A118119 -id:A118119 - cp's OEIS frontend

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

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

A186710 a(n) = gcd(k^n + 1, (k+1)^n + 1) for the smallest k at which the GCD exceeds 1.

Original entry on oeis.org

5, 7, 17, 11, 5, 29, 17, 19, 25, 23, 17, 53, 145, 61, 353, 137, 5, 191, 41, 43, 5, 47, 97, 11, 265, 19, 337, 59, 25, 5953, 257, 67, 5, 29, 17, 223, 5, 157, 17, 83, 145, 173, 89, 19, 5, 283, 353, 29, 12625, 307, 17, 107, 5, 121, 1921, 229, 5, 709, 241, 367, 5, 817, 769, 521, 5, 269, 137, 139, 725, 853, 55969, 293, 745, 61, 17, 29, 265

Offset: 2

Michel Lagneau, Feb 26 2011

nonn

For k=0, the GCD equals 1. Increasing k, the GCD first exceeds 1 at k = A118119(n), and that GCD is a(n).

			a(2) = 5 because 2^2 + 1 = 5 and 3^2+1 = 2*5;
a(3) = 7 because 5^3 + 1 = 2*3^2*7 and 6^3 + 1 = 7*31.

Cf. A118119.

A186710 := proc(n) local k ,g; for k from 1 do g := igcd(k^n+1,(k+1)^n+1) ; if g>1 then return g ; end if; end do: end proc: # R. J. Mathar, Mar 07 2011

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

A010034 Numbers k such that gcd(k^17 + 9, (k+1)^17 + 9) > 1.

Original entry on oeis.org

8424432925592889329288197322308900672459420460792433, 17361015163508605989239159575667846308252873717727992, 26297597401424322649190121829026791944046326974663551, 35234179639340039309141084082385737579839780231599110

Offset: 1

Ilan Vardi, Stan Wagon

In other words, let f(n) = gcd(n^17 + 9, (n+1)^17 + 9). Then f(n) = 1 for all n <= 8424432925592889329288197322308900672459420460792432, but f(8424432925592889329288197322308900672459420460792433) > 1.

In fact f(8424432925592889329288197322308900672459420460792433) = 8936582237915716659950962253358945635793453256935559.

M. F. Hasler, Table of n, a(n) for n = 1..100
Tanya Khovanova, Recursive Sequences
Stan Wagon, Macalester College Problem of the week # 805, MacPOW archive on MathForum.org. Spring 1996.
Péter E. Frenkel, József Pelikán, On the greatest common divisor of the value of two polynomials, Amer. Math. Monthly 124:5 (2017), 446-450. arXiv:1608.07936 [math.NT]
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A118119, A255859.

Table[8424432925592889329288197322308900672459420460792433+ 8936582237915716659950962253358945635793453256935559(n-1),{n,5}] (* or *) LinearRecurrence[{2,-1},{8424432925592889329288197322308900672459420460792433,17361015163508605989239159575667846308252873717727992},5] (* Harvey P. Dale, Jun 12 2014 *)

A010034(n)=8936582237915716659950962253358945635793453256935559*n-512149312322827330662764931050044963334032796143126 \\ M. F. Hasler, Mar 17 2015

\\ The values (a(1),p) can also be found using:
{p=polresultant(x^17+9,(x+1)^17+9);s=vector(2,i,Mod(-9,p)^(1/17));(u=s[2]/s[1])!=1&&until(setsearch(Set(s=concat(s,s[#s]*u)),s[#s]+1),)}
\\ Then the last element s[#s] equals Mod(a(1),p). - M. F. Hasler, Mar 26 2015

a(n) = 8424432925592889329288197322308900672459420460792433 + 8936582237915716659950962253358945635793453256935559*(n-1). - Max Alekseyev, Jul 26 2009

a(1) = A255859(17). - M. F. Hasler, Mar 17 2015

More terms from Max Alekseyev, Jul 26 2009

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

Original entry on oeis.org

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

cp's OEIS Frontend

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

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A186710 a(n) = gcd(k^n + 1, (k+1)^n + 1) for the smallest k at which the GCD exceeds 1.

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Maple

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

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A010034 Numbers k such that gcd(k^17 + 9, (k+1)^17 + 9) > 1.

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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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