A036570 -id:A036570 - cp's OEIS frontend

A005383 Primes p such that (p+1)/2 is prime.

3, 5, 13, 37, 61, 73, 157, 193, 277, 313, 397, 421, 457, 541, 613, 661, 673, 733, 757, 877, 997, 1093, 1153, 1201, 1213, 1237, 1321, 1381, 1453, 1621, 1657, 1753, 1873, 1933, 1993, 2017, 2137, 2341, 2473, 2557, 2593, 2797, 2857, 2917, 3061, 3217, 3253

Offset: 1

N. J. A. Sloane

Also, n such that sigma(n)/2 is prime. - Joseph L. Pe, Dec 10 2001; confirmed by Vladeta Jovovic, Dec 12 2002
Primes that are followed by twice a prime, i.e., are followed by a semiprime. (For primes followed by two semiprimes, see A036570.) - Zak Seidov, Aug 03 2013, Dec 31 2015
If A005382(n) is in A168421 then a(n) is a twin prime with a Ramanujan prime, A104272(k) = a(n) - 2. - John W. Nicholson, Jan 07 2016
Starting with 13 all terms are congruent to 1 mod 12. - Zak Seidov, Feb 16 2017
Numbers n such that both n and n+12 are terms are 61, 661, 1201, 4261, 5101, 6121, 6361 (all congruent to 1 mod 60). - Zak Seidov, Mar 16 2017
Primes p for which there exists a prime q < p such that 2q == 1 (mod p). Proof: q = (p + 1)/2. - David James Sycamore, Nov 10 2018
Prime numbers n such that phi(sigma(2n)) = phi(2n), excluding n=3 and n=5; as well as phi(sigma(3n)) = phi(3n), excluding n=3 only. - Richard R. Forberg, Dec 22 2020

			Both 3 and (3+1)/2 = 2 are primes, both 5 and (5+1)/2 = 3 are primes. - _Zak Seidov_, Nov 19 2012

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..10000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
R. P. Boas & N. J. A. Sloane, Correspondence, 1974
Benoit Cloitre, On the fractal behavior of primes, 2011.

Cf. A005382, A057326, A057327, A057328, A057329, A057330, A005603.
A subsequence of A000040 which has A036570 as subsequence.
Cf. A005385, A010051, A065091, A048161, A036570.

a005383 n = a005383_list !! (n-1)
a005383_list = [p | p <- a065091_list, a010051 ((p + 1) `div` 2) == 1]
-- Reinhard Zumkeller, Nov 06 2012

LIMIT = 8000 % Find all members of A005383 less than LIMIT A = primes(LIMIT); n = length(A); %n is number of primes less than LIMIT B = 2*A - 1; C = ones(n, 1)*A; %C is an n X n matrix, with C(i, j) = j-th prime D = B'*ones(1, n); %D is an n X n matrix, with D(i, j) = (i-th prime)*2 - 1 [i, j] = find(C == D); A(j)

[n: n in [1..3300] | IsPrime(n) and IsPrime((n+1) div 2) ]; // Vincenzo Librandi, Sep 25 2012

for n to 300 do
  X := ithprime(n);
Y := ithprime(n+1);
Z := 1/2 mod Y;
  if isprime(Z) then print(Y);
end if:
end do:
# David James Sycamore, Nov 11 2018

Select[Prime[Range[1000]], PrimeQ[(# + 1)/2] &] (* Zak Seidov, Nov 19 2012 *)

A005383_list(n) = select(m->isprime(m\2+1),primes(n)[2..n]) \\ Charles R Greathouse IV, Sep 25 2012

from sympy import isprime
[n for n in range(3, 5000) if isprime(n) and isprime((n + 1)//2)]
# Indranil Ghosh, Mar 17 2017

[n for n in prime_range(3, 1000) if is_prime((n + 1) // 2)]
# F. Chapoton, Dec 17 2019

a(n) = A129521(n)/A005382(n). - Reinhard Zumkeller, Apr 19 2007
A000035(a(n))*A010051(a(n))*A010051((a(n)+1)/2) = 1. - Reinhard Zumkeller, Nov 06 2012
a(n) = 2*A005382(n) - 1. - Zak Seidov, Nov 19 2012
a(n) = A005382(n) + phi(A005382(n)) = A005382(n) + A000010(A005382(n)). - Torlach Rush, Mar 10 2014

More terms from David Wasserman, Jan 18 2002

Name changed by Jianing Song, Nov 27 2021

A163573 Primes p such that (p+1)/2, (p+2)/3 and (p+3)/4 are also primes.

