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.

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A162175 Primes classified by weight.

Original entry on oeis.org

11, 17, 29, 41, 59, 67, 71, 79, 83, 89, 101, 103, 107, 109, 137, 149, 167, 179, 191, 193, 197, 227, 229, 239, 241, 251, 269, 277, 281, 283, 311, 331, 347, 349, 359, 367, 379, 383, 409, 419, 431, 433, 439, 443, 449, 461, 463, 467, 487, 491, 499, 503, 521, 557
Offset: 1

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Author

Rémi Eismann, Jun 27 2009

Keywords

Comments

Conjecture: primes classified by level are rarefying among prime numbers.
A000040(n) = 2, 3, 7, A162174(n), a(n). - Rémi Eismann, Jun 27 2009
By definition, primes classified by weight have a prime gap g(n) < sqrt(p(n)) (or more precisely, for primes classified by weight, we have A001223(n) <= sqrt(A118534(n)) - 1 ). So by definition, prime numbers classified by weight follow Legendre's conjecture and Andrica's conjecture - Rémi Eismann, Aug 26 2013

Examples

			For prime(5)=11, A117078(5)=3 <= A117563(5)=3 ; prime(5)=11 is classified by weight. For prime(170)=1013, A117078(170)=19 <= A117563(170)=53 ; prime(170)=1013 is classified by weight.
		

Crossrefs

Formula

If for prime(n), A117078(n) (the weight) <= A117563(n) (the level) and A117078(n) <> 0 then prime(n) is classified by weight. If for prime(n), A117078(n) (the weight) > A117563(n) (the level) then prime(n) is classified by level.

A179620 a(n) = largest k such that A002808(n+1) = A002808(n) + (A002808(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 4, 7, 8, 8, 10, 13, 14, 14, 16, 19, 20, 20, 23, 24, 25, 26, 26, 28, 31, 32, 33, 34, 34, 37, 38, 38, 40, 43, 44, 44, 47, 48, 49, 50, 50, 53, 54, 55, 56, 56, 58, 61, 62, 63, 64, 64, 67, 68, 68, 70, 73, 74, 75, 76, 76, 79, 80, 80, 83, 84, 85, 86, 86, 89
Offset: 1

Views

Author

Rémi Eismann, Jan 09 2011

Keywords

Comments

a(n) = A002808(n) - A073783(n) if A002808(n) - A073783(n) > A073783(n), 0 otherwise.
A002808(n): composite numbers; A073783(n): first difference of composite numbers.

Examples

			For n = 1 we have A002808(n) = 4, A002808(n+1) = 6; there is no k such that 6 - 4 = 2 = (4 mod k), hence a(1) = 0.
For n = 3 we have A002808(n) = 8, A002808(n+1) = 9; 7 is the largest k such that 9 - 8 = 1 = (8 mod k), hence a(3) = 7; a(3) = A002808(3) - A073783(3) = 8 - 1 = 7.
For n = 24 we have A002808(n) = 36, A002808(n+1) = 38; 34 is the largest k such that 38 - 36 = 2 = (36 mod k), hence a(24) = 34; a(24) = A002808(24) - A073783(24) = 34.
		

Crossrefs

A184218 a(n) = largest k such that A000217(n+1) = A000217(n) + (A000217(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377
Offset: 1

Views

Author

Rémi Eismann, Jan 10 2011

Keywords

Comments

From the definition, a(n) = A000217(n) - (n + 1) if A000217(n) - (n + 1) > (n + 1), or 0 otherwise, where A000217 are the triangular numbers.

Examples

			For n = 3 we have A000217(3) = 6, A000217(4) = 10; there is no k such that 10 - 6 = 4 = (6 mod k), hence a(3) = 0.
For n = 5 we have A000217(5) = 15, A000217(6) = 21; 9 is the largest k such that 21 - 15 = 6 = (15 mod k), hence a(5) = 9; a(5) = A000217(5) - (5 + 1) = 15 - 6 = 9.
For n = 24 we have A000217(24) = 300, A000217(25) = 325; 275 is the largest k such that 325 - 300 = 25 = (300 mod k), hence a(24) = 275; a(24) = A000217(24) - (24 + 1) = 275.
		

Crossrefs

Cf. essentially the same as A000096, A000217, A000027, A130703, A184219, A118534, A117078, A117563, A001223.

