A002407 -id:A002407 - cp's OEIS frontend

A275878 Standard Jacobi primes.

7, 61, 331, 547, 1951, 2437, 3571, 4219, 7351, 8269, 9241, 10267, 13669, 23497, 25117, 55897, 60919, 74419, 89269, 92401, 102121, 112327, 115837, 126691, 145861, 170647, 202021, 231019, 241117, 246247, 251431, 267307, 283669, 329677, 347821, 360187, 372769

Offset: 1

N. J. A. Sloane, Aug 17 2016

nonn

From Peter Bala, Feb 20 2022: (Start)
Primes of the form (3*k + 2)^3 - (3*k + 1)^3 = 27*k^2 + 27*k + 7.
Equivalently, primes p such that 4*p = 27*x^2 + 1, where x is odd.
Primes p of the form 6*m + 1, where 8*m + 1 is an odd square.
A prime p is in this list iff binomial(2*(p-1)/3,(p-1)/3) == -1 (mod p). See Cosgrave and Dilcher, Theorem 5, Corollary 3. (End)
Subsequence of cuban primes (A002407). - Bernard Schott, Jul 28 2022

Dana Jacobsen, Table of n, a(n) for n = 1..10000
John B. Cosgrave and Karl Dilcher, An Introduction to Gauss Factorials, Amer. Math. Monthly, Vol. 118, No. 9 (November 2011), pp. 812-829.
J. B. Cosgrave and Karl Dilcher, A role for generalized Fermat numbers, Math. Comp., to appear 2016; see also Paper #10, See Table 7.1.

Cf. A002407, A275879.

use ntheory ":all"; forprimes { if (($%3)==1) { my $z = znorder(factorial(($-1)/3),$); $z/=3 unless $z%3; say if $z==1; } } 1e6; # _Dana Jacobsen, Aug 18 2016

use ntheory ":all"; for (0..1000) { my $p = 27*$*$ + 27*$ + 7; say $p if is_prime($p); } # _Dana Jacobsen, Aug 18 2016

Terms a(21) and beyond from Dana Jacobsen, Aug 18 2016

A376907 a(n) is the least n-digit cuban prime.

Original entry on oeis.org

7, 19, 127, 1657, 10267, 102121, 1021417, 10052191, 100381321, 1000556719, 10000510297, 100025541019, 1000011191887, 10000028937841, 100000062634561, 1000001305386991, 10000001240507791, 100000021541868691, 1000000084213608427, 10000000012591553221, 100000000159478313337

Offset: 1

Stefano Spezia, Oct 08 2024

a(n) - A011557(n-1) is a multiple of 3.

Robert Israel, Table of n, a(n) for n = 1..996

Cf. A002407, A003215, A003617, A011557, A111251, A113478, A145203, A376933.

nextcuban:= proc(n)
  local k,y;
  for k from ceil((sqrt(12*n-3)-3)/6) do
    y:= (k+1)^3 - k^3;
    if isprime(y) then return y fi
  od
end proc:
seq(nextcuban(10^i), i = 0 .. 25); # Robert Israel, Nov 08 2024

a[n_]:=Module[{k=1},While[!PrimeQ[m=3k^2+3k+1]||IntegerLength[m]

from itertools import count
from math import isqrt
from sympy import isprime
def A376907(n):
    for k in count(isqrt((((a:=10**(n-1))<<2)-1)//12)):
        m = 3*k*(k+1)+1
        if m >= a and isprime(m):
            return m # Chai Wah Wu, Oct 13 2024

Conjecture: a(n+1)/a(n) ~ 10.

A127854 Largest number k such that k^2 divides A007781(6n+1).