Original entry on oeis.org

12721, 16921, 19441, 24481, 49681, 61561, 104161, 229321, 255361, 259681, 266401, 291721, 298201, 311041, 331921, 419401, 423481, 436801, 446881, 471241, 525241, 532801, 539401, 581521, 600601, 663601, 704161, 709921, 783721, 867001, 904801

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jul 31 2009

Are all terms == 1 (mod 10)?
Subsequence of A005383, of A091180 and of A036570. - R. J. Mathar, Aug 01 2009
Since (p+2)/3 and (p+3)/4 must be integer, the Chinese remainder theorem shows that all terms are == 1 (mod 12). - R. J. Mathar, Aug 01 2009
All terms are of the form 120k+1: a(n)=120*A163625(n)+1. - Zak Seidov, Aug 01 2009
Each term is congruent to 1 mod 120, so the last digits are always '1': For all four values to be integers it must be that p = 1 (mod 12). As p is prime, it must be that p = 1, 13, 37, 49, 61, 73, 97, or 109 (mod 120). In all but the first case either (p+3)/4 is even or one of the three expressions gives a value divisible by 5 (or both, and possibly the same expression). - Rick L. Shepherd, Aug 01 2009
{6*a(n)}A050498.%20Proof:%20with%20p%20=%20a(n)%20the%20arithmetic%20progression%20with%20four%20terms%20of%20difference%206%20and%20constant%20value%20of%20Euler's%20phi,%20namely%202*(p-1),%20is%206*(p,%202*(p+1)/2,%203*(p+2)/3,%204*(p+3)/4).%20Use%20phi(n,%20prime)%20=%20phi(n)*(prime-1)%20if%20gcd(n,%20prime)%20=%201.%20Here%20n%20=%206,%2012,%2018,%2024%20and%20prime%20%3E%203%20for%20p%20%3E=%20a(1).%20Thanks%20to%20_Hugo%20Pfoertner">{n >= 1} is a subsequence of A050498. Proof: with p = a(n) the arithmetic progression with four terms of difference 6 and constant value of Euler's phi, namely 2*(p-1), is 6*(p, 2*(p+1)/2, 3*(p+2)/3, 4*(p+3)/4). Use phi(n, prime) = phi(n)*(prime-1) if gcd(n, prime) = 1. Here n = 6, 12, 18, 24 and prime > 3 for p >= a(1). Thanks to _Hugo Pfoertner for a link to the present sequence in connection with A339883. - Wolfdieter Lang, Jan 11 2021

Vincenzo Librandi and Chai Wah Wu, Table of n, a(n) for n = 1..10001 (First 1000 terms from Vincenzo Librandi)

Cf. A005383, A091180, A036570, A050498, A163623, A163624, A163625, A278583, A278585, A339883.

[p: p in PrimesInInterval(6, 1200000) | IsPrime((p+1) div 2) and IsPrime((p+2) div 3) and IsPrime((p+3) div 4)]; // Vincenzo Librandi, Apr 09 2013

lst={};Do[p=Prime[n];If[PrimeQ[(p+1)/2]&&PrimeQ[(p+2)/3]&&PrimeQ[(p+3)/ 4],AppendTo[lst,p]],{n,2*9!}];lst

is(n)=n%120==1 && isprime(n) && isprime(n\2+1) && isprime(n\3+1) && isprime(n\4+1) \\ Charles R Greathouse IV, Nov 30 2016

from sympy import prime, isprime
A163573_list = [4*q-3 for q in (prime(i) for i in range(1,10000)) if isprime(4*q-3) and isprime(2*q-1) and (not (4*q-1) % 3) and isprime((4*q-1)//3)] # Chai Wah Wu, Nov 30 2016

Slightly edited by R. J. Mathar, Aug 01 2009

A350675 Numbers k such that tau(k) + tau(k+1) + tau(k+2) = 10, where tau is the number of divisors function A000005.

Original entry on oeis.org

6, 11, 13, 17, 21, 37, 57, 157, 177, 381, 501, 541, 717, 877, 1201, 1317, 1381, 1437, 1621, 1821, 2017, 2557, 2577, 2857, 2901, 3061, 3117, 3777, 4281, 4357, 4441, 4677, 4701, 5077, 5097, 5581, 5637, 5701, 5937, 6337, 6637, 6661, 6717, 6997, 7417, 8221, 8781

