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 1-9 of 9 results.

A292939 Odd numbers k such that A292270((k-1)/2) is a square.

Original entry on oeis.org

1, 3, 5, 7, 11, 13, 15, 19, 27, 29, 31, 37, 53, 59, 61, 63, 67, 83, 95, 101, 107, 127, 131, 139, 149, 163, 173, 179, 181, 197, 211, 227, 255, 269, 277, 293, 305, 317, 347, 349, 373, 379, 389, 419, 421, 443, 461, 467, 491, 509, 511, 519, 523, 541, 547, 557, 563, 587, 613, 619, 653, 659, 661, 677, 701, 709, 757, 773
Offset: 1

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Keywords

Crossrefs

Union of A000225 and A292940.
Cf. A001122, A292379 (subsequences).
Cf. also A292270, A292938.

Formula

a(n) = 1 + 2*A292938(n).

A293218 a(n) = A007913(A292270(n)).

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 1, 38, 1, 3, 26, 31, 1, 1, 1, 103, 73, 1, 42, 14, 7, 91, 3, 58, 14, 1, 170, 303, 1, 1, 1, 66, 1, 385, 91, 93, 301, 65, 563, 1093, 1, 11, 355, 38, 118, 83, 1, 1254, 763, 1, 1043, 39, 1, 249, 141, 238, 19, 71, 43, 133, 11, 781, 1, 649, 1, 554, 1081, 614, 1, 1633, 5, 317, 1398, 1, 269, 626, 10, 527, 1285, 1191, 1
Offset: 0

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Author

Antti Karttunen, Oct 06 2017

Keywords

Crossrefs

Cf. A292938 (gives the positions of ones).

Programs

Formula

a(n) = A007913(A292270(n)).

A292938 Numbers n for which A292270(n) is a square.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 9, 13, 14, 15, 18, 26, 29, 30, 31, 33, 41, 47, 50, 53, 63, 65, 69, 74, 81, 86, 89, 90, 98, 105, 113, 127, 134, 138, 146, 152, 158, 173, 174, 186, 189, 194, 209, 210, 221, 230, 233, 245, 254, 255, 259, 261, 270, 273, 278, 281, 293, 306, 309, 326, 329, 330, 338, 350, 354, 378, 386, 393, 398, 410
Offset: 1

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Comments

Positions of ones in A293218.

Crossrefs

Cf. A292270, A292939 (corresponding odd numbers).
Cf. A000225 (a subsequence), A293218.

A292947 a(n) = A007947(A292270(n)).

Original entry on oeis.org

1, 1, 2, 1, 13, 5, 6, 1, 38, 3, 6, 26, 62, 11, 14, 1, 103, 73, 6, 42, 14, 35, 91, 21, 58, 14, 26, 170, 303, 29, 30, 1, 66, 33, 385, 182, 93, 301, 195, 563, 1093, 41, 22, 355, 38, 118, 83, 22, 1254, 763, 10, 1043, 78, 53, 498, 282, 238, 133, 71, 129, 665, 22, 1562, 1, 649, 65, 554, 1081, 614, 69, 1633, 85, 634, 1398, 74, 269, 626, 70, 1054, 1285, 1191, 3
Offset: 0

Views

Author

Antti Karttunen, Oct 06 2017

Keywords

Crossrefs

Cf. A000225 (gives the positions of ones).

Programs

A293219 a(n) = A000188(A292270(n)).

Original entry on oeis.org

1, 1, 2, 1, 1, 5, 6, 1, 1, 9, 2, 1, 2, 11, 14, 1, 1, 1, 18, 1, 4, 5, 1, 7, 2, 1, 26, 1, 1, 29, 30, 1, 2, 33, 1, 2, 1, 1, 3, 1, 1, 41, 2, 1, 2, 1, 1, 22, 1, 1, 50, 1, 2, 53, 2, 2, 2, 7, 1, 3, 5, 4, 2, 1, 1, 65, 1, 1, 2, 69, 1, 17, 2, 1, 74, 1, 1, 7, 2, 1, 1, 81, 1, 1, 6, 7, 86, 1, 1, 89, 90, 26, 2, 1, 1, 1, 1, 1, 98, 3, 3, 6, 2, 1, 1, 105
Offset: 0

Views

Author

Antti Karttunen, Oct 07 2017

Keywords

Crossrefs

Cf. A163782 (after a(1) = 1 gives the other positions where a(n) = n).

Programs

Formula

a(n) = A000188(A292270(n)).

A292940 Odd numbers k for which A292270((k-1)/2) is a square larger than one.