Original entry on oeis.org

19, 61, 127, 217, 331, 469, 631, 817, 1027, 1261, 1519, 1801, 2107, 2437, 2791, 3169, 3571, 3997, 4447, 4921, 5419, 5941, 6487, 7057, 7651, 8269, 8911, 9577, 10267, 10981, 11719, 12481, 13267, 14077, 14911, 15769, 16651, 17557, 18487, 19441

Offset: 1

Alexander Adamchuk, Apr 05 2007

nonn

A007781(n) = (n+1)^(n+1) - n^n. A007781(6n+1) is not squarefree for n > 0. a(n) is the largest square divisor of A007781(6n+1). All terms belong to A003215 Hex (or centered hexagonal) numbers: 3n(n+1)+1 (crystal ball sequence for hexagonal lattice). It appears that a(n) = A003215(2n) = 6n(2n+1)+1. A007781(6n+1)/A003215(2n)^2 = ((6n+2)^(6n+2)-(6n+1)^(6n+1))/(6n(2n+1)+1)^2 = {44193, 2904899682603, 6378521938392937343349, 128847538453506016002947264859159, 13183819636551142123977274666051092130410345, ...}. Prime terms of a(n) belong to A002407. Factorizations of the terms of a(n) are {19, 61, 127, 7*31, 331, 7*67, 631, 19*43, 13*79, 13*97, 7*7*31, 1801, 7*7*43, 2437, 2791, 3169, 3571, 7*571, 4447, 7*19*37, 5419, 13*457, 13*499, 7067, 7*1093, 8269, 7*19*67, 61*157, 10267, 79*139, ...}. All prime factors of a(n) are of the form 6k+1.

Cf. A007781 = (n+1)^(n+1) - n^n. Cf. A000312, A068955, A003215, A002407.

Conjecture: a(n) = 12n^2 + 6n + 1.
Conjecture: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); g.f.: x*(19 + 4*x + x^2)/(1-x)^3. - Colin Barker, Mar 16 2012
These conjectures are false. For n=74, 12*n^2 + 6*n + 1 = 66157 but A007781(6*74+1) is divisible by 5491031^2. - Robert Israel, Nov 19 2017

a(24) corrected by T. D. Noe, Mar 14 2008

A334520 Primes that are the sum of two cubes.

Original entry on oeis.org

2, 7, 19, 37, 61, 127, 271, 331, 397, 547, 631, 919, 1657, 1801, 1951, 2269, 2437, 2791, 3169, 3571, 4219, 4447, 5167, 5419, 6211, 7057, 7351, 8269, 9241, 10267, 11719, 12097, 13267, 13669, 16651, 19441, 19927, 22447, 23497, 24571, 25117, 26227

Offset: 1

N. J. A. Sloane, May 07 2020

nonn

Union of 2 and A002407. Believed to be infinite.

			2 = 1^3 + 1^3.
7 = 2^3 + (-1)^3.
19 = 3^3 + (-2)^3.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Andrew Sutherland, Sums of three cubes, Slides of a talk given May 07 2020 on the Number Theory Web.
Fernando Rodriguez Villegas and Don Zagier, Which primes are sums of two cubes?, CMS Conference Proceedings 15 (1995), pp. 295-306.

Cf. A002407.

Union[{2},Select[Table[3n^2+3n+1,{n,93}],PrimeQ]] (* Paul F. Marrero Romero, Oct 21 2024 *)

A159961 Cuban composites: composite numbers equal to the difference of two consecutive cubes.

Original entry on oeis.org

91, 169, 217, 469, 721, 817, 1027, 1141, 1261, 1387, 1519, 2107, 2611, 2977, 3367, 3781, 3997, 4681, 4921, 5677, 5941, 6487, 6769, 7651, 7957, 8587, 8911, 9577, 9919, 10621, 10981, 11347, 12481, 12871, 14077, 14491, 14911, 15337, 15769, 16207, 17101, 17557

Offset: 1

Giacomo Fecondo, Apr 28 2009

nonn

Analogous to the cuban primes A002407, but select the composite numbers rather than the primes.
Cuban composites are a subset of hexagonal centered numbers.
A cuban composite has an integer divisor of the form 6*k+1 other than 1 and itself.
Also, composite numbers of the form (n^2 + nm + m^2) where n and m are consecutive numbers. - K. D. Bajpai, Jun 12 2014

			a(1) = 91 = 1+3t*(t+1) with t = 5 is the smallest cuban composite number. Note that 91 = 7*13, so its factors have the form 6k+1, in fact 7 = 6*1+1.

