A255832 -id:A255832 - cp's OEIS frontend

A255860 Least m > 0 such that gcd(m^n+10, (m+1)^n+10) > 1, or 0 if there is no such m.

1, 0, 20, 3, 2, 3, 320, 874, 6, 33, 1, 124, 465, 23433448460229, 81920, 3, 2, 82, 65, 2101, 1, 3, 3, 2398892314, 7270, 3, 11, 21, 2, 97546469, 1, 765170730, 6, 15, 3, 3, 23, 370460325141871548, 29206018, 3, 1

Offset: 0

M. F. Hasler, Mar 08 2015

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

			For n=0, gcd(m^0+10, (m+1)^0+10) = gcd(11, 11) = 11 for any m > 0, therefore a(0)=1 is the smallest possible positive value.
For n=1, gcd(m^n+10, (m+1)^n+10) = gcd(m+10, m+11) = 1, therefore a(1)=0.
For n=2, we have gcd(20^2+10, 21^2+10) = gcd(410, 451) = 41, and the pair (m,m+1)=(20,21) is the smallest which yields a GCD > 1, therefore a(2)=20.

Cf. A118119, A255832, A255852-A255869

A255860[n_] := Module[{m = 1}, While[GCD[m^n + 10, (m + 1)^n + 10] <= 1, m++]; m]; Join[{1, 0}, Table[A255860[n], {n, 2, 12}]] (* Robert Price, Oct 16 2018 *)

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

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

a(37)-a(40) from Max Alekseyev, Aug 06 2015

A255861 Least m > 0 such that gcd(m^n+11, (m+1)^n+11) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 1, 2, 1, 23, 1, 19010820161, 1, 7, 1, 360, 1, 41953103, 1, 4, 1, 638386957517954762853, 1, 38884, 1, 2, 1, 2852, 1, 23, 1, 102, 1, 8384, 1, 36556, 1, 33, 1, 37, 1, 336, 1, 2, 1, 1123, 1, 19734, 1, 9, 1, 135356, 1, 399351, 1, 33, 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(m^n+11, (m+1)^n+11) = gcd(m+11, m+12) = 1, therefore a(1)=0.
For n=2, we have gcd(2^2+11, 3^2+11) = gcd(15, 20) = 5, and the pair (m,m+1)=(2,3) is the smallest which yields a GCD > 1, therefore a(2)=2.

Cf. A118119, A255832, A255852-A255869

A255861[n_] := Module[{m = 1}, While[GCD[m^n + 11, (m + 1)^n + 11] <= 1, m++]; m]; Join[{1, 0}, Table[A255861[n], {n, 2, 6}]] (* Robert Price, Oct 16 2018 *)

a(n,c=11,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)+11, 2^(2k)+11) = gcd(12, 4^k-1) = 3.

a(7)-a(48) from Hiroaki Yamanouchi, Mar 12 2015

a(49)-a(52) from Max Alekseyev, Aug 06 2015

A255862 Least m > 0 such that gcd(m^n+12, (m+1)^n+12) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 3, 1926, 96, 6, 2, 26, 3, 320, 538, 27, 1, 145, 3, 6, 393216, 982, 3, 2557, 3, 2, 30, 18781248, 1, 6, 3, 188, 14, 145, 3, 2808, 3, 16, 24340653915, 6, 1

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+12, (m+1)^0+12) = gcd(13, 13) = 13, therefore a(1)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+12, (m+1)^n+12) = gcd(m+12, m+13) = 1, therefore a(1)=0.
For n=2, see formula with k=0.

Cf. A118119, A255832, A255852-A255869

A255862[n_] := Module[{m = 1}, While[GCD[m^n + 12, (m + 1)^n + 12] <= 1, m++]; m]; Join[{1, 0}, Table[A255862[n], {n, 2, 22}]] (* Robert Price, Oct 16 2018 *)

a(n,c=12,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

a(6k+2) = 3 for k>=0, because 3^(6k+2) = 9^(3k+1), 4^(6k+2) = 16^(3k+1), and 9 = 16 = 2 (mod 7), 2^3 = 1 (mod 7) and 12 = -2 (mod 7), therefore 3^(6k+2)+12 = 4^(6k+2)+12 = 0 (mod 7) and gcd(3^(6k+2)+12, 4^(6k+2)+12) >= 7.

a(23)-a(33) from Hiroaki Yamanouchi, Mar 13 2015

a(34)-a(36) from Max Alekseyev, Aug 07 2015

A255863 Least m > 0 such that gcd(m^n+13, (m+1)^n+13) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 26, 1, 5, 24308100, 1, 329, 71, 1, 6, 59, 1, 135, 5, 1, 23, 7711, 1, 82, 6, 1, 8, 320594291825643656342, 1, 45, 10, 1, 755, 1107, 1, 4279, 30269, 1, 5, 205961, 1, 259, 8, 1, 9, 101975, 1, 6491, 5, 1, 8

Offset: 0

M. F. Hasler, Mar 10 2015

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

			For n=1, gcd(m^n+13, (m+1)^n+13) = gcd(m+13, m+14) = 1, therefore a(1)=0.
For n=2, gcd(26^2+13, 27^2+13) = 53, and (m, m+1) = (26, 27) is the smallest pair which yields a GCD > 1 here.
For n=0, n=3, n=6,... see formula.

