A255832 -id:A255832 - cp's OEIS frontend

A255852 Least k > 0 such that gcd(k^n+2, (k+1)^n+2) > 1, or 0 if there is no such k.

1, 0, 1, 51, 1, 40333, 1, 434, 1, 16, 1, 1234, 1, 78607, 1, 8310, 1, 817172, 1, 473, 1, 116, 1, 22650, 1, 736546059, 1, 22, 1, 1080982, 1, 252, 1, 7809, 1, 644, 1, 1786225573, 1

Offset: 0

M. F. Hasler, Mar 08 2015

See A118119, which is the main entry for this class of sequences.
a(39) <= 8105110304875691067. - Max Alekseyev, Aug 06 2015
a(41) = 34290868, a(49) <= 2002111070, a(47) = 32286649814088452353414982038778088771611290478685407234712300075870593693164721\
99455164873287615636327176797646292254029648497024652505965417768073756378034012\
80883965289152013363422286845290874810700297549641281106223286199677401563701715\
56997846264124867393209579875386439424082082891813462700417531719383529314983727. - Hiroaki Yamanouchi, Mar 10 2015
a(43) = 3585, a(45) = 5, a(51) = 16, a(57) = 22, a(59) = 4495, a(63) = 1291, a(65) = 108, a(67) = 220, a(69) = 218039, a(71) = 2112. - Chai Wah Wu, May 08 2024

			For n=1, gcd(k^n+2,(k+1)^n+2) = gcd(k+2,k+3) = 1, therefore a(1)=0.
For n=2k, see formula.
For n=3, we have gcd(51^3+2,52^3+2) = 109, and the pair (k,k+1)=(51,52) is the smallest which yields a GCD > 1, therefore a(3)=51.

Cf. A118119, A255832, A255853-A255869

A255852[n_] := Module[{m = 1}, While[GCD[m^n + 2, (m + 1)^n + 2] <= 1, m++]; m];
Join[{1, 0}, Table[A255852[n], {n, 2, 24}]]

a(n, c=2, L=10^7, S=1)={n!=1 && for(a=S, L, gcd(a^n+c, (a+1)^n+c)>1&&return(a))}

from sympy import primefactors,resultant, nthroot_mod
from sympy.abc import x
def A255852(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+2,(x+1)**n+2)):
        for d in (a for a in sorted(nthroot_mod(-2,n,p,all_roots=True)) if pow(a+1,n,p)==-2%p):
            k = min(d,k) if k else d
            break
    return k # Chai Wah Wu, May 08 2024

a(2k)=1 for k>=0, because gcd(1^(2k)+2,2^(2k)+2) = gcd(3,4^k-1) = 3.

a(2k+1) = A255832(k).

a(25),a(37),a(41),a(47) conjectured by Hiroaki Yamanouchi, Mar 10 2015; confirmed by Max Alekseyev, Aug 06 2015

A255869 Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 3, 2408, 1, 3976, 608, 28, 1, 88, 23, 464658, 1, 319924724, 3, 7, 1, 1628, 138, 2219409, 1, 6, 5, 594, 1, 872, 3, 92, 1, 392, 65, 2278155, 1, 3755866, 4793, 13, 1, 7873, 3, 614294, 1, 448812437, 5

Offset: 0

M. F. Hasler, Mar 09 2015

See A118119, which is the main entry for this class of sequences.

a(43) <= 8153777984244162781089834. - Max Alekseyev, Aug 06 2015

			For n=0 and n=4, see formula with k=0 resp. k=1.
For n=1, gcd(m^n+19, (m+1)^n+19) = gcd(m+19, m+20) = 1, therefore a(1)=0.
For n=2, gcd(3^2+19, 4^2+19) = 7 and (m,m+1) = (3,4) is the smallest pair which yields a GCD > 1 here.

