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 1-2 of 2 results.

A172143 a(n) = (A172126(n) - 1)/3.

Original entry on oeis.org

0, 1, 3, 5, 13, 21, 28, 53, 85, 113, 213, 227, 341, 453, 853, 909, 1365, 1813, 1820, 3413, 3637, 5461, 7253, 7281, 13653, 14549, 14563, 21845, 29013, 29125, 54613, 58197, 58253, 87381, 116053, 116501, 116508, 218453, 232789, 233013, 349525, 464213, 466005, 466033
Offset: 1

Views

Author

Ralf Stephan, Nov 19 2010

Keywords

Comments

Conjecture: sequence consists of an infinite number of subsequences S(m,0) = A172241(n) = (1/18)[8^n-(-1)^n-9], m>0, S(m,n+1) = 4*S(m,n)+1. The first subsequences
S(1,n) = A002450(n) = (4^n-1)/3 = 0, 1, 5, 21, 85, ...,
S(2,n) = A072197(n) = (10*4^n-1)/3 = 3, 13, 53, 213, ...,
S(3,n) = (85*4^n-1)/3 = 28, 113, 453, ...,
S(4,n) = (682*4^n-1)/3 = 227, 909, 3637, ..., and generally,
S(m,n) = [(3*A172241(m) + 1) * 4^n - 1]/3.

Links

Crossrefs

Cf. A002450, A072197, A172126, A172241.

Programs

  • Mathematica
    seq[max_] := Module[{kmax = Floor[Log[4, 3*max+1]], s = {}, s1, odd},Do[odd = (4^k-1)/3; s1 = 2^Range[0, Floor[Log2[max/odd]]] * odd; s = Join[s, s1], {k, 1, kmax}]; Select[(Union[s] - 1)/3, IntegerQ]]; seq[10^7] (* Amiram Eldar, Sep 01 2024 *)
  • PARI
    for(n=1, 300000, o=3*n/2^valuation(n, 2)+1; b=ispower(o); if(b&&b%2==0&&round(sqrtn(o, b/2))==4&&(n-1)%3==0, print1((n-1)/3, ", ")))

Extensions

More terms from Amiram Eldar, Sep 01 2024

A368415 Array read by ascending antidiagonals. A(n, k) = floor((n^k + 3)*(n/(2*n + 2))).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 4, 3, 1, 3, 7, 11, 6, 1, 3, 11, 26, 31, 11, 1, 4, 16, 53, 103, 92, 22, 1, 4, 22, 93, 261, 410, 274, 43, 1, 5, 29, 151, 556, 1303, 1639, 821, 86, 1, 5, 37, 228, 1051, 3333, 6511, 6554, 2461, 171, 1, 6, 46, 329, 1821, 7354, 19996, 32553, 26215, 7382, 342, 1, 6, 56, 455, 2953
Offset: 1

Views

Author

Thomas Scheuerle, Dec 23 2023

Keywords

Comments

Let p be an odd prime number, then A(p, k) is the number of distinct quadratic residues mod p^k. Let m = p1^k1^*p2^k2*..*pz^kz with p1..pz odd primes, then A(p1, k1)*A(p2, k2)*..*A(pz, kz) is the number of distinct quadratic residues mod m. For 2^t*m is floor((2^t+10)*(1/6))*A(p1, k1)*A(p2, k2)*..*A(pz, kz) the number of distinct quadratic residues mod 2^t*m.

Examples

			The array A(n, k) begins:
1,  1,   1,    1,     1,      1,       1,        1,         1,          1
1,  2,   3,    6,    11,     22,      43,       86,       171,        342
2,  4,  11,   31,    92,    274,     821,     2461,      7382,      22144
2,  7,  26,  103,   410,   1639,    6554,    26215,    104858,     419431
3, 11,  53,  261,  1303,   6511,   32553,   162761,    813803,    4069011
3, 16,  93,  556,  3333,  19996,  119973,   719836,   4319013,   25914076
4, 22, 151, 1051,  7354,  51472,  360301,  2522101,  17654704,  123582922
4, 29, 228, 1821, 14564, 116509,  932068,  7456541,  59652324,  477218589
5, 37, 323, 2953, 26573, 239149, 2152337, 19371025, 174339221, 1569052981
5, 46, 455, 4546, 45455, 454546, 4545455, 45454546, 454545455, 4545454546

Crossrefs

Cf. A000124, A005578, A039300, A039302, A039304, A172241, A363773.

Programs

  • PARI
    A(n, k) = (n^(k+1)+n*3)\(2*n+2)

Formula

A(n, k) = n*A(n, k-1) + A(n, k-2) - n*A(n, k-3), for k > 2 and A(n, 0) = 1.
A(1, k) = 1.
A(2, k) = A005578(k).
A(3, k) = A039300(k).
A(4, k) = A363773(k).
A(5, k) = A039302(k).
A(7, k) = A039304(k).
A(8, k) = A172241(k+1)+1.
A(n, 2) = A000124(n-1), for n > 0.
Showing 1-2 of 2 results.

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