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.

User: Vassilis Papadimitriou

Vassilis Papadimitriou's wiki page.

Vassilis Papadimitriou has authored 12 sequences. Here are the ten most recent ones:

A342481 Prime numbers whose binary expansion involves powers of 2 with only composite (or zero) exponents.

Original entry on oeis.org

17, 257, 337, 577, 593, 769, 1297, 1361, 1553, 1601, 1873, 4177, 4673, 5393, 5441, 5953, 16657, 17489, 17681, 17729, 17921, 18257, 20753, 21313, 21521, 21569, 21841, 22273, 32833, 33601, 33617, 33809, 33857, 34129, 34369, 36929, 37201, 37441, 37633, 37649, 37889, 38737
Offset: 1

Views

Author

Vassilis Papadimitriou, Mar 13 2021

Keywords

Examples

			17 = 2^4 + 2^0 is a term.
337 = 2^8 + 2^6 + +2^4 + 2^0 is a term.

Crossrefs

Cf. A342475.

A342475 Prime numbers whose binary expansion contains only prime powers of 2 and the zeroth power.

Original entry on oeis.org

5, 13, 37, 41, 137, 173, 2053, 2081, 2089, 2213, 2221, 8233, 8237, 8329, 8353, 10253, 10273, 10369, 131113, 131213, 133121, 133153, 133157, 133253, 133261, 139273, 139297, 139301, 139309, 139393, 139397, 139429, 141353, 141481, 524429, 524453, 526373, 526381, 526501
Offset: 1

Views

Author

Vassilis Papadimitriou, Mar 13 2021

Keywords

Comments

The numbers m = 2^e(0) + 2^e(1) + 2^e(2) + ... where all e(i) are either 0 or prime are 1, 4, 5, 8, 9, 12, 13, 32, 33, 36, 37, 40, 41, 44, 45, 128, 129, 132, 133, 136, 137, 140, 141, 160, 161, 164, ... The sequence contains the m which are primes. - R. J. Mathar, Apr 21 2021

Examples

			5 = 2^2 + 2^0 is a term.
7 = 2^2 + 2^1 + 2^0 is not a term, because the exponent 1 is not a prime.
11 = 2^3 + 2^1 + 2^0 is not a term, because the exponent 1 is not a prime.
13 = 2^3 + 2^2 + 2^0 is a term.

Crossrefs

Cf. A004676, A326782, A342481.

Programs

  • Mathematica
    Select[Array[1 + Total@ MapIndexed[#1*2^Prime[#2] & @@ {#1, First[#2]} &, Reverse@ IntegerDigits[#, 2]] &, 140], PrimeQ] (* Michael De Vlieger, Mar 13 2021 *)
  • PARI
    isok(p) = if (isprime(p), my(b=Vecrev(binary(p))); sum(i=1, #b, b[i]*((i!=1) && !isprime(i-1))) == 0); \\ Michel Marcus, Apr 22 2021

A182109 Records in A094593.

Original entry on oeis.org

1, 2, 4, 5, 6, 12, 16, 39, 84, 156, 350, 358, 589, 984, 2030, 2682, 3312, 4364, 19152, 61320, 61632, 142066, 353998, 702794, 1063044, 2056526, 2866334, 5479152, 8751462, 43544486
Offset: 1

Views

Author

Vassilis Papadimitriou, Apr 12 2012

Keywords

Comments

It is for A094593 what A152598 is for A001917.

Examples

			First few terms of A094593 are 1, 1, 2, 4, 1, 1, 2, 1, 1, 2, 5, 1, 2, 1, 2, 6, 3, 2, 6, 1, 2, 1, 2, 1, 3, 2, 4, 1, 1, 2, 1, 1, 1, 3, 2, 1, 2, 1, 2, 4, 2, 12, so a(1) to a(6) are 1, 2, 4, 5, 6, 12.

Crossrefs

Cf. A094593.

Extensions

a(26)-a(30) from Chai Wah Wu, Jan 15 2020

A162730 Semiprimes n = pq such that q = kp - k + 1, where p,q primes and k > 1.

Original entry on oeis.org

6, 10, 14, 15, 21, 22, 26, 33, 34, 38, 39, 46, 51, 57, 58, 62, 65, 69, 74, 82, 85, 86, 87, 91, 93, 94, 106, 111, 118, 122, 123, 129, 133, 134, 141, 142, 145, 146, 158, 159, 166, 177, 178, 183, 185, 194, 201, 202, 205, 206, 213, 214, 217, 218, 219, 226, 237, 249, 254
Offset: 1

Views

Author

Vassilis Papadimitriou, Jul 12 2009, Jul 13 2009

Keywords

Comments

It seems that most of the semiprimes of this form (but not all, only those satisfying an additional property) can be factored very quickly (e.g. numbers with up to 1200 decimal digits can be factored in a couple of seconds) using a very simple method.
Squarefree semiprimes n such that lpf(n)-1 divides n-1. Semiprimes n = pq with primes p < q such that p-1 divides q-1. If n is such a semiprime, then q^n == q (mod n). - Thomas Ordowski, Sep 18 2018

Crossrefs

Subsequence of A006881 (squarefree semiprimes).