Cf. A118119, A255832, A255852-A255868

A255869[n_] := Module[{m = 1}, While[GCD[m^n + 19, (m + 1)^n + 19] <= 1, m++]; m]; Join[{1, 0}, Table[A255869[n], {n, 2, 12}]] (* Robert Price, Oct 16 2018 *)

a(n,c=19,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255869(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+19,(x+1)**n+19)):
        for d in (a for a in nthroot_mod(-19,n,p,all_roots=True) if pow(a+1,n,p)==-19%p):
            k = min(d,k) if k else d
    return k # Chai Wah Wu, May 07 2024

a(4k) = 1 for k>=0, because gcd(1^(4k)+19, 2^(4k)+19) = gcd(20, 16^k-1) >= 5 since 16 = 1 (mod 5).

a(13)-a(40) from Hiroaki Yamanouchi, Mar 12 2015

a(41)-a(42) from Max Alekseyev, Aug 06 2015

A255853 Least k > 0 such that gcd(k^n+3, (k+1)^n+3) > 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 6, 56, 3, 29, 96, 1159823, 384, 9, 3, 1994117680, 13, 247, 6, 15, 3, 1256, 4, 25211925041, 15, 5785, 3, 93602696971, 24, 11, 6, 182, 3, 4644, 92, 12506, 9, 13, 3, 484, 2, 420, 6, 130, 3, 16032496, 12

Offset: 0

M. F. Hasler, Mar 08 2015

See A118119, which is the main entry for this class of sequences.

a(43) <= 291613846670877. - Max Alekseyev, Aug 07 2015

			For n=1, gcd(k^n+3, (k+1)^n+3) = gcd(k+3, k+4) = 1, therefore a(1)=0.
For n=2, we have gcd(6^2+3, 7^2+3) = gcd(39, 52) = 13, and the pair (k,k+1)=(6,7) is the smallest which yields a GCD > 1, therefore a(2)=6.

Cf. A118119, A255832.

Cf. A255852, A255854, A255855, A255856, A255857, A255858, A255869.

A255853[n_] := Module[{m = 1}, While[GCD[m^n + 3, (m + 1)^n + 3] <= 1, m++]; m]; Join[{1, 0}, Table[A255853[n], {n, 2, 10}]] (* Robert Price, Oct 15 2018 *)

a(n,c=3,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255853(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+3,(x+1)**n+3)):
        for d in (a for a in sorted(nthroot_mod(-3,n,p,all_roots=True)) if pow(a+1,n,p)==-3%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 08 2024

For k>=0, a(6k+4)=3 because gcd(3^(6k+4)+3, 4^(6k+4)+3) = gcd(9^(3k+2)+3, 16^(3k+2)+3) and 9 = 16 = 2 (mod 7) and 2^(3k+2)+3 = 2^2+3 = 0 (mod 7), so the GCD is a positive multiple of 7.

a(11)-a(40) from Hiroaki Yamanouchi, Mar 12 2015

a(41)-a(42) from Max Alekseyev, Aug 06 2015

A255856 Least k > 0 such that gcd(k^n+6, (k+1)^n+6) > 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 2, 1, 48, 9, 1, 19, 14, 1, 2, 77364635, 1, 1943, 2, 1, 5, 299788383228819788, 1, 1578270389554680057141787800241971645032008710129107338825798, 9, 1, 2, 6676, 1, 415, 2, 1, 39, 168338080349, 1, 305, 6806, 1, 2, 9, 1

Offset: 0

M. F. Hasler, Mar 08 2015

See A118119, which is the main entry for this class of sequences.

			For n=1, gcd(k^n+6, (k+1)^n+6) = gcd(k+6, k+7) = 1, therefore a(1)=0.
For n=2, we have gcd(2^2+6, 3^2+6) = gcd(10, 15) = 5, and the pair (k,k+1)=(2,3) is the smallest which yields a gcd > 1, therefore a(2)=2.
For n=3k, see formula.