Cf. A118119, A255832, A255852-A255869

A255863[n_] := Module[{m = 1}, While[GCD[m^n + 13, (m + 1)^n + 13] <= 1, m++]; m]; Join[{1, 0}, Table[A255863[n], {n, 2, 22}]] (* Robert Price, Oct 16 2018 *)

a(n,c=13,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

a(3k) = 1 for k>=0, because 1^(3k)+13 = 14, 2^(3k)+13 = 8^k+13 = 14 (mod 7), therefore gcd(1^(3k)+13, 2^(3k)+13) >= 7.

a(5)-a(46) from Hiroaki Yamanouchi, Mar 12 2015

A255864 Least m > 0 such that gcd(m^n+14, (m+1)^n+14) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 1, 12, 1, 15, 1, 2, 1, 1929501, 1, 13228907223310811104028677, 1, 94, 1, 11, 1, 85364353, 1, 1563, 1, 49, 1, 9258095644888888790279763522646107297983, 1, 23, 1, 66, 1

Offset: 0

M. F. Hasler, Mar 09 2015

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

a(29) with 141 decimal digits is too large to include here (see b-file).

			For n=1, gcd(m^n+14, (m+1)^n+14) = gcd(m+14, m+15) = 1, therefore a(1)=0.
For n=0 and n=2, see formula with k=0 and k=1.
For n=3, gcd(12^3+14, 13^3+14) = 67, and (m, m+1) = (12, 13) is the smallest pair which yields a GCD > 1 here.

Max Alekseyev, Table of n, a(n) for n = 0..40

Cf. A118119, A255832, A255852-A255869

A255864[n_] := Module[{m = 1}, While[GCD[m^n + 14, (m + 1)^n + 14] <= 1, m++]; m]; Join[{1, 0}, Table[A255864[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)

a(n,c=14,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)+14, 2^(2k)+14) = gcd(15, 4^k-1) >= 3, since 4^k-1 = 1-1 = 0 (mod 3).

a(11)-a(40) from Max Alekseyev, Aug 06 2015

A255865 Least m > 0 such that gcd(m^n+15, (m+1)^n+15) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 30, 5, 120, 133, 15, 14, 11, 5, 7680, 968, 18, 243, 26, 5, 9, 1844434621273219148118716000949433592399169477194046126, 8, 22173201293492286974730770140, 51, 5, 593, 5885, 41, 112, 15, 5, 23

Offset: 0

M. F. Hasler, Mar 09 2015

nonn

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

			For n=0, gcd(m^0+15, (m+1)^0+15) = gcd(16, 16) = 16, therefore a(0)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+15, (m+1)^n+15) = gcd(m+15, m+16) = 1, therefore a(1)=0.
For n=2, gcd(30^2+15, 31^2+15) = 61 and (m, m+1) = (30, 31) is the smallest pair which yields a GCD > 1 here.
For n=3, see formula with k=0.

Table of n, a(n) for n = 0..42

Cf. A118119, A255832, A255852-A255869

A255865[n_] := Module[{m = 1}, While[GCD[m^n + 15, (m + 1)^n + 15] <= 1, m++]; m]; Join[{1, 0}, Table[A255865[n], {n, 2, 16}]] (* Robert Price, Oct 16 2018 *)

a(n,c=15,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

a(6k+3) = 5 for k>=0, because 5^(6k+3) = 125^(2k+1), 6^(6k+3) = 216^(2k+1), and 125 = 216 = -1 (mod 7), therefore gcd(5^(6k+3)+15, 6^(6k+3)+15) >= 7.

a(17)-a(36) from Hiroaki Yamanouchi, Mar 12 2015

a(37)-a(42) from Max Alekseyev, Aug 07 2015

A255866 Least m > 0 such that gcd(m^n+16, (m+1)^n+16) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 2, 22, 128, 12, 2, 81, 1, 5982, 2, 11417025, 32768, 70471611388086, 2, 26, 1, 1019, 2, 12168420936538713481747, 48, 128, 2, 788, 1, 131711329, 2, 91, 13, 2920553219286322570768516629247, 2, 237, 1, 22, 2, 108, 27, 9404578, 2, 2859801, 1, 41772125, 2

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+16, (m+1)^0+16) = gcd(16, 16) = 16, therefore a(0)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+16, (m+1)^n+16) = gcd(m+15, m+16) = 1, therefore a(1)=0.
For n=2, see formula with k=0.
For n=3, gcd(22^3+16, 23^3+16) = 31 and (m, m+1) = (22, 23) is the smallest pair which yields a GCD > 1 here.