Programs

  • PARI
    isok(n) = {if ((bigomega(n) == 2) && (omega(n) == 2), my(p = factor(n)[1, 1], q = factor(n)[2, 1]); (q-1) % (p-1) == 0;);} \\ Michel Marcus, Sep 18 2018

Extensions

More terms from R. J. Mathar, Aug 06 2009

A174240 The multiplicative order of 2 mod n, where n an odd squarefree semiprime (A046388).

Original entry on oeis.org

4, 6, 10, 12, 12, 8, 20, 18, 12, 22, 30, 8, 28, 12, 10, 36, 36, 44, 24, 20, 14, 18, 46, 60, 28, 20, 52, 33, 58, 60, 36, 40, 66, 84, 20, 90, 70, 28, 15, 18, 24, 92, 78, 36, 82, 110, 36, 52, 22, 60, 48, 116, 132, 42, 100, 60, 102, 140, 106, 72, 36, 69, 132, 28
Offset: 1

Views

Author

Vassilis Papadimitriou, Mar 13 2010

Keywords

Examples

			For n = 1, a(1) = 4 is the multiplicative order of 2 mod 15 (15 is the first odd squarefree semiprime).

Links

Crossrefs

Cf. A002326, A046388.

Programs

  • Mathematica
    MultiplicativeOrder[2, #] & /@ Select[Range[1, 350, 2], PrimeOmega[#] == 2 && PrimeNu[#] == 2 &] (* Amiram Eldar, Feb 24 2021 *)
  • PARI
    lista(nn) = {forstep(n=1, nn, 2, if ((bigomega(n) == 2) && (omega(n)==2), print1(znorder(Mod(2, n)), ", ")););} \\ Michel Marcus, Feb 05 2015

Formula

a(n) = A002326((A046388(n)+1)/2). - Amiram Eldar, Feb 24 2021

Extensions

Edited by Ray Chandler, Jan 09 2012

A174435 lambda(y)/x, where y an odd squarefree semiprime and x = ord(2,y) the smallest positive integer such that 2^x == 1 mod y (the multiplicative order of 2 mod y).

Original entry on oeis.org

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

Views

Author

Vassilis Papadimitriou, Mar 19 2010

Keywords

Examples

			For n=1 the a(1)= 1, as the first odd squarefree semiprime is 15, lambda(15)=4 and ord(2,15)=4

Links

Crossrefs

Cf. A002322, A002326, A046388, A174240.

Programs

  • Mathematica
    (CarmichaelLambda[#]/MultiplicativeOrder[2, #]) & /@ Select[Range[1, 530, 2], PrimeOmega[#] == 2 && PrimeNu[#] == 2 &] (* Amiram Eldar, Feb 24 2021 *)

Formula

Equals lambda(A046388)/ord(2, A046388), or lambda(A046388)/A174240.
a(n) = A002322(A046388(n))/A002326((A046388(n)+1)/2). - Amiram Eldar, Feb 24 2021

A162409 Semiprimes of the form p*(k*p-1) where k > 1 (and p prime).

Original entry on oeis.org

6, 10, 14, 15, 22, 26, 33, 34, 38, 46, 51, 58, 62, 69, 74, 82, 86, 87, 91, 94, 95, 106, 118, 122, 123, 134, 141, 142, 145, 146, 158, 159, 166, 177, 178, 194, 202, 206, 213, 214, 218, 226, 249, 254, 262, 267, 274, 278, 287, 295, 298, 302, 303, 314, 321, 326, 334
Offset: 1

Views

Author

Vassilis Papadimitriou, Jul 02 2009

Keywords

Comments

Regarding k = 1: 3 is the only prime p such that p-1 is prime, so 3*(1*3-1) = 6. But 6 is a term for p = 2 and k = 2 (see example), therefore the sequence does not change if k = 1 is allowed in the definition.

Examples

			For p = 2 and k = 2 we have 2*(2*2-1) = 6, so 6 is a term. For p = 3 and k = 6 we have 3*(6*3-1) = 51, so 51 is a term.

Crossrefs

Subsequence of A006881 (product of two distinct primes).

Programs

  • Magma
    m:=170; { s: p, q in PrimesUpTo(m) | s le 2*m and exists(t){ k: k in [2..p*q div 2] | q eq p*k-1 } where s is p*q };

Extensions

Edited, corrected and extended by Klaus Brockhaus, Jul 06 2009

A174437 Successive maximal values of A174435.