Cf. A118119, A255832, A255852-A255869

A255856[n_] := Module[{m = 1}, While[GCD[m^n + 6, (m + 1)^n + 6] <= 1, m++]; m]; Join[{1, 0}, Table[A255856[n], {n, 2, 10}]] (* Robert Price, Oct 15 2018 *)

a(n,c=6,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255856(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+6,(x+1)**n+6)):
        for d in (a for a in sorted(nthroot_mod(-6,n,p,all_roots=True)) if pow(a+1,n,p)==-6%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 09 2024

a(3k)=1 for k>=0, because gcd(1^(3k)+6, 2^(3k)+6) = gcd(7, 8^k-1) = 7.

a(11)-a(36) from Hiroaki Yamanouchi, Mar 13 2015

A255857 Least k > 0 such that gcd(k^n+7,(k+1)^n+7) > 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 14, 320, 56, 675113, 224, 13, 5, 283, 33, 192, 26, 242, 5, 2, 10, 140, 5, 50, 142, 29, 18, 605962, 11, 97, 234881024, 951, 5, 3332537854, 14

Offset: 0

M. F. Hasler, Mar 08 2015

nonn

See A118119, which is the main entry for this class of sequences.

			For n=0, gcd(k^0+7, (k+1)^0+7) = gcd(8, 8) = 8 for any k > 0, therefore a(0)=1 is the smallest possible positive value.
For n=1, gcd(k^n+7, (k+1)^n+7) = gcd(k+7, k+8) = 1, therefore a(1)=0.
For n=2, we have gcd(14^2+7, 15^2+7) = gcd(203, 232) = 29, and the pair (k,k+1)=(14,15) is the smallest which yields a gcd > 1, therefore a(2)=14.

Hiroaki Yamanouchi, Table of n, a(n) for n = 0..36

Cf. A118119, A255832, A255852-A255869

A255857[n_] := Module[{m = 1}, While[GCD[m^n + 7, (m + 1)^n + 7] <= 1, m++]; m]; Join[{1, 0}, Table[A255857[n], {n, 2, 25}]] (* Robert Price, Oct 15 2018 *)

a(n,c=7,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255857(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+7,(x+1)**n+7)):
        for d in (a for a in sorted(nthroot_mod(-7,n,p,all_roots=True)) if pow(a+1,n,p)==-7%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 09 2024

a(26)-a(36) from Hiroaki Yamanouchi, Mar 12 2015

A255858 Least k > 0 such that gcd(k^n + 8, (k+1)^n + 8) > 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 1, 5, 1, 5238149, 1, 24747, 1, 5, 1, 1042068640211853141849723, 1, 28913777, 1, 5, 1, 9380, 1, 87940, 1, 5, 1, 677083547, 1, 1584, 1, 5, 1, 5355, 1, 257, 1, 5, 1, 118, 1, 3250, 1, 5, 1, 78609080, 1, 1807714890, 1, 5, 1

Offset: 0

M. F. Hasler, Mar 08 2015

See A118119, which is the main entry for this class of sequences.

			For n=1, gcd(k^n + 8, (k+1)^n + 8) = gcd(k+8, k+9) = 1, therefore a(1)=0.
For n=2*k, see formula.
For n=3, we have gcd(5^3 + 8, 6^3 + 8) = gcd(133, 224) = 7, and the pair (k,k+1)=(5,6) is the smallest which yields a GCD > 1, therefore a(3)=5.

Cf. A118119, A255832, A255852-A255869.

