A247967
a(n) is the smallest k such that prime(k+i) (mod 6) takes successively the values 5, 5, ... for i = 0, 1, ..., n-1.
Original entry on oeis.org
3, 9, 15, 54, 290, 987, 4530, 21481, 58554, 60967, 136456, 136456, 673393, 1254203, 1254203, 7709873, 21357253, 21357253, 25813464, 25813464, 39500857, 39500857, 947438659, 947438659, 947438659, 5703167678, 5703167678, 16976360924, 68745739764, 117327812949
Offset: 1
a(1)= 3 => prime(3) == 5 (mod 6).
a(2)= 9 => prime(9) == 5 (mod 6), prime(10) == 5 (mod 6).
a(3)= 15 => prime(15) == 5 (mod 6), prime(16) == 5 (mod 6), prime(17) == 5 (mod 6).
From _Michel Marcus_, Sep 30 2014: (Start)
The resulting primes are:
5;
23, 29;
47, 53, 59;
251, 257, 263, 269;
1889, 1901, 1907, 1913, 1931;
7793, 7817, 7823, 7829, 7841, 7853;
43451, 43457, 43481, 43487, 43499, 43517, 43541;
243161, 243167, 243197, 243203, 243209, 243227, 243233, 243239;
... (End)
-
N = 2*10^8; % to use primes up to N
P = mod(primes(N),6);
P5 = find(P==5);
n5 = numel(P5);
a(1) = P5(1);
for k = 2:100
r = find(P5(k:n5) == P5(1:n5+1-k)+k-1,1,'first');
if numel(r) == 0
break
end
a(k) = P5(r);
end
a % Robert Israel, Sep 02 2016
-
for n from 1 to 22 do :
ii:=0:
for k from 3 to 10^5 while (ii=0)do :
s:=0:
for i from 0 to n-1 do:
r:=irem(ithprime(k+i),6):
if r = 5
then
s:=s+1:
else
fi:
od:
if s=n and ii=0
then
printf ( "%d %d \n",n,k):ii:=1:
else
fi:
od:
od:
-
Table[k = 1; While[Times @@ Boole@ Map[Mod[Prime[k + #], 6] == 5 &, Range[0, n - 1]] == 0, k++]; k, {n, 10}] (* Michael De Vlieger, Sep 02 2016 *)
-
a(n) = {k = 1; ok = 0; while (!ok, nb = 0; for (i=0, n-1, if (prime(k+i) % 6 == 5, nb++, break);); if (nb == n, ok=1, k++);); k;} \\ Michel Marcus, Sep 28 2014
-
m=c=i=0;forprime(p=1,, i++;p%6!=5&&(!c||!c=0)&&next; c++>m||next; print1(1+i-m=c,",")) \\ M. F. Hasler, Sep 02 2016
A276414
Index of the first prime which starts a run of n consecutive primes all congruent to each other mod 3 (or mod 6).
Original entry on oeis.org
1, 9, 15, 54, 271, 271, 2209, 11199, 13717, 13717, 34369, 136456, 172146, 1254203, 1254203, 4308948, 12762142, 21357253, 25813464, 25813464, 39500857, 39500857, 947438659, 947438659, 947438659, 5703167678, 5703167678, 16976360924, 57446769091, 57446769091, 57446769091
Offset: 1
prime(9) = 23 starts the first run of 2 consecutive primes, {23, 29}, which are congruent to each other (mod 6). Therefore a(2) = 9.
prime(15) = 47 starts the first run of 3 consecutive primes, {47, 53, 59}, which are congruent to each other (mod 6). Therefore a(3) = 15.
prime(54) = 251 starts the first run of 4 consecutive primes, {251, 257, 263, 269}, which are congruent to each other (mod 6). Therefore a(4) = 54.
prime(271) = 1741 starts the first run of 5 consecutive primes, {1741, 1747, 1753, 1759, 1777}, which are congruent to each other (mod 6). Therefore a(5) = 271. This is the first case where the primes are of the form 3k+1.
prime(271) = 1741 also starts the first run of 6 consecutive primes, {1741, 1747, 1753, 1759, 1777, 1783}, which are all congruent to each other (mod 6). Therefore a(6) = 271, too.
Subsequence of
A270190 (after discarding 1 and duplicates of other terms).
-
m=c=i=o=0;print1(1);forprime(p=1,,i++;(o-o=p)%3&&(!c||!c=0)&&next;c++>m||next;print1(",",i-m=c))
a(30)-a(31) from and name clarified by
Jinyuan Wang, Feb 24 2020
A247969
a(n) is the smallest k such that prime(k+i) (mod 6) takes successively the values 1,5,1,5,... for i = 0, 1,...,n-1 ending with 1 or 5.
Original entry on oeis.org
4, 4, 4, 4, 4, 4, 25, 25, 59, 141, 141, 141, 141, 141, 141, 141, 141, 141, 141, 141, 280230, 280230, 981960, 981960, 981960, 4505195, 4505195, 7438440, 15658002, 15658002, 15658002, 15658002, 2628111621, 4671618380, 4671618380, 5803722576, 5803722576, 5803722576
Offset: 1
a(1)= 4 => prime(4) (mod 6)= 1, and not for k = 1, 2, 3.
a(2)= 4 => prime(4) (mod 6)= 1, prime(5) (mod 6) = 5;
a(3)= 4 => prime(4) (mod 6)= 1, prime(5) (mod 6)= 5, prime(6) (mod 6)= 1.
The corresponding primes are for
n= 6: 7, 11, 13, 17, 19, 23;
n= 8: 97, 101, 103, 107, 109, 113, 127, 131;
n= 9: 277, 281, 283, 293, 307, 311, 313, 317, 331;
n= 20: 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941. - _Michel Marcus_, Sep 29 2014
-
for n from 1 to 21 do :
ii:=0:
for k from 3 to 10^5 while (ii=0)do :
s:=0:
for i from 0 to n-1 do:
r:=irem(ithprime(k+i),6):
if r = irem(5^i,6)
then
s:=s+1:
else
fi:
od:
if s=n and ii=0
then
printf ( "%d %d \n",n,k):ii:=1:
else
fi:
od:
od:
-
a(n) = {k = 1; ok = 0; while (! ok, m = 1; nb = 0; for (i=0, n-1, if ((prime(k+i) % 6) == m, nb++, break); m = 5*m % 6;); if (nb == n, ok = 1, k++);); k;} \\ Michel Marcus, Sep 29 2014
-
See Links section.
A247970
a(n) is the smallest k such that prime(k+i) (mod 6) takes successively the values 5,1,5,1... for i = 0, 1,...,n-1 ending with 1 or 5.
Original entry on oeis.org
3, 3, 3, 3, 3, 3, 3, 24, 24, 140, 140, 140, 140, 140, 140, 140, 140, 140, 140, 140, 140
Offset: 1
a(1)= 3 => prime(3) (mod 6)= 5;
a(2)= 3 => prime(3) (mod 6)= 5, prime(4) (mod 6) = 1;
a(3)= 3 => prime(3) (mod 6)= 5, prime(4) (mod 6)= 1, prime(5) (mod 6)= 5.
-
with(numtheory):
for n from 1 to 35 do :
ii:=0:
for k from 3 to 10^5 while (ii=0)do :
s:=0:
for i from 0 to n-1 do:
r:=irem(ithprime(k+i),6):
if r = irem(5^(i+1),6)
then
s:=s+1:
else
fi:
od:
if s=n and ii=0
then
printf ( "%d %d \n",n,k):ii:=1:
else
fi:
od:
od:
Showing 1-4 of 4 results.
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