cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 372 results. Next

A080148 Positions of primes of the form 4*k+3 (A002145) among all primes (A000040).

Original entry on oeis.org

2, 4, 5, 8, 9, 11, 14, 15, 17, 19, 20, 22, 23, 27, 28, 31, 32, 34, 36, 38, 39, 41, 43, 46, 47, 48, 49, 52, 54, 56, 58, 61, 63, 64, 67, 69, 72, 73, 75, 76, 81, 83, 85, 86, 90, 91, 92, 93, 94, 95, 96, 99, 101, 103, 105, 107, 109, 111, 114, 115, 117, 118, 120, 124, 125, 128
Offset: 1

Views

Author

Antti Karttunen, Feb 11 2003

Keywords

Comments

It appears that a(n) = k such that binomial(prime(k),3) mod 2 = 1. See Maple code. - Gary Detlefs, Dec 06 2011
The above is correct (work mod 4). - Charles R Greathouse IV, Dec 06 2011
The asymptotic density of this sequence is 1/2 (by Dirichlet's theorem). - Amiram Eldar, Mar 01 2021

Crossrefs

Almost complement of A080147 (1 is excluded from both).

Programs

Formula

a(n) = A049084(A002145(n)). - R. J. Mathar, Oct 06 2008

A385165 Let p = A002145(n) be the n-th prime == 3 (mod 4); a(n) is the multiplicative order of 2+-i modulo p in Gaussian integers.

Original entry on oeis.org

8, 48, 30, 180, 528, 96, 1848, 2208, 1740, 1496, 360, 1560, 2296, 10608, 11448, 5376, 4290, 1932, 11400, 8856, 27888, 16020, 1216, 3300, 3710, 49728, 51528, 14280, 3150, 69168, 7344, 80088, 8568, 48360, 13695, 40136, 6444, 44896, 7980, 146688, 29260, 92880, 48180
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

A002145 are precisely the rational primes in the ring of Gaussian integers.
From the representation of complex numbers as 2 X 2 matrices, a(n) is also the multiplicative order of the matrix [2,-1;1,2] or [2,1;-1,2] modulo p.
a(n) is divisible by ord(5,p): If (2+-i)^n == 1 (mod p), then 5^n == 1 (mod p).
a(n) divides (p+1) * ord(5,p), since we have (2+-i)^(p+1) == 5 (mod p).
If 5 is a quadratic residue modulo p, then ord(5,p) divides (p-1)/2, and so a(n) divides (p^2-1)/2. Conversely, if a(n) divides (p^2-1)/2, then (x+-y*i)^2 == 2+-i (mod p) for some integers x, y, and so (x^2+y^2)^2 == 5 (mod p), which means that 5 is a quadratic residue modulo p.

Examples

			The multiplicative order of 2+-i modulo A002145(3) = 11 is a(3) = 30, since (2+-i)^30 == 1 (mod 11), and 30 is the smallest such exponent.
		

Crossrefs

Cf. A002145, A211241, A385163 (multiplicative order of 1+-i), A385166.

Programs

  • PARI
    ord(p) = my(d = divisors((p+1)*znorder(Mod(5,p)))); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i,p)
    forprime(p=3, 1e3, if(p%4==3, print1(ord(p), ", ")))

A334426 Decimal expansion of Product_{k>=1} (1 + 1/A002145(k)^3).

Original entry on oeis.org

1, 0, 4, 1, 1, 5, 8, 0, 7, 2, 8, 2, 3, 4, 4, 4, 5, 8, 0, 3, 3, 8, 3, 6, 0, 5, 6, 9, 9, 2, 5, 6, 1, 5, 6, 6, 9, 3, 7, 6, 0, 7, 1, 3, 5, 1, 1, 3, 4, 9, 3, 5, 4, 1, 7, 3, 9, 4, 9, 8, 8, 6, 6, 6, 1, 7, 8, 5, 4, 1, 3, 5, 5, 8, 5, 6, 1, 3, 5, 0, 3, 5, 3, 5, 6, 0, 4, 7, 4, 5, 5, 4, 6, 7, 1, 0, 8, 7, 4, 3, 1, 5, 3, 6, 3
Offset: 1

Views

Author

Vaclav Kotesovec, Apr 30 2020

Keywords

Examples

			1.041158072823444580338360569925615669376071...
		

References

  • B. C. Berndt, Ramanujan's notebook part IV, Springer-Verlag, 1994, p. 64-65.

Crossrefs

Formula

A334426 / A334427 = 28*zeta(3)/Pi^3.
A334424 * A334426 = 840*zeta(3)/Pi^6.

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A334427 Decimal expansion of Product_{k>=1} (1 - 1/A002145(k)^3).