Programs

  • Magma
    [0,0,0,0] cat [(n+1)*(n-2)/2: n in [5..60]]; // Vincenzo Librandi, Jun 22 2016
  • Mathematica
    Join[{0, 0, 0, 0}, LinearRecurrence[{3, -3, 1}, {9, 14, 20}, 100]] (* G. C. Greubel, Jun 22 2016 *)
    lim = 10^4; Table[SelectFirst[Reverse@ Range@ lim, Function[k, PolygonalNumber[n + 1] == # + Mod[#, k] &@ PolygonalNumber@ n]], {n, 53}] /. {k_ /; MissingQ@ k -> 0, k_ /; k == lim -> 0} (* Michael De Vlieger, Jun 30 2016, Version 10.4 *)

Formula

a(n) = (n+1)*(n-2)/2 = A000096(n-2) for n >= 5 and a(n) = 0 for n <= 4. - M. F. Hasler, Jan 10 2011
From Chai Wah Wu, Jun 21 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 7.
G.f.: x^5*(5*x^2 - 13*x + 9)/(1 - x)^3. (End)

A184220 a(n) = largest k such that A000290(n+1) = A000290(n) + (A000290(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 14, 23, 34, 47, 62, 79, 98, 119, 142, 167, 194, 223, 254, 287, 322, 359, 398, 439, 482, 527, 574, 623, 674, 727, 782, 839, 898, 959, 1022, 1087, 1154, 1223, 1294, 1367, 1442, 1519, 1598, 1679, 1762
Offset: 1

Views

Author

Rémi Eismann, Jan 10 2011

Keywords

Comments

From the definition, a(n) = A000290(n) - A005408(n) if A000217(n) - A005408(n) > A005408(n), 0 otherwise, where A000290 are the squares and A005408 are the gaps between squares: 2n + 1.

Examples

			For n = 3 we have A000290(3) = 9, A000290(4) = 16; there is no k such that 16 - 9 = 7 = (9 mod k), hence a(3) = 0.
For n = 5 we have A000290(5) = 25, A000290(6) = 36; 14 is the largest k such that 36 - 25 = 11 = (25 mod k), hence a(5) = 14; a(5) = A000290(5) - A005408(5) = 25 - 11 = 14.
For n = 25 we have A000217(25) = 625, A000217(26) = 676; 574 is the largest k such that 676 - 625 = 51 = (625 mod k), hence a(25) = 574; a(25) = A000290(25) - A005408(25) = 574.
		

Crossrefs

Cf. essentially the same as A008865, A000290, A005408, A133150, A184221, A118534, A117078, A117563, A001223.

Formula

a(n) = (n-1)^2-2 = A008865(n-1) for n >= 5 and a(n) = 0 for n <= 4.

A106752 Numbers of prime factors of k, k defined in A117078.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Rémi Eismann, Jun 22 2007, Feb 14 2008

Keywords

Comments

a(n) = 0 only for n = 1, 2 and 4.

Examples

			For a(1), k=0 thus a(1)=0,
For a(3), k=3 thus a(3)=1,
For a(11), k=25=5*5 thus a(11)=2.
		

Crossrefs

Formula

a(n) = numbers of factors of A117078(n). A117078(n) : smallest k such that prime(n+1) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

A117877 Least p=prime(k) for which A118123(k)=n.

Original entry on oeis.org

2, 5, 11, 17, 67, 101, 109, 107, 227, 569, 499, 821, 1163, 2153, 1489, 1487, 1579, 4111, 6841, 10739, 5783, 21383, 4729, 3467, 34183, 58741, 19319, 22283, 22279, 22277, 16069, 16067, 17333, 91583, 20479, 20477, 82223, 158363, 31189, 70877, 45061
Offset: 0

Views

Author

Keywords

Examples

			a(0)=2 because no k exists and it is the least of the three {2, 3 & 7} in A117563 or A117078.
a(1)=5 because 5 + 5 (mod 3) = 7,
a(2)=11 because 11 + 11 (mod 3) = 11 + 11 (mod 9) = 13.
a(3)=17 because 17 + 17 (mod 3) = 17 + 17 (mod 5) = 17 + 17 (mod 15) = 19,
a(4)=67 because 67 + 67 (mod 7) = 67 + 67 (mod 9) = 67 + 67 (mod 21) = 67 + 67 (mod 63) = 71,
a(5)=101 because 101 + 101 (mod 3) = 101 + 101 (mod 9) = 101 + 101 (mod 11) = 101 + 101 (mod 33) = 101 + 101 (mod 99), etc.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{p = Prime@n, np = Prime[n + 1]}, Length@ Select[ Divisors[2p - np], # >= np - p &]]; t = Table[0, {50}]; Do[ a = f@n; If[a < 50 && t[[a + 1]] == 0, t[[a + 1]] = n; Print[{a, n, Prime@n}]], {n, 100000}]
  • PARI
    A117877(n)={ for( k=n+1, 1e9, n==A118123(k) & return(prime(k)))}

Extensions

Edited by M. F. Hasler, Nov 07 2009

A118741 Primes for which the weight as defined in A117078 is 7.