Offset: 1

Jon E. Schoenfield, Jan 10 2022

nonn

Since tau(k) + tau(k+1) + tau(k+2) = 10 and no three consecutive integers include more than one square, the triple (tau(k), tau(k+1), tau(k+2)) must consist of three even numbers, so it must be one of (2, 2, 6), (2, 4, 4), (2, 6, 2), (4, 2, 4), (4, 4, 2), and (6, 2, 2). Of these, (2, 2, 6) and (6, 2, 2) are impossible. Of the remaining patterns:
(2, 4, 4) requires that k be an odd prime other than 3, followed by two semiprimes, so k is a prime p such that (p+1)/2 and (p+2)/3 are also prime, and such primes are 13, 37, 157, 541, ... (A036570);
(2, 6, 2) requires that (k, k+2) be a twin prime pair whose average has exactly 6 divisors, and is thus either 12 or 18, so k is 11 or 17;
(4, 2, 4) requires that k+1 be an odd prime, with both k and k+2 having exactly 4 divisors, even though one of them is a multiple of 4, so that one is k+2 = 2^3 = 8, so k = 6;
(4, 4, 2) requires that k+2 be an odd prime > 3, preceded by two semiprimes, so k+2 is a prime p such that (p-1)/2 and (p-2)/3 are also prime, so k+2 is in {23, 59, 179, 383, ...} (which is A181841, after its first two terms, 7 and 11), so k is in {A181841(n) - 2} \ {5, 9}, i.e., k is in {21, 57, 177, 381, ...}.
Tau(k) + tau(k+1) + tau(k+2) >= 10 for all sufficiently large k; the only numbers k for which tau(k) + tau(k+1) + tau(k+2) < 10 are 1..5, 7, and 9.

			Each of the patterns (tau(k), ..., tau(k+2)) that appears repeatedly for large k corresponds to one of the two possible orders in which the multipliers m=1..3 can appear among 3 consecutive integers of the form m*prime. E.g., k=37 begins a run of 3 consecutive integers having the form (p, 2*q, 3*r), where p, q, and r are distinct primes > 3; k=57 begins a similar run, but there the 3 consecutive integers have the form (3*p, 2*q, r).
For each of the patterns of tau values that does not occur repeatedly for large k, one or more of the three consecutive integers in k..k+2 has no prime factor > 3; in the table below, each such integer appears in parentheses in the columns on the right.
.
                              factorization as
              # divisors of    m*(prime > 3)
  n  a(n)=k     k  k+1 k+2      k   k+1  k+2
  -  ------    --- --- ---    ---- ---- ----
  1      6      4   2   4      (6)   q   (8)
  2     11      2   6   2       p  (12)   r
  3     13      2   4   4       p   2q   3r
  4     17      2   6   2       p  (18)   r
  5     21      4   4   2      3p   2q    r
  6     37      2   4   4       p   2q   3r
  7     57      4   4   2      3p   2q    r

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A000005, A036570, A181841, A307120, A317670.

Numbers k such that Sum_{j=0..N-1} tau(k+j) = 2*Sum_{k=1..N} tau(k): A000040 (N=1), A350593 (N=2), (this sequence) (N=3), A350686 (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^4], 3, 1], 10] // Flatten (* Amiram Eldar, Jan 11 2022 *)

isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) == 10; \\ Michel Marcus, Jan 16 2022

{ k : tau(k) + tau(k+1) + tau(k+2) = 10 }.
UNION({6}, {11, 17}, A036570, {A181841(n) - 2} \ {5, 9}).
a(n) = A317670(n) - 1.

A278583 Numbers k such that k+1 is a prime, k+2 is twice a prime, and k+3 is three times a prime.

Original entry on oeis.org

12, 36, 156, 540, 876, 1200, 1380, 1620, 2016, 2556, 2856, 3060, 4356, 4440, 5076, 5580, 5700, 6336, 6636, 6660, 6996, 7416, 8220, 9180, 9660, 9900, 10836, 11496, 12456, 12600, 12720, 12756, 13680, 14436, 15240, 16920, 17076, 18216, 18300, 18396, 19440, 21000, 21576, 22620, 23556, 24480

Offset: 1

N. J. A. Sloane, Nov 30 2016

nonn

All terms are divisible by 12. - Daniel Poveda Parrilla, Dec 12 2016

R. K. Guy, Posting to Number Theory Mailing List, Nov 30 2016

Antti Karttunen and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 563 terms from Antti Karttunen)

Equals A036570(n) - 1.
Positions of terms >= 3 in A278500.
Cf. A074200.

Select[Range[12,25000,12],AllTrue[{#+1,(#+2)/2,(#+3)/3},PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Mar 26 2020 *)

list(lim)=my(v=List()); forprime(p=2,lim+1, if(p%6==1 && isprime(p\2+1) && isprime(p\3+1), listput(v,p-1))); Vec(v) \\ Charles R Greathouse IV, Dec 03 2016

A181841 Supersafe primes.