Original entry on oeis.org

9, 5, 9, 1, 4, 2, 7, 1, 1, 0, 4, 3, 2, 0, 7, 3, 4, 4, 9, 9, 9, 7, 0, 5, 9, 1, 3, 7, 5, 0, 2, 0, 9, 8, 1, 5, 3, 6, 5, 4, 2, 3, 6, 5, 9, 7, 7, 4, 4, 5, 7, 1, 0, 6, 3, 4, 8, 6, 2, 6, 6, 4, 3, 2, 8, 0, 6, 8, 5, 4, 9, 8, 8, 3, 8, 6, 4, 2, 2, 3, 8, 9, 3, 4, 1, 2, 3, 9, 3, 7, 7, 5, 3, 7, 4, 3, 9, 7, 1, 3, 5, 8, 1, 1, 1, 3
Offset: 0

Views

Author

Vaclav Kotesovec, Apr 30 2020

Keywords

Examples

			0.959142711043207344999705913750209815365423...
		

References

  • B. C. Berndt, Ramanujan's notebook part IV, Springer-Verlag, 1994, p. 64-65.

Crossrefs

Formula

A334426 / A334427 = 28*zeta(3)/Pi^3.
A334425 * A334427 = 8/(7*zeta(3)).

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A348746 Fully multiplicative with a(2) = 3, a(3) = 5, a(A002144(n)) = A002144(1+n) and a(A002145(1+n)) = a(A002145(1+n)) for all n >= 1, where A002144 and A002145 give the primes of the form 4k+1 and 4k+3 respectively.

Original entry on oeis.org

1, 3, 5, 9, 13, 15, 7, 27, 25, 39, 11, 45, 17, 21, 65, 81, 29, 75, 19, 117, 35, 33, 23, 135, 169, 51, 125, 63, 37, 195, 31, 243, 55, 87, 91, 225, 41, 57, 85, 351, 53, 105, 43, 99, 325, 69, 47, 405, 49, 507, 145, 153, 61, 375, 143, 189, 95, 111, 59, 585, 73, 93, 175, 729, 221, 165, 67, 261, 115, 273, 71, 675, 89, 123
Offset: 1

Views

Author

Antti Karttunen, Nov 02 2021

Keywords

Comments

Permutation of odd numbers. Preserves the prime signature.

Crossrefs

Cf. A000720, A002144, A002145, A348744, A348747 (left inverse).
Cf. also A003961, A332818 for similar maps.

Programs

  • PARI
    A348746(n) = { my(f=factor(n)); for(k=1,#f~, if(2==f[k,1], f[k,1]=3, if(3==f[k,1], f[k,1]=5, if(1==(f[k,1]%4), for(i=1+primepi(f[k,1]),oo,if(1==(prime(i)%4), f[k,1]=prime(i); break)))))); factorback(f); };

Formula

Fully multiplicative with a(p) = A348744(A000720(p)), where A348744 is the lexicographically earliest bijection from primes to odd primes where each prime of the form 4k+1 is mapped to the next larger prime of the same form.

A095008 Number of 4k+3 primes (A002145) in range ]2^n,2^(n+1)].

Original entry on oeis.org

1, 1, 1, 3, 3, 7, 13, 22, 37, 71, 128, 231, 440, 807, 1519, 2872, 5371, 10204, 19341, 36759, 70179, 134241, 256856, 492936, 947272, 1822615, 3513691, 6781495, 13103816, 25348667, 49092241, 95168205, 184661253, 358636497, 697094872, 1356052491, 2639893495, 5142817901
Offset: 1

Views

Author

Antti Karttunen and Labos Elemer, Jun 01 2004

Keywords

Crossrefs

Formula

a(n) = A036378(n) - A095007(n) = A095010(n) + A095012(n) = A095092(n) + A095093(n).

Extensions

a(34)-a(38) from Amiram Eldar, Jun 12 2024

A107978 Products of two primes of the form 4n+3 (A002145).

Original entry on oeis.org

9, 21, 33, 49, 57, 69, 77, 93, 121, 129, 133, 141, 161, 177, 201, 209, 213, 217, 237, 249, 253, 301, 309, 321, 329, 341, 361, 381, 393, 413, 417, 437, 453, 469, 473, 489, 497, 501, 517, 529, 537, 553, 573, 581, 589, 597, 633, 649, 669, 681, 713, 717, 721, 737
Offset: 1

Views

Author

Jonathan Vos Post, Jun 12 2005

Keywords

Comments

Every odd semiprime must be in one of three disjoint sets: the product of two primes of the form 4n+1 (A121387), the product of two primes of the form 4n+3 (A107978), or the product of a prime of the form 4n+1 and a prime of the form 4n+3 (A080774).

Crossrefs

Union of A131574 and A080109.
Third row of A121388.

Programs

  • Mathematica
    p = Select[ Prime@ Range@ 60, Mod[ #, 4] == 3 &]; Take[ Sort@ Flatten@ Table[ p[[i]] p[[j]], {j, 30}, {i, j}], 54] (* or *)
    fQ[n_] := Block[{fi = FactorInteger@ n}, Plus @@ Last /@ fi == 2 && Union@ Mod[ First /@ fi, 4] == {3}]; Select[ Range@ 748, fQ@# &] (* Robert G. Wilson v, May 20 2010 *)

Formula

{a(n)} = {p*q: p and q both elements of A002145}.