A255858[n_] := Module[{m = 1}, While[GCD[m^n + 8, (m + 1)^n + 8] <= 1, m++]; m]; Join[{1, 0}, Table[A255858[n], {n, 2, 10}]] (* Robert Price, Oct 15 2018 *)

a(n,c=8,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

a(2k)=1 for k >= 0, because gcd(1^(2k) + 8, 2^(2k) + 8) = gcd(9, 4^k-1) >= 3.

a(11)-a(40) from Hiroaki Yamanouchi, Mar 13 2015

a(41)-a(46) from Max Alekseyev, Aug 06 2015

A255854 Least k > 0 such that gcd(k^n+4, (k+1)^n+4) > 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 8, 210, 1, 82, 128, 4763358550, 1, 22, 8, 4050643070777669523228, 1, 1010633974733, 7784, 100, 1, 26627469676193276478340, 8, 179, 1, 4082, 48, 1293523748876425462850, 1, 173, 8, 5, 1, 2423, 320, 342, 1, 1162, 8, 93, 1, 455207, 128, 22, 1, 11383, 8, 58768, 1, 91, 96, 306824898, 1, 187751, 8, 84, 1

Offset: 0

M. F. Hasler, Mar 08 2015

nonn

See A118119, which is the main entry for this class of sequences.

			For n=1, gcd(k^n+4, (k+1)^n+4) = gcd(k+4, k+5) = 1, therefore a(1)=0.
For n=2, we have gcd(8^2+4, 9^2+4) = gcd(68, 85) = 17, and the pair (k,k+1)=(8,9) is the smallest with this property, therefore a(2)=8.
More generally, a(8k+2)=8 because gcd(8^(8k+2)+4, 9^(8k+2)+4) = gcd(64^(4k+1)+4, 81^(4k+1)+4) >= 17, since 64 = 81 = 13 (mod 17) and 13^4 = 1 (mod 17).
Also a(4k)=1, because gcd(1^(4k)+4, 2^(4k)+4) = gcd(5, 16^k-1) = 5.

Cf. A118119, A255832.

Cf. A255852, A255853, A255855, A255856, A255857, A255858, A255869.

A255854[n_] := Module[{m = 1}, While[GCD[m^n + 4, (m + 1)^n + 4] <= 1, m++]; m]; Join[{1, 0}, Table[A255854[n], {n, 2, 6}]] (* Robert Price, Oct 15 2018 *)

a(n,c=4,L=10^6,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255854(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+4,(x+1)**n+4)):
        for d in (a for a in sorted(nthroot_mod(-4,n,p,all_roots=True)) if pow(a+1,n,p)==-4%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 08 2024

a(4k)=1, a(8k+2)=8 (k>=0), cf. examples.

a(7)-a(46) from Hiroaki Yamanouchi, Mar 13 2015

a(47)-a(52) from Max Alekseyev, Aug 06 2015

A255855 Least k > 0 such that gcd(k^n+5, (k+1)^n+5) > 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 1, 5, 1, 533360, 1, 55, 1, 7, 1, 796479131355665831357, 1, 41, 1, 5, 1, 3775, 1, 42296, 1, 7, 1, 653246700175064613889, 1, 21, 1, 5, 1, 1619, 1, 42842, 1, 7, 1, 2945, 1, 323371, 1, 5, 1, 1102221, 1, 633524110177, 1, 7, 1

Offset: 0

M. F. Hasler, Mar 08 2015

See A118119, which is the main entry for this class of sequences.

			For n=1, gcd(k^n+5, (k+1)^n+5) = gcd(k+5, k+6) = 1, therefore a(1)=0.
For n=2k, see formula.
For n=3, we have gcd(5^3+5, 6^3+5) = 13, and the pair (k,k+1)=(5,6) is the smallest which yields a GCD > 1, therefore a(3)=5.

Cf. A118119, A255832.

Cf. A255852, A255853, A255854, A255856, A255857, A255858, A255869.