Cf. A118119, A255832, A255852-A255869

A255866[n_] := Module[{m = 1}, While[GCD[m^n + 16, (m + 1)^n + 16] <= 1, m++]; m]; Join[{1, 0}, Table[A255866[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)

a(n,c=16,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

a(4k+2) = 2 for k>=0, because 2^(4k+2) = 4^(2k+1), 3^(4k+2) = 9^(2k+1), and 4 = 9 = -1 (mod 5), therefore gcd(2^(4k+2)+16, 3^(4k+2)+16) >= 5.

a(11)-a(42) from Max Alekseyev, Aug 06 2015

A255867 Least m > 0 such that gcd(m^n+17, (m+1)^n+17) > 1, or 0 if there is no such m.

Original entry on oeis.org

1, 0, 1, 1925, 1, 189812175, 1, 2, 1, 116, 1, 55508752881180794569675021, 1, 337276, 1, 230, 1, 162, 1, 2628, 1, 15, 1, 3604979675443168377172749, 1, 53, 1, 248, 1, 254, 1, 5998484614, 1, 1323, 1, 2, 1, 42750021, 1, 51, 1, 17870, 1, 108, 1, 87, 1, 8274, 1, 2, 1, 35, 1, 4049, 1, 308, 1, 8885, 1, 2805086, 1

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+17, (m+1)^0+17) = gcd(18, 18) = 18, therefore a(0)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+17, (m+1)^n+17) = gcd(m+17, m+18) = 1, therefore a(1)=0.
For n=2, see formula with k=0.
For n=3, gcd(1925^3+17, 1926^3+17) = 1951 and (m, m+1) = (1925, 1926) is the smallest pair which yields a GCD > 1 here.

Cf. A118119, A255832, A255852-A255869

f:= proc(n) local q1, q2, r, m, bestm,p,A;
  q1:= m^n + 17;
  q2:= (m+1)^n + 17;
  r:= resultant(q1,q2, m);
  bestm:= infinity;
  for p in numtheory:-factorset(r) do
    A:= [msolve(q1, p)];
    A:= select(s -> eval(q2, s) mod p = 0, A);
    bestm:= min(bestm, op(map(s -> subs(s,m), A)));
  od;
  if bestm = infinity then -1 else bestm fi
end proc:
f(0):= 1: f(1):=0:
map(f, [$1..26]); # Robert Israel, May 31 2019

A255867[n_] := Module[{m = 1}, While[GCD[m^n + 17, (m + 1)^n + 17] <= 1, m++]; m]; Join[{1, 0}, Table[A255867[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)

a(n,c=17,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 m
def A255867(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(m**n+17,(m+1)**n+17)):
        for d in (a for a in nthroot_mod(-17,n,p,all_roots=True) if pow(a+1,n,p)==-17%p):
            k = min(d,k) if k else d
    return k  # Chai Wah Wu, May 07 2024

a(2k) = 1 for k>=0, because gcd(1^(2k)+17, 2^(2k)+17) = gcd(18, 4^k-1) >= 3 since 4 = 1 (mod 3).

a(5)-a(22) from Hiroaki Yamanouchi, Mar 12 2015

a(23)-a(60) from Max Alekseyev, Aug 06 2015

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A255860 Least m > 0 such that gcd(m^n+10, (m+1)^n+10) > 1, or 0 if there is no such m.

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A255861 Least m > 0 such that gcd(m^n+11, (m+1)^n+11) > 1, or 0 if there is no such m.

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A255862 Least m > 0 such that gcd(m^n+12, (m+1)^n+12) > 1, or 0 if there is no such m.

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A255863 Least m > 0 such that gcd(m^n+13, (m+1)^n+13) > 1, or 0 if there is no such m.

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A255864 Least m > 0 such that gcd(m^n+14, (m+1)^n+14) > 1, or 0 if there is no such m.

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A255865 Least m > 0 such that gcd(m^n+15, (m+1)^n+15) > 1, or 0 if there is no such m.

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A255866 Least m > 0 such that gcd(m^n+16, (m+1)^n+16) > 1, or 0 if there is no such m.

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