Extensions

Edited by N. J. A. Sloane, May 20 2010

A267101 2 followed by permutation of odd primes, where each n-th prime of the form 4k+1 (A002144) has been replaced with the n-th prime of the form 4k+3 (A002145) and vice versa.

Original entry on oeis.org

2, 5, 3, 13, 17, 7, 11, 29, 37, 19, 41, 23, 31, 53, 61, 43, 73, 47, 89, 97, 59, 101, 109, 67, 71, 79, 113, 137, 83, 103, 149, 157, 107, 173, 127, 181, 131, 193, 197, 139, 229, 151, 233, 163, 167, 241, 257, 269, 277, 179, 191, 281, 199, 293, 211, 313, 223, 317, 227, 239, 337, 251, 349, 353, 263, 271, 373, 283, 389
Offset: 1

Views

Author

Antti Karttunen, Feb 01 2016

Keywords

Comments

After 2, for each n >= 1, swap the places of primes A002144(n) and A002145(n) in A000040.

Examples

			For n=2, for which A000040(2) = 3, the first prime of the form 4k+3, we select the first prime of the form 4k+1, which is 5, thus a(2) = 5.
For n=3, for which A000040(3) = 5, the first prime of the form 4k+1, we select the first prime of the form 4k+3, which is 3, thus a(3) = 3.
For n=4, for which A000040(4) = 7, the second prime of the form 4k+3, we select the second prime of the form 4k+1, which is 13, thus a(4) = 13.
For n=5, for which A000040(5) = 11, the third prime of the form 4k+3, we select the third prime of the form 4k+1, which is 17, thus a(5) = 17.
		

Crossrefs

Programs

Formula

a(1) = 2; after which, if prime(n) modulo 4 = 1, a(n) = A002145(A267097(n)), otherwise a(n) = A002144(A267098(n)).
a(n) = A000040(A267100(n)).
a(n) = A267099(A000040(n)).

A334448 Decimal expansion of Product_{k>=1} (1 - 1/A002145(k)^4).

Original entry on oeis.org

9, 8, 7, 1, 6, 2, 6, 2, 5, 4, 2, 2, 2, 2, 6, 8, 5, 6, 4, 8, 2, 7, 0, 1, 2, 6, 4, 5, 7, 7, 3, 7, 0, 8, 2, 7, 7, 2, 4, 0, 3, 2, 7, 9, 7, 2, 9, 2, 8, 2, 4, 1, 4, 7, 4, 3, 4, 8, 3, 2, 6, 5, 0, 8, 5, 5, 7, 3, 0, 8, 9, 4, 7, 5, 6, 6, 7, 0, 0, 1, 8, 8, 9, 0, 8, 4, 1, 5, 0, 4, 9, 9, 8, 9, 0, 7, 3, 3, 4, 7, 7, 0, 3, 5, 3, 6
Offset: 0

Views

Author

Vaclav Kotesovec, Apr 30 2020

Keywords

Comments

In general, for s>1, Product_{k>=1} (1 + 1/A002145(k)^s)/(1 - 1/A002145(k)^s) = 2^s * (2^s - 1) * zeta(s) / (zeta(s, 1/4) - zeta(s, 3/4)).

Examples

			0.98716262542222685648270126457737082772403279729282414743483...
		

References

  • B. C. Berndt, Ramanujan's notebook part IV, Springer-Verlag, 1994, p. 64-65.

Crossrefs

Formula

A334447 / A334448 = 1/(PolyGamma(3, 1/4)/(8*Pi^4) - 1).
A334446 * A334448 = 96/Pi^4.

Extensions

More digits from Vaclav Kotesovec, Jun 27 2020

A082075 First differences of primes of the form 4*k+3 (A002145).

Original entry on oeis.org

4, 4, 8, 4, 8, 12, 4, 12, 8, 4, 8, 4, 20, 4, 20, 4, 8, 12, 12, 4, 12, 12, 8, 12, 12, 4, 12, 12, 12, 8, 12, 24, 4, 20, 16, 12, 8, 12, 4, 36, 12, 8, 4, 20, 4, 12, 8, 4, 8, 4, 20, 24, 16, 8, 16, 12, 8, 12, 12, 12, 4, 12, 24, 8, 28, 8, 12, 4, 8, 36, 24, 12, 4, 12, 20, 4, 20, 4, 20, 4, 8, 28, 20, 4
Offset: 1

Views

Author

Labos Elemer, Apr 07 2003

Keywords

Examples

			The first and second primes of the form 4*k+3 are 3 and 7, so a(1) = 7-3 = 4.
		

Crossrefs

Programs

  • Mathematica
    k=0; m=4; r=3; Do[s=Mod[Prime[n], m]; If[Equal[s, r], rp=ep; k=k+1; ep=Prime[n]; Print[ep-rp]; ], {n, 1, 1000}]

Formula

a(n) = A002145(n+1) - A002145(n).
Showing 1-10 of 372 results. Next