Original entry on oeis.org

67, 83, 167, 193, 251, 277, 433, 487, 503, 587, 601, 613, 727, 823, 907, 1063, 1217, 1231, 1553, 1663, 1777, 1861, 1873, 1973, 1987, 2083, 2281, 2293, 2351, 2377, 2393, 2797, 2897, 3217, 3343, 3541, 3847, 4073, 4283, 4451, 4507, 4591, 4813, 4871, 5081
Offset: 1

Views

Author

Rémi Eismann, May 22 2006; May 27 2006; May 04 2007

Keywords

Comments

The gap as defined in A001223 of this prime numbers is 4 or 6.
The prime numbers in this sequence are of the form (14i-1) (if gap=6) or (14i-3) (if gap=4) with i=(level(n)+1)/2, level(n) defined in A117563.

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{a, p = Prime[n], np = Prime[n + 1]}, a = Min[Select[Divisors[2*p - np], #1 > np - p & ]]; If[a == Infinity, 0, a]]; Prime@ Select[ Range@695, f@# == 7 &] (* Robert G. Wilson v, May 26 2006 *)

Formula

A117078 : a(n) = smallest k such that prime(n+1) = prime(n) + (prime(n) mod k), or 0 if no such k exists. prime(n) for which k=7.

Extensions

More terms from Robert G. Wilson v, May 26 2006

A119593 Primes for which the weight as defined in A117078 is 7 and the gap as defined in A001223 is 4.

Original entry on oeis.org

67, 193, 277, 487, 613, 823, 907, 1663, 1873, 2083, 2293, 2377, 2797, 3217, 3343, 3847, 4813, 5233, 5527, 5653, 5737, 6577, 6997, 7207, 7753, 8677, 8803, 9433, 11113, 11617
Offset: 1

Views

Author

Rémi Eismann, Jun 01 2006, May 04 2007

Keywords

Comments

The prime numbers in this sequence are of the form (14i-3) with i=(level(n)+1)/2, level(n) defined in A117563.

Crossrefs

Programs

  • PARI
    forprimestep(p=67,1e4,14, t=p%5; if((t==2 || t==3) && isprime(p+4), print1(p", "))) \\ Charles R Greathouse IV, Sep 17 2022
    
  • PARI
    p=67; forprime(q=p+2,1e4, if(q-p==4 && (p%70==53 || p%70==67), print1(p", ")); p=q) \\ Charles R Greathouse IV, Sep 17 2022

Formula

Primes p such that (1) p = 53 or 67 mod 70 and (2) p+4 is prime. - Charles R Greathouse IV, Sep 17 2022
a(n) = Omega(n log^2 n). - Charles R Greathouse IV, Sep 17 2022

A119594 Primes for which the weight as defined in A117078 is 9 and the gap as defined in A001223 is 4.

Original entry on oeis.org

13, 103, 463, 643, 877, 967, 1093, 1597, 1867, 1993, 2137, 2857, 3037, 3163, 3253, 3613, 3793, 4153, 4513, 4783, 5413, 5503, 5647, 6007, 6043, 6547, 6907, 7537, 7573, 7933, 8167, 8293, 9157, 9337, 9463, 9787
Offset: 1

Views

Author

Rémi Eismann, Jun 01 2006, May 04 2007

Keywords

Comments

The prime numbers in this sequence are of the form (18i-5) with i=(level(n)+1)/2, level(n) defined in A117563.

Examples

			a(1)=13 because of 17=13+mod(13;9) and 17-13=4.
18*1-5=13, level=1
a(2)=103 because of 107=103+mod(103;9) and 107-103=4
18*((11+1)/2)-5=103, level=11
		

Crossrefs

A133346 a(n) = smallest k such that prime(n+2) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 0, 7, 11, 0, 15, 21, 21, 31, 7, 11, 35, 9, 17, 17, 61, 9, 21, 23, 23, 77, 7, 19, 97, 101, 91, 19, 13, 41, 25, 127, 47, 139, 21, 17, 31, 11, 167, 13, 37, 11, 61, 25, 39, 7, 13, 73, 9, 227, 25, 239, 35, 15, 9, 29, 271, 269, 37, 25, 7, 61, 59, 27, 21, 13, 11, 113, 113
Offset: 1

Views

Author

Rémi Eismann, Oct 20 2007

Keywords

Examples

			For n = 1 we have prime(n) = 2, prime(n+2) = 5; there is no k such that 5 - 2 = 3 = (2 mod k), hence a(1) = 0.
For n = 6 we have prime(n) = 13, prime(n+2) = 19; 7 is the smallest k such that 19 - 13 = 6 = (13 mod k), hence a(6) = 7.
For n = 30 we have prime(n) = 113, prime(n+2) = 131; 19 is the smallest k such that 131 - 113 = 18 = (113 mod k), hence a(30) = 19.
		

Crossrefs

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