Original entry on oeis.org

7, 11, 23, 59, 179, 383, 503, 719, 1319, 1439, 1823, 2579, 2903, 3119, 3779, 4283, 4679, 4703, 5099, 5639, 5939, 6719, 8783, 8819, 10079, 12659, 12983, 13523, 15299, 15683, 16223, 17483, 18743, 19079, 21059, 21383, 21563, 22643, 23099, 23603, 24659, 25343

Offset: 1

Peter Luschny, Nov 17 2010

nonn

p is a supersafe prime iff p is a safe prime (A005385) and floor(p/3) is a prime.
Each prime p > 7 is preceded by two semiprimes; a third semiprime is not possible. See A036570. - Zak Seidov, Sep 30 2012
Terms > 7 are congruent to 11 (mod 12). - Zak Seidov, Feb 09 2015

			11 is a supersafe prime because floor(11/2) = 5 and floor(11/3) = 3 are primes.

Zak Seidov, Table of n, a(n) for n = 1..2000
Peter Luschny, Strong coprimality.

Cf. A005385, A036570.

A181841_list := n->select(i->isprime(iquo(i,3)),
select(i->isprime(iquo(i,2)), select(i->isprime(i),[$1..n]))):

Join[{7}, Select[Table[Prime[n], {n, 4000}], PrimeQ[(# - 1)/2] && PrimeQ[(# - 2)/3] &]] (* Vladimir Joseph Stephan Orlovsky, Jun 21 2011 *)

A278585 Numbers k such that k+1 is a prime, k+2 is twice a prime, k+3 is three times a prime, and k+4 is four times a prime.

Original entry on oeis.org

12720, 16920, 19440, 24480, 49680, 61560, 104160, 229320, 255360, 259680, 266400, 291720, 298200, 311040, 331920, 419400, 423480, 436800, 446880, 471240, 525240, 532800, 539400, 581520, 600600, 663600, 704160, 709920, 783720, 867000, 904800, 908040, 918360

Offset: 1

N. J. A. Sloane, Nov 30 2016

nonn

a(n) == 0 mod 120 (see comment in A163573). - Chai Wah Wu, Nov 30 2016

Chai Wah Wu, Table of n, a(n) for n = 1..10001

Equals A163573(n) - 1.
Cf. A036570, A074200.
Positions of terms >= 4 in A278500, thus a subsequence of A278583, A089965 and A006093.

Select[Range[920000],AllTrue[{#+1,(#+2)/2,(#+3)/3,(#+4)/4},PrimeQ]&] (* Harvey P. Dale, Aug 08 2021 *)

is(k)=k%120==0 && isprime(k+1) && isprime(k/2+1) && isprime(k/3+1) && isprime(k/4+1) \\ Charles R Greathouse IV, Dec 03 2016

from sympy import prime, isprime
A278585_list = [4*q-4 for q in (prime(i) for i in range(1,10000)) if isprime(4*q-3) and isprime(2*q-1) and (not (4*q-1) % 3) and isprime((4*q-1)//3)] # Chai Wah Wu, Nov 30 2016

A147615 a(n) = 13 + Sum_{j=4..n+3} j!.

Original entry on oeis.org

13, 37, 157, 877, 5917, 46237, 409117, 4037917, 43954717, 522956317, 6749977117, 93928268317, 1401602636317, 22324392524317, 378011820620317, 6780385526348317, 128425485935180317, 2561327494111820317, 53652269665821260317, 1177652997443428940317

Offset: 0

Ricky Pollack (rickypollack(AT)gmail.com), Apr 30 2009

nonn

Are there infinitely many terms in the intersection of this sequence and A036570?

Answer from Don Reble, Mar 25 2016: No, after the ninth term, all terms are multiples of 13.

			a(0) = 13, a(1) = 13 + 4!, a(2) = 13 + 4! + 5!, ....

G. C. Greubel, Table of n, a(n) for n = 0..445

Cf. A003422, A036570.

[n eq 0 select 13 else 13 + (&+[Factorial(j+4): j in [0..n-1]]): n in [0..30]]; // G. C. Greubel, Oct 24 2022

A147615[n_]:= 13 +Sum[j!, {j,4,n+3}]; Table[A147615[n], {n,0,30}] (* G. C. Greubel, Oct 24 2022 *)

def A147615(n): return 13 + sum(factorial(j+4) for j in range(n))
[A147615(n) for n in range(30)] # G. C. Greubel, Oct 24 2022

a(n) = 13 + !(n+4) - !4 = 13 + A003422(n+4) - A003422(4). - G. C. Greubel, Oct 24 2022

A386540 Primes p such that 2p - 1, 3p - 2, (p + 1)/2, and (p + 2)/3 are also prime.