K. D. Bajpai, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)

Cf. A003215, A002407, A221717.

nn = 200; Select[Table[3 x^2 + 3 x + 1, {x, nn}], ! PrimeQ[#] &] (* T. D. Noe, Jan 30 2013 *)
Select[Table[m=n+1;( n^2 + n m + m^2),{n,100}],!PrimeQ[#]&] (* K. D. Bajpai, Jun 12 2014 *)
Select[Differences[Range[80]^3],CompositeQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 07 2018 *)

a(1)=1+3t*(t+1) with t=5, a(2)=1+3t*(t+1) with t=7.

A268426 Primes of the form p^2 + 12*q^2, p, q primes.

Original entry on oeis.org

73, 97, 157, 229, 277, 337, 349, 397, 409, 421, 577, 613, 661, 709, 757, 829, 877, 1009, 1069, 1117, 1429, 1549, 1621, 1669, 1741, 1789, 2053, 2269, 2293, 2389, 2437, 2557, 2797, 2857, 2917, 3109, 3301, 3517, 3529, 3637

Offset: 1

Zak Seidov, Feb 04 2016

nonn

A variant of the cuban primes.

Green & Sawhney prove that this sequence is infinite. - Charles R Greathouse IV, Oct 08 2024

			73=5^2+12*2^2, 97= 7^2+12* 2^2, 157= 7^2+12*3^2.

Zak Seidov, Table of a(n), p, q for n=1..116.
Ben Green and Mehtaab Sawhney, Primes of the form p^2 + nq^2, arXiv preprint (2024). arXiv:2410.04189 [math.NT]

Cf. A002407.

list(lim)=my(v=List()); lim\=1; forprime(q=2, sqrtint((lim-25)\12), my(t=12*q^2); forprime(p=5, sqrtint(lim-t), my(r=t+p^2); if(isprime(r), listput(v, r)))); Set(v) \\ Charles R Greathouse IV, Oct 08 2024

A293990 a(n) = (3*n + ((n-2) mod 4))/2.

Original entry on oeis.org

1, 3, 3, 5, 7, 9, 9, 11, 13, 15, 15, 17, 19, 21, 21, 23, 25, 27, 27, 29, 31, 33, 33, 35, 37, 39, 39, 41, 43, 45, 45, 47, 49, 51, 51, 53, 55, 57, 57, 59, 61, 63, 63, 65, 67, 69, 69, 71, 73, 75, 75, 77, 79, 81, 81, 83, 85, 87, 87, 89, 91, 93, 93

Offset: 0

Dimitris Valianatos, Oct 21 2017

The product (2/3) * (4/3) * (6/5) * (6/7) * (8/9) * (10/9) * (12/11) * (12/13) * ... = Pi/(2*sqrt(3)). The denominators are a(n) for n >= 1 and numerators are a(n-1) + A093148(n) for n >= 1 -> [2, 4, 6, 6, 8, 10, 12, 12, ...].
Let r(n) = (a(n)-1)/(a(n)+1) if a(n) mod 4 = 1, (a(n)+1)/(a(n)-1) otherwise; then Product_{n>=1} r(n) = (2/1) * (2/1) * (2/3) * (4/3) * (4/5) * (4/5) * (6/5) * (6/7) * ... = Pi*sqrt(3)/2 = 2.72069904635132677...
The odd numbers of partial sums this sequence, are identified with the A003215 sequence. Also the prime numbers that appear in partial sums in this sequence, are identified with the A002407 sequence.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A007310, A047298, A063305, A093148, A134967, A157932, A162330, A168329, A214549, A285869, A003215, A002407.

[(3*n+((n-2) mod 4))/2 : n in [0..100]]; // Wesley Ivan Hurt, Oct 29 2017

A293990:=n->(3*n+((n-2) mod 4))/2: seq(A293990(n), n=0..100); # Wesley Ivan Hurt, Oct 29 2017

Table[(3*n + Mod[(n - 2), 4])/2, {n, 0, 100}] (* Wesley Ivan Hurt, Oct 29 2017 *)
f[n_] := (3n + Mod[n - 2, 4])/2; Array[f, 65, 0] (* or *)
LinearRecurrence[{1, 0, 0, 1, -1}, {1, 3, 3, 5, 7}, 65] (* or *)
CoefficientList[ Series[(x^4 + 2x^3 + 2x + 1)/((x - 1)^2 (x^3 + x^2 + x + 1)), {x, 0, 64}], x] (* Robert G. Wilson v, Nov 28 2017 *)