Original entry on oeis.org

5, 11, 13, 19, 27, 29, 37, 53, 59, 61, 67, 83, 95, 101, 107, 131, 139, 149, 163, 173, 179, 181, 197, 211, 227, 269, 277, 293, 305, 317, 347, 349, 373, 379, 389, 419, 421, 443, 461, 467, 491, 509, 519, 523, 541, 547, 557, 563, 587, 613, 619, 653, 659, 661, 677, 701, 709, 757, 773, 787, 797, 821, 827, 829, 853
Offset: 1

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Crossrefs

Subsequence of A292939.
Cf. A001122 (a subsequence).
Cf. A292270.

A292379 Odd numbers k such that A292270((k-1)/2) is a square < ((k-1)/2)^2.

Original entry on oeis.org

7, 15, 27, 31, 63, 95, 127, 255, 277, 305, 511, 519, 1023, 1921, 2047, 2349, 3575, 4095, 8191, 8743, 10525, 16383, 17955, 24661, 26585, 27305, 32767, 36611, 40677, 65535, 85173, 95571, 122915, 131071, 262143, 360213, 386019, 524287, 667879, 736251, 806603, 1020963
Offset: 1

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Crossrefs

Subsequence of A292939.
Doesn't contain any terms of A001122.
Cf. A000225 (subsequence from its term a(3) = 7 onward).
Cf. A292270.

Programs

  • PARI
    A000265(n) = n >> valuation(n, 2);
    A292270(n) = { my(x = n+n+1, m = A000265(1+x), z = m); while(m!=1, z += A000265(x+m); m = A000265(x+m)); z; };
    n=1; k=0; while(k < 120, m = A292270(n); if(issquare(m) && (m < n^2), k++; write("b292379.txt", k, " ", 1+(2*n))); n++);

Extensions

More terms from Jinyuan Wang, Apr 29 2025

A001122 Primes with primitive root 2.

Original entry on oeis.org

3, 5, 11, 13, 19, 29, 37, 53, 59, 61, 67, 83, 101, 107, 131, 139, 149, 163, 173, 179, 181, 197, 211, 227, 269, 293, 317, 347, 349, 373, 379, 389, 419, 421, 443, 461, 467, 491, 509, 523, 541, 547, 557, 563, 587, 613, 619, 653, 659, 661, 677, 701, 709, 757, 773, 787, 797
Offset: 1

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Comments

Artin conjectured that this sequence is infinite.
Conjecture: sequence contains infinitely many pairs of twin primes. - Benoit Cloitre, May 08 2003
Pieter Moree writes (Oct 20 2004): Assuming the Generalized Riemann Hypothesis, it can be shown that the density of primes p such that a prescribed integer g has order (p-1)/t, with t fixed, exists and, moreover, it can be computed. This density will be a rational number times the so-called Artin constant. For 2 and 10 the density of primitive roots is A, the Artin constant itself.
It seems that this sequence consists of A050229 \ {1,2}.
Primes p such that 1/p, when written in base 2, has period p-1, which is the greatest period possible for any integer.
Positive integer 2*m-1 is in the sequence iff A179382(m)=m-1. - Vladimir Shevelev, Jul 14 2010
These are the odd primes p for which the polynomial 1+x+x^2+...+x^(p-1) is irreducible over GF(2). - V. Raman, Sep 17 2012 [Corrected by N. J. A. Sloane, Oct 17 2012]
Prime(n) is in the sequence if (and conjecturally only if) A133954(n) = prime(n). - Vladimir Shevelev, Aug 30 2013
Pollack shows that, on the GRH, that there is some C such that a(n+1) - a(n) < C infinitely often (in fact, 1 can be replaced by any positive integer). Further, for any m, a(n), a(n+1), ..., a(n+m) are consecutive primes infinitely often. - Charles R Greathouse IV, Jan 05 2015
From Jianing Song, Apr 27 2019: (Start)
All terms are congruent to 3 or 5 modulo 8. If we define
Pi(N,b) = # {p prime, p <= N, p == b (mod 8)};
Q(N) = # {p prime, p <= N, p in this sequence},
then by Artin's conjecture, Q(N) ~ C*N/log(N) ~ 2*C*(Pi(N,3) + Pi(N,5)), where C = A005596 is Artin's constant.
Conjecture: if we further define
Q(N,b) = # {p prime, p <= N, p == b (mod 8), p in this sequence},
then we have:
Q(N,3) ~ (1/2)*Q(N) ~ C*Pi(N,3);
Q(N,5) ~ (1/2)*Q(N) ~ C*Pi(N,5). (End)
Conjecture: for a prime p > 5, p has primitive root 2 iff p == +-3 (mod 8) divides 2^k + 3 for some k < p - 1 and divides 2^m + 5 for some m < p - 1. It seems that all primes of the form 2^k + 3 for k <> 2 (A057732) have primitive root 2. - Thomas Ordowski, Nov 27 2023