Original entry on oeis.org

37, 2557, 3061, 5581, 88741, 124021, 157081, 178537, 216217, 216757, 217057, 330661, 344821, 352081, 387577, 423481, 459397, 477577, 521137, 790861, 806521, 865957, 869521, 1369657, 1517881, 1673401, 1704397, 1710661, 1970257, 2132797, 2292781, 2361781, 2680141

Offset: 1

Holger Wallenta, Jul 25 2025

All terms are congruent to 1 modulo 12.

Let q = (p + 1)/2 and r = (p + 2)/3; then 3r = 2q + 1.

			37 is a term, since it is prime and 2*37 - 1 = 73, 3*37 - 2 = 109, (37 + 1)/2 = 19 and (37 + 2)/3 = 13 are all prime.

Robert Israel, Table of n, a(n) for n = 1..1000

Intersection of A036570 and A174734.

select(p -> andmap(isprime,[p, 2*p-1,3*p-2,(p+1)/2,(p+2)/3]), [seq(1+12*i,i=1..10^6)]); # Robert Israel, Jul 25 2025

Select[Prime[Range[2*10^5]],AllTrue[{2#-1,3#-2,(#+1)/2,(#+2)/3},PrimeQ]&] (* James C. McMahon, Jul 25 2025 *)

from gmpy2 import is_prime
def ok(p): return p&1 and p%3 == 1 and all(is_prime(q) for q in [p, 2*p-1, 3*p-2, (p+1)//2, (p+2)//3])
print([k for k in range(1, 10**7, 12) if ok(k)]) # Michael S. Branicky, Jul 25 2025

A306467 Let S(n)_k be the smallest positive integer t that t!k is a multiple of n (t!k is k-tuple factorial of t); then a(n) is the smallest k for which S(n)_k = n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 3, 5, 1, 4, 1, 5, 3, 2, 1, 9, 4, 2, 7, 7, 1, 6, 1, 9, 3, 2, 5, 13, 1, 2, 3, 15, 1, 6, 1, 11, 10, 2, 1, 15, 6, 8, 3, 13, 1, 14, 5, 21, 3, 2, 1, 15, 1, 2, 14, 17, 5, 6, 1, 17, 3, 10, 1, 35, 1, 2, 12, 19, 7, 6, 1, 25

Offset: 1

Lechoslaw Ratajczak, Feb 17 2019

nonn

If p is prime, a(p) = 1.
Conjecture: consecutive primes p satisfying the equation a(p+1) = 2 are consecutive elements of A005383 (primes p such that (p+1)/2 are also primes, for p > 3). The conjecture was checked for all primes < 10^4.
Conjecture: consecutive primes p satisfying the equations a(p+1) = 2 and a(p+2) = 3 are consecutive elements of A036570 (primes p such that (p+1)/2 and (p+2)/3 are also primes). The conjecture was checked for all primes < 10^4.
The first six solutions of the equation a(n) = a(n+1) are 1, 2, 3, 4, 9, 27. Is there a larger n? If such a number n exists, it is larger than 4000.

			a(8) = 3 because:
- for k = 1 is: 1!1, 2!1, 3!1 are not multiples of 8 and 4!1 is a multiple of 8, then (t = 4 = S(8)_1) <> (n = 8);
- for k = 2 is: 1!2, 2!2, 3!2 are not multiples of 8 and 4!2 is a multiple of 8, then (t = 4 = S(8)_2) <> (n = 8);
- for k = 3 is: 1!3, 2!3, 3!3, 4!3, 5!3, 6!3, 7!3 are not multiples of 8 and 8!3 is a multiple of 8, then (t = 8 = S(8)_3) = (n = 8), hence a(8) = k = 3.

J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function

Cf. A002034, A005383, A007922, A036570, A063917.

cp's OEIS Frontend

A005383 Primes p such that (p+1)/2 is prime.

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A163573 Primes p such that (p+1)/2, (p+2)/3 and (p+3)/4 are also primes.

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A350675 Numbers k such that tau(k) + tau(k+1) + tau(k+2) = 10, where tau is the number of divisors function A000005.

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A278583 Numbers k such that k+1 is a prime, k+2 is twice a prime, and k+3 is three times a prime.

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A181841 Supersafe primes.

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A278585 Numbers k such that k+1 is a prime, k+2 is twice a prime, k+3 is three times a prime, and k+4 is four times a prime.

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A147615 a(n) = 13 + Sum_{j=4..n+3} j!.