PARI
```
a(n) = (3*n + (n-2)%4) / 2
```

Vec(x*(1 + 2*x + 2*x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^30)) \\ Colin Barker, Oct 21 2017

first(n) = my(start=[1,3,3,5,7,9,9,11]); if(n<=8, return(start)); my(res=vector(n)); for (i=1, 8, res[i] = start[i]); for(i = 1, n-8 ,res[i+8] = res[i] + 12); res \\ David A. Corneth, Oct 21 2017

Sum_{n>=0} 1/a(n)^2 = 5*Pi^2/36 = 1.3707783890401886970... = 10*A086729.
(a(n) - n) * (-1)^(n+1) = A134967(n) for n >= 0.
a(n) - n = A162330(n) for n >= 0.
a(n) - n = A285869(n+1) for n >= 0.
a(n) + a(n+1) = A157932(n+2) for n >= 0.
a(n) + (2*n+1) = A047298(n+1) for n >= 0.
From Colin Barker, Oct 21 2017: (Start)
G.f.: x*(1 + 2*x + 2*x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
(End)
a(n + 8) = a(n) + 12. - David A. Corneth, Oct 21 2017
a(4*k+4) * a(4*k+3) - a(4*k+2) * a(4*k+1) = 2*A063305(k+3) for k >= 0.
Sum_{n>=0} 1/(a(n) + a(n+2))^2 = (4*Pi^2 - 27) / 108 = (A214549 - 1) / 4.

A307585 Positive sums of two distinct cubes (of arbitrary sign).

Original entry on oeis.org

1, 7, 8, 9, 19, 26, 27, 28, 35, 37, 56, 61, 63, 64, 65, 72, 91, 98, 117, 124, 125, 126, 127, 133, 152, 169, 189, 208, 215, 216, 217, 218, 224, 243, 271, 279, 280, 296, 316, 331, 335, 341, 342, 343, 344, 351, 370, 386, 387, 397, 407, 448, 468, 469, 485, 488, 504, 511, 512, 513, 520, 539, 547, 559

Offset: 1

Robert Israel, Apr 15 2019

nonn

All terms == 0, 1, 2, 7 or 8 (mod 9).

			a(3) = 8 = 0^3 + 2^3.
a(4) = 9 = 1^3 + 2^3.
a(5) = 19 = (-2)^3 + 3^3.

Robert Israel, Table of n, a(n) for n = 1..10000
Index to sequences related to sums of cubes

Contained in A045980.  Contains A024670.
Primes in this sequence: A002407.
Cf. A060464.

filter:= proc(n) local d, dp, r;
   for d in numtheory:-divisors(n) do
     dp:= n/d;
     r:= 12*dp - 3*d^2;
     if r > 0 and issqr(r) and (sqrt(r)/6 + d/2)::integer then return true fi
   od;
   false
end proc:
select(filter, [$0..1000]);

filterQ[n_] := Module[{d, dp, r}, Catch[Do[dp = n/d; r = 12 dp - 3 d^2; If[r > 0 && IntegerQ[Sqrt[r]] && IntegerQ[Sqrt[r]/6 + d/2], Throw[True]], {d, Divisors[n]}]; False]];
Select[Range[1000], filterQ] (* Jean-François Alcover, Oct 17 2020, after Maple *)

A338610 Integers m such that there exist one prime p and one positive integer k, for which the expression k^3 + k^2*p is a perfect cube m^3.

Original entry on oeis.org

2, 12, 36, 80, 252, 810, 1100, 1452, 2366, 2940, 5202, 12696, 14400, 16250, 20412, 22736, 27900, 33792, 40460, 52022, 56316, 70602, 75852, 93150, 112896, 120050, 143312, 169400, 198476, 242172, 254016, 291852, 305252, 410700, 518400, 538002, 643452, 689216, 737100

Offset: 1

Bernard Schott, Nov 03 2020

nonn

This concerns Problem 131 of Project Euler (see link).
For each such term m with this property, the values of k and of p are unique.
The solution to the Diophantine equation is: (q^3)^3 + (q^3)^2 * ((q+1)^3 - q^3) = ((q+1) * q^2)^3, where
- the prime p is the cuban prime (q+1)^3 - q^3 = A002407(n),
- corresponding to q = A111251(n),
- the positive integer k = q^3, and,
- the resulting m = (q+1)*q^2 = (A111251(n)+1)*(A111251(n))^2.

