n - 1); \\ David Wasserman, Mar 07 2005

a(n) = A007913(n^2-1).

a(n) = A005563(n-1) / A008833(n^2 - 1). - Reinhard Zumkeller, Nov 26 2011; corrected by Georg Fischer, Dec 10 2022

Edited by Dean Hickerson, Mar 19 2002

Entry revised by N. J. A. Sloane, Apr 27 2007

A128972 n^3 - 1 divided by its largest cube divisor.

Original entry on oeis.org

7, 26, 63, 124, 215, 342, 511, 91, 37, 1330, 1727, 2196, 2743, 3374, 4095, 614, 17, 254, 7999, 9260, 10647, 12166, 13823, 1953, 17575, 19682, 813, 24388, 26999, 29790, 32767, 4492, 39303, 42874, 46655, 1876, 54871, 59318, 63999, 8615, 74087, 79506

Offset: 2

Jonathan Vos Post, Apr 28 2007

In other words, cubefree part of n^3-1, or cubefree kernel of n^3-1. Cube analog of A068310.

			a(9) = (9^3-1)/8 = (2^3 * 7 * 13)/(2^3) = 728/8 = 91.
a(10) = (10^3-1)/27 = (3^3 * 37)/(3^3) = 999/27 = 37.
a(18) = (18^3-1)/343 = (7^3 * 17)/(7^3) = 5831/343 = 17.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A002350, A004709, A007948, A062378, A067872, A033314, A068310.

a:= n -> mul(f[1]^(f[2] mod 3), f = ifactors(n^3-1)[2]):
seq(a(n),n=2..100); # Robert Israel, Sep 24 2014

a(n) = A062378(A068601(n)) = A062378(n^3-1).

More terms from Carl R. White, Nov 09 2010

A266236 Least m > 0 such that m*n^3 + 1 is a cube.

Original entry on oeis.org

1, 7, 91, 37, 4291, 16003, 1801, 17, 263683, 19927, 1003003, 1775557, 111169, 506115, 17145, 423001, 16789507, 24152311, 1261657, 3266062, 64024003, 5080, 113411851, 148072393, 7082497, 244187503, 1922636, 14355469, 3132736, 594896491, 27009001, 8341522, 1073840131

Offset: 0

Alex Ratushnyak, Dec 25 2015

nonn

Least m>0 for which x^3 - m*y^3 = 1 has a solution with y = n.

			17*7^3+1 = 18^3, and 17 is the smallest positive m such that m*7^3+1 is a cube, so a(7)=17.

Robert G. Wilson v, Table of n, a(n) for n = 0..1000

Cf. A000578, A035096, A067872.

f[n_] := Block[{x = 2, n3 = n^3}, While[ Mod[x^3 - 1, n3] != 0, x++]; (x^3 - 1)/n3]; f[0] = 1; Array[f, 34, 0] (* Robert G. Wilson v, Mar 24 2016 *)

a(n) = {my(m = 1, cn = n^3); while (!ispower(m*cn + 1, 3), m++); m;} \\ Michel Marcus, Feb 09 2016

a(n) = A076947(n^3). - Robert Israel, Dec 25 2015

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

A128251 n^4 - 1 divided by its largest fourth power divisor.

Original entry on oeis.org

15, 5, 255, 39, 1295, 150, 4095, 410, 9999, 915, 20735, 1785, 38415, 3164, 65535, 5220, 104975, 8145, 159999, 12155, 234255, 17490, 331775, 24414, 456975, 33215, 614655, 44205, 809999, 57720, 1048575, 74120, 1336335, 93789, 1679615, 117135

Offset: 2

Jonathan Vos Post, May 03 2007

In other words, biquadratefree part of n^4-1, or biquadratefree kernel of n^4-1. Fourth power analog of what A128972 is to cubes and A068310 is to squares. A046100 Biquadratefree numbers. A008835 Largest 4th power dividing n.

			a(3) = 5 because (3^4 - 1)/16 = 80/16 = (2^4 * 5)/(2^4) = 5.
a(5) = 39 because (5^4 - 1)/16 = 624/16 = (2^4 * 3 * 13)/(2^4) = 39.
a(7) = 150 because (7^4 - 1)/16 = 2400/16 = (2^5 * 3 * 5^2)/(2^4) = 150.
a(9) = 410 because (9^4 - 1)/16 = 6560/16 = (2^5 * 5 * 41)/(2^4) = 410.
a(63) = 61535 because (63^4 - 1)/256 = 15752960/256 = (2^8 * 5 * 31 * 397)/(2^8) = 61535.

Eric Weisstein's World of Mathematics, Biquadratefree.

Cf. A000188, A000583, A002350, A004709, A007948, A008835, A062378, A067872, A033314, A068310, A128972.

a(n) = (n^4 - 1)/A008835(n^4 - 1) = (A000583(n)-1)/A008835((A000583(n)-1)).

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A068310 n^2 - 1 divided by its largest square divisor.

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A128972 n^3 - 1 divided by its largest cube divisor.

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A266236 Least m > 0 such that m*n^3 + 1 is a cube.

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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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A128251 n^4 - 1 divided by its largest fourth power divisor.

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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.