A255855[n_] := Module[{m = 1}, While[GCD[m^n + 5, (m + 1)^n + 5] <= 1, m++]; m]; Join[{1, 0}, Table[A255855[n], {n, 2, 10}]]

a(n,c=5,L=10^7,S=1)->for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255855(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+5,(x+1)**n+5)):
        for d in (a for a in sorted(nthroot_mod(-5,n,p,all_roots=True)) if pow(a+1,n,p)==-5%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 08 2024

a(2k)=1 for k>=0, because gcd(1^(2k)+5,2^(2k)+5) = gcd(6,4^k-1) = 3.

a(11)-a(46) from Hiroaki Yamanouchi, Mar 12 2015

A255859 Least m > 0 such that gcd(m^n+9,(m+1)^n+9) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 18, 533, 1, 32, 288, 484, 1, 364, 6, 176427, 1, 31239, 533, 8, 1, 8424432925592889329288197322308900672459420460792433, 30, 16561, 1, 4, 6, 349, 1, 32, 546, 2579, 1, 375766, 11, 5061867704425915, 1, 5620, 6, 8, 1

Offset: 0

M. F. Hasler, Mar 08 2015

See A118119, which is the main entry for this class of sequences.

			For n=1, gcd(m^n+9, (m+1)^n+9) = gcd(m+9, m+10) = 1, therefore a(1)=0.
For n=2, we have gcd(18^2+9, 19^2+9) = gcd(333, 370) = 37, and the pair (m,m+1)=(18,19) is the smallest which yields a GCD > 1, therefore a(2)=37.
For n=4k, see formula.

Will Wei, Patterns that appear to hold, but don't - 8424432925592889329288197322308900672459420460792433, video (2020)

Cf. A118119, A255832, A255852-A255869

A255859[n_] := Module[{m = 1}, While[GCD[m^n + 9, (m + 1)^n + 9] <= 1, m++]; m]; Join[{1, 0}, Table[A255859[n], {n, 2, 16}]] (* Robert Price, Oct 16 2018 *)

a(n,c=9,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

a(4k)=1 for k>=0, because gcd(1^(4k)+9, 2^(4k)+9) = gcd(10, 16^k-1) = 5.

a(17)-a(30) from Hiroaki Yamanouchi, Mar 12 2015

a(31)-a(36) from Max Alekseyev, Aug 06 2015

A255868 Least m > 0 such that gcd(m^n+18, (m+1)^n+18) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 36, 5, 8, 193801631, 7, 16280817091929, 5, 4, 9216, 815167161742047217904392262, 7, 46, 20, 5, 19, 1837, 1, 224, 8, 7, 56, 13215457, 5, 130689, 221, 4, 5, 1167507, 7, 9708, 65, 7, 20, 63, 1, 4248, 5, 5, 5, 527010, 7

Offset: 0

M. F. Hasler, Mar 09 2015

See A118119, which is the main entry for this class of sequences.

			For n=0, gcd(m^0+18, (m+1)^0+18) = gcd(19, 19) = 19, therefore a(0)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+18, (m+1)^n+18) = gcd(m+18, m+19) = 1, therefore a(1)=0.
For n=2, gcd(36^2+18, 37^2+18) = 73 and (m, m+1) = (36, 37) is the smallest pair which yields a GCD > 1 here.

Cf. A118119, A255832, A255852-A255869

A255868[n_] :=  Module[{m = 1}, While[GCD[m^n + 18, (m + 1)^n + 18] <= 1, m++]; m]; Join[{1, 0}, Table[A255868[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)

a(n,c=18,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

a(5)-a(42) from Max Alekseyev, Aug 06 2015

cp's OEIS Frontend

A255852 Least k > 0 such that gcd(k^n+2, (k+1)^n+2) > 1, or 0 if there is no such k.

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A255869 Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m.

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A255853 Least k > 0 such that gcd(k^n+3, (k+1)^n+3) > 1, or 0 if there is no such k.

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A255856 Least k > 0 such that gcd(k^n+6, (k+1)^n+6) > 1, or 0 if there is no such k.

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A255857 Least k > 0 such that gcd(k^n+7,(k+1)^n+7) > 1, or 0 if there is no such k.

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A255858 Least k > 0 such that gcd(k^n + 8, (k+1)^n + 8) > 1, or 0 if there is no such k.

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A255854 Least k > 0 such that gcd(k^n+4, (k+1)^n+4) > 1, or 0 if there is no such k.

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