			For n=1, q=A111251(1)=1 and 1^3 + 1^2*(2^3 - 1^3) = 1+1*7 = 2^3, hence, k=1^3, cuban prime=7, and a(1)=m=2.
For n=3, q=A111251(3)=3 and (3^3)^3 + (3^3)^2*(4^3 - 3^3) = 27^3 + 27^2*37 = 46656 = 36^3, hence, k=3^3, cuban prime=37, and a(3)=m=36.

Robert Israel, Table of n, a(n) for n = 1..10000
Project Euler, Problem 131: Prime cube partnership.

Cf. A002407, A111251.

Subsequence of A011379.

for q from 1 to 90 do
p:=3*q^2+3*q+1;
if isprime(p) then print((q+1)*q^2); else fi; od:

f[n_] := n^2*(n+1); f /@ Select[Range[100], PrimeQ[3*#^2 + 3*# + 1] &] (* Amiram Eldar, Nov 05 2020 *)

lista(nn) =  apply(x->x^2*(x+1), select(x->isprime(3*x^2 + 3*x + 1), [1..nn])); \\ Michel Marcus, Nov 05 2020

a(n) = (A111251(n) + 1)*(A111251(n))^2.

a(n) = A011379(A111251(n)).

A370519 Intersection of A002061 and A016105.

Original entry on oeis.org

21, 57, 133, 381, 553, 813, 993, 1057, 1333, 1561, 1641, 1893, 1981, 2653, 2757, 3193, 3661, 5257, 5853, 6973, 8373, 8557, 9121, 9313, 10713, 10921, 12657, 13341, 15253, 15501, 16257, 18633, 19741, 22053, 24493, 29413, 30801, 32221, 32581, 33673, 35157, 39801

Offset: 1

Marius A. Burtea, Feb 27 2024

nonn

If p is a cuban prime (A002407) and p == 3 (mod 4) (A002145), then m = 3*p is a term. Indeed, there is k for which p = 1 + 3*k*(k + 1) and m = 3*p = 3 + 9*k*(k + 1) = (3*k + 2)^2 - (3*k + 2) + 1, so m is a term.

The sequence also includes terms that do not have this form: 133 = 12^2 - 12 + 1 = 7*19, 553 = 24^2 - 24 + 1 = 7*79, 1057 = 33^2 - 33 + 1 = 7*151, 1333 = 37^2 - 37 + 1= 31*43 and others.

			A002061(5) = 21 = A016105(1), so 21 is a term.
A002061(8) = 57 = A016105(3), so 57 is a term.

Robert Israel, Table of n, a(n) for n = 1..572

Cf. A002061, A002145, A002407, A016105.

pd:=PrimeDivisors; blum:=func; [n:n in [s^2-s+1:s in [2..2000]]|blum(n)];

N:= 10^5: # for terms <= N
P:= select(isprime, [seq(i,i=3..N/3,4)]):
sort(select(t -> t <= N and issqr(4*t-3), [seq(seq(P[i]*P[j],i=1..j-1),j=1..nops(P))])); # Robert Israel, Feb 27 2025

TR=40000; R1=Ceiling[(1+Sqrt[1-4(1-TR)])/2]; R2=TR/4; Intersection[Table[n^2-n+1, {n, 0, R1}], Select[4Range[5, R2]+1, PrimeNu[#]==2&&MoebiusMu[#]==1&&Mod[FactorInteger[#][[1, 1]], 4]!=1&]] (* James C. McMahon, Feb 27 2024 *)

cp's OEIS Frontend

A275878 Standard Jacobi primes.

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A376907 a(n) is the least n-digit cuban prime.

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A127854 Largest number k such that k^2 divides A007781(6n+1).

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A334520 Primes that are the sum of two cubes.

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Mathematica

A159961 Cuban composites: composite numbers equal to the difference of two consecutive cubes.

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Mathematica

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A268426 Primes of the form p^2 + 12*q^2, p, q primes.

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PARI

A293990 a(n) = (3*n + ((n-2) mod 4))/2.

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Magma

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A307585 Positive sums of two distinct cubes (of arbitrary sign).

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