Original entry on oeis.org

1, 2, 3, 4, 9, 10, 16, 31, 36, 48, 105, 129, 315, 316, 387, 538, 1285, 1542, 2048, 3855, 4599, 4864, 7760, 13797, 18166, 24417, 60787, 104694
Offset: 1

Views

Author

Vassilis Papadimitriou, Mar 19 2010

Keywords

Examples

			First terms of A174435 are: 1,1,1,1,1,2,1,1,1,1,1,2,1,1,3,1,1,1,2,2,3,1,1,1,1,3,1,2,1,1,1,2,1,1,2,1,1,3,2,4,2,1,1,1,1,1,1,1,4,2,2,1,1,1,1,1,1,1,1,2,3,2,1,4,3,1,2,1,1,9, so a(1)=1, a(2)=2, a(3)=3, a(4)=4, a(5)=9.

Crossrefs

It is for A174435 what A152598 is for A001917.

A145756 a(n) = ((2^prime(n+2)-2)/prime(n+2))/3, where n >= 1.

Original entry on oeis.org

2, 6, 62, 210, 2570, 9198, 121574, 6170930, 23091222, 1238188770, 17878237850, 68186767614, 998138215286, 56649051916610, 3256840408493918, 12600235023025650, 734198769102867726, 11085367330679918342
Offset: 1

Views

Author

Vassilis Papadimitriou, Oct 17 2008

Keywords

Examples

			For n = 1, a(1) = ((2^prime(1+2)-2)/prime(1+2))/3 = ((2^prime(3)-2)/prime(3))/3 = ((2^5-2)/5)/3 = 30/5/3 = 2

Crossrefs

Equals A064535/3

A120451 Number of ways a prime number p can be expressed as 2*(p1-p2) + 3*p3, where p1, p2, p3 are primes or 1, less than or equal to p.

Original entry on oeis.org

0, 3, 4, 7, 9, 12, 13, 16, 18, 20, 23, 30, 32, 32, 33, 42, 43, 51, 50, 57, 64, 61, 69, 83, 84, 93, 89, 92, 110, 115, 114, 123, 133, 133, 153, 143, 157, 154, 163, 176, 179, 211, 197, 220, 233, 216, 227, 230, 233, 269, 278, 268, 310, 274, 314
Offset: 1

Views

Author

Vassilis Papadimitriou, Jul 20 2006

Keywords

Comments

At least for the first 200 primes, it is true that every prime p > 2 can be expressed as 2*(p1-p2) + 3*p3, where p1, p2, p3 are primes or 1, less than or equal to p (the proof would be straightforward if both (a) Levy's conjecture and (b) the conjecture saying that every prime p > 3 can be expressed as 2*p1 + 3*p2, where p1, p2 are primes, were true). It would be interesting to study how the sequence changes when we remove the restriction for p1, p2, p3 to be less than or equal to p.

Examples

			a(12)=30 because 37 (the 12th prime) can be expressed as
2*(1 - 2) + 3*13
OR 2*(1 - 11) + 3*19
OR 2*(1 - 17) + 3*23
OR 2*(1 - 29) + 3*31
OR 2*(2 - 3) + 3*13
OR 2*(3 - 1) + 3*11
OR 2*(3 - 13) + 3*19
OR 2*(3 - 19) + 3*23
OR 2*(3 - 31) + 3*31
OR 2*(5 - 3) + 3*11
OR 2*(7 - 5) + 3*11
OR 2*(7 - 17) + 3*19
OR 2*(7 - 23) + 3*23
OR 2*(11 - 3) + 3*7
OR 2*(13 - 2) + 3*5
OR 2*(13 - 5) + 3*7
OR 2*(13 - 11) + 3*11
OR 2*(13 - 23) + 3*19
OR 2*(13 - 29) + 3*23
OR 2*(17 - 3) + 3*3
OR 2*(19 - 2) + 3*1
OR 2*(19 - 5) + 3*3
OR 2*(19 - 11) + 3*7
OR 2*(19 - 17) + 3*11
OR 2*(19 - 29) + 3*19
OR 2*(31 - 17) + 3*3
OR 2*(31 - 23) + 3*7
OR 2*(31 - 29) + 3*11
OR 2*(37 - 23) + 3*3
OR 2*(37 - 29) + 3*7.

Programs

  • PARI
    a(n) = {my(vp = concat(1, primes(n)), nb=0, p=prime(n), p1, p2, p3); for (i=1, #vp, p1 = vp[i]; for (j=1, #vp, p2 = vp[j]; for (k=1, #vp, p3 = vp[k]; if (2*(p1-p2) + 3*p3 == p, nb++);););); nb;} \\ Michel Marcus, Jan 26 2021

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