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 864.
  • E. Bach and Jeffrey Shallit, Algorithmic Number Theory, I; see p. 221.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, New York, 1996; see p. 169.
  • M. Kraitchik, Recherches sur la Théorie des Nombres. Gauthiers-Villars, Paris, Vol. 1, 1924, Vol. 2, 1929, see Vol. 1, p. 56.
  • Lehmer, D. H. and Lehmer, Emma; Heuristics, anyone? in Studies in mathematical analysis and related topics, pp. 202-210, Stanford Univ. Press, Stanford, Calif., 1962.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 20.
  • D. Shanks, Solved and Unsolved Problems in Number Theory, 2nd. ed., Chelsea, 1978, p. 81.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A002326 for the multiplicative order of 2 mod 2n+1. (Alternatively, the least positive value of m such that 2n+1 divides 2^m-1).
Cf. A216838 (Odd primes for which 2 is not a primitive root).

Programs

  • Mathematica
    Select[ Prime@Range@200, PrimitiveRoot@# == 2 &] (* Robert G. Wilson v, May 11 2001 *)
    pr = 2; Select[Prime[Range[200]], MultiplicativeOrder[pr, # ] == # - 1 &] (* N. J. A. Sloane, Jun 01 2010 *)
  • PARI
    forprime(p=3, 1000, if(znorder(Mod(2, p))==(p-1), print1(p,", "))); \\ [corrected by Michel Marcus, Oct 08 2014]
    
  • Python
    from itertools import islice
    from sympy import nextprime, is_primitive_root
    def A001122_gen(): # generator of terms
        p = 2
        while (p:=nextprime(p)):
            if is_primitive_root(2,p):
                yield p
    A001122_list = list(islice(A001122_gen(),30)) # Chai Wah Wu, Feb 13 2023

Formula

Delta(a(n),2^a(n)*x) = a(n)*Delta(a(n),2*x), where Delta(k,x) is the difference between numbers of evil(A001969) and odious(A000069) integers divisible by k in interval [0,x). - Vladimir Shevelev, Aug 30 2013
For n >= 2, a(n) = 1 + 2*A163782(n-1). - Antti Karttunen, Oct 07 2017

A293220 Sum of all partial fractions in the algorithm used for calculation of A053446(n).

Original entry on oeis.org

1, 1, 1, 7, 12, 1, 8, 10, 1, 30, 12, 91, 108, 8, 6, 44, 157, 1, 45, 271, 300, 73, 164, 91, 162, 234, 1, 125, 588, 122, 175, 225, 684, 368, 65, 919, 373, 45, 512, 443, 206, 630, 300, 196, 506, 213, 118, 550, 303, 510, 459, 679, 2028, 1, 208, 941, 286, 1218, 201, 2611, 62, 691, 751, 724, 1575, 1374, 540, 3367, 1004, 36
Offset: 1

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Comments

This sequence gives an additional insight (cf. A292270) into the algorithm for the calculation of A053446(m), where m=A001651(n). Let us estimate how many steps are required before (the first) 1 will appear. Note that all partial fractions (which are indeed, integers) are residues modulo A001651(n) not divisible by 3 from the interval [1, A001651(n)-1]. So, if there is no repetition, then the number of steps does not exceed n-1. Suppose then that there is a repetition before the appearance of 1. Then for a not divisible by 3 residue k from [1, A001651(n)-1], 3^m_1 == 3^m_2 == k (mod A001651(n)) such that m_2 > m_1. But then 3^(m_2-m_1) == 1 (mod A001651(n)). So, since m_2 - m_1 < m_2, it means that 1 should appear earlier than the repetition of k, which is a contradiction. So the number of steps <= n-1. For example, for n=12, A001651(12) = 17, we have exactly n-1 = 11 steps with all other not divisible by 3 residues <= 17 - 1 = 16 modulo 17 appearing before the final 1: 2, 4, 7, 8, 14, 16, 11, 5, 13, 10 , 1.

Crossrefs

Cf. A038754 (seems to give the positions of ones).

Formula

Let n = 12. According to the comment, a(12) = 2 + 4 + 7 + 8 + 14 + 16 + 11 + 5 + 13 + 10 + 1 = 91.
Showing 1-9 of 9 results.