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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A299107 Probable primes in sequence {s_k(4)}, where s_k(4) = 4*s_{k-1}(4) - s_{k-2}(4), k >= 2, s_0(4) = 1, s_1(4) = 5.

Original entry on oeis.org

5, 19, 71, 3691, 191861, 138907099, 26947261171, 436315574686414344004975231616076636245689199862837798457639364993981991744926792179
Offset: 1

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Keywords

Comments

From a problem in A269254. For detailed theory, see [Hone].
Subsequent terms have too many digits to display.

Crossrefs

Formula

a(n) = s_{A299100(n)}(4) = A001834(A299100(n)).

A298878 Union_{p prime, n >= 0} {T_p(n)}, where T_m(x) = x*T_{m-1}(x) - T_{m-2}(x), m >= 2, T_0(x) = 2, T_1(x) = x (dilated Chebyshev polynomials of the first kind).

Original entry on oeis.org

-2, -1, 0, 1, 2, 7, 14, 18, 23, 34, 47, 52, 62, 79, 98, 110, 119, 123, 142, 167, 194, 198, 223, 254, 287, 322, 359, 398, 439, 482, 488, 527, 574, 623, 674, 702, 724, 727, 782, 839, 843, 898, 959, 970, 1022, 1087, 1154, 1223, 1294, 1298, 1367, 1442, 1519, 1598
Offset: 1

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Author

Keywords

Comments

From a problem in A269254. For detailed theory, see [Hone].

Crossrefs

A269251 a(n) = smallest prime in the sequence s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n - 1 (or a(n) = -1 if no such prime exists).

Original entry on oeis.org

-1, -1, 2, 3, 19, 5, 41, 7, 71, 89, 109, 11, 2003, 13, 3121, 239, 271, 17, 729962708557509701, 19, 419, 461, 11593, 23, 599, 11356201, 701, 11546481261621528160662473705515857458665002781273993, 811, 29, 929
Offset: 1

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Author

Arkadiusz Wesolowski, Jul 09 2016

Keywords

Comments

For n >= 2, smallest prime of the form (x^y + 1/x^y)/(x + 1/x), where x = (sqrt(n+2) +- sqrt(n-2))/2 and y is an odd positive integer, or -1 if no such prime exists.
If a(34) > 0 then a(34) > 10^1000. - Robert Israel, Feb 06 2018
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
Values of n where a(n) might need more than 1000 digits: 34, 52, 123, 254, 275, 285, 322, 371, 401, 413, 437, 460, 508, 518, 535, 540, 629, 643, 653, 691, 723, 724, 753, 797, 837, 843, 876, 881, 898, 913, 960, 970, 981, 986, 987, ... - Jean-François Alcover, Mar 01 2018

Crossrefs

Programs

  • Magma
    lst:=[]; for n in [1..31] do if n le 2 then Append(~lst, 0); else a:=1; c:=1; repeat b:=n*a-c; c:=a; a:=b; until IsPrime(a); Append(~lst, a); end if; end for; lst;
  • Maple
    f:= proc(n) local a,b,t;
    a:= 1; b:= n-1;
    do
      if isprime(b) then return b fi;
      t:= n*b-a;
      a:= b;
      b:= t;
    od
    end proc:
    f(1):= -1: f(2):= -1:
    map(f, [$1..33]); # Robert Israel, Feb 06 2018
  • Mathematica
    max = 10^1000; a[1] = a[2] = -1; a[n_] := Module[{s}, s[0] = 1; s[1] = n-1; s[k_] := s[k] = n s[k-1] - s[k-2]; For[k = 1, s[k] <= max, k++, If[PrimeQ[s[k]], Return[s[k]]]]] /. Null -> -1; Table[a[n], {n, 1, 33}] (* Jean-François Alcover, Mar 01 2018 *)

Formula

If n is prime then a(n+1) = n.

Extensions

Changed the value for the exceptional case from 0 to -1 for consistency with other sequences. - N. J. A. Sloane, Jan 19 2018

A269252 Define a sequence by s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n - 1. a(n) is the smallest index k such that s(k) is prime, or -1 if no such k exists.

Original entry on oeis.org

-1, -1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 3, 2, 2, 1, 14, 1, 2, 2, 3, 1, 2, 5, 2, 36, 2, 1, 2, 1, 15, -1, 6, 2, 3, 1, 2, 2, 6, 1, 3, 1, 2, 2, 2, 1, 2, 3, 2, -1, 3, 1, 2, 2, 2, 6, 3, 1, 2, 1, 30, 3, 2, 2, 2, 1, 2, 5, 2, 1, 5, 1, 6, 3, 2, 6, 3, 1, 8, 6, 14, 1, 3
Offset: 1

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Author

Arkadiusz Wesolowski, Jul 09 2016

Keywords

Comments

For n >= 2, positive integer k yielding the smallest prime of the form (x^y + 1/x^y)/(x + 1/x), where x = (sqrt(n+2) +/- sqrt(n-2))/2 and y = 2*k + 1, or -1 if no such k exists.
Every positive term belongs to A005097.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Examples

			Let b(k) be the recursive sequence defined by the initial conditions b(0) = 1, b(1) = 10, and the recursive equation b(k) = 11*b(k-1) - b(k-2). a(11) = 2 because b(2) = 109 is the smallest prime in b(k).
Let c(k) be the recursive sequence defined by the initial conditions c(0) = 1, c(1) = 12, and the recursive equation c(k) = 13*c(k-1) - c(k-2). a(13) = 3 because c(3) = 2003 is the smallest prime in c(k).
		

Crossrefs

Programs

  • Magma
    lst:=[]; for n in [1..85] do if n in [1, 2, 34, 52] then Append(~lst, -1); else a:=1; c:=1; t:=0; repeat b:=n*a-c; c:=a; a:=b; t+:=1; until IsPrime(a); Append(~lst, t); end if; end for; lst;
  • Mathematica
    s[k_, m_] := s[k, m] = Which[k == 0, 1, k == 1, 1 + m, True, m s[k - 1, m] - s[k - 2, m]]; Table[SelectFirst[Range[120], PrimeQ@ Abs@ s[#, -n] &] /. k_ /; MissingQ@ k -> -1, {n, 85}] (* Michael De Vlieger, Feb 03 2018 *)

Formula

If n is prime then a(n+1) = 1.

A331478 Irregular triangle T(n,k) = n - (s - k + 1)^2 for 1 <= k <= s, with s = floor(sqrt(n)).

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 2, 5, 3, 6, 4, 7, 0, 5, 8, 1, 6, 9, 2, 7, 10, 3, 8, 11, 4, 9, 12, 5, 10, 13, 6, 11, 14, 0, 7, 12, 15, 1, 8, 13, 16, 2, 9, 14, 17, 3, 10, 15, 18, 4, 11, 16, 19, 5, 12, 17, 20, 6, 13, 18, 21, 7, 14, 19, 22, 8, 15, 20, 23, 0, 9, 16, 21, 24, 1
Offset: 1

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Author

Michael De Vlieger, Jan 17 2020

Keywords

Comments

Row n begins with n - floor(sqrt(n)).
Zero appears in row n for n that are perfect squares. Let r = sqrt(n). For perfect square n, there exists a partition of n that consists of a run of r parts that are each r themselves; e.g., for n = 4, we have {2, 2}, for n = 9, we have {3, 3, 3}. It is clear through the Ferrers diagram of these partitions that they are equivalent to their Durfee square, thus n - s^2 = 0.
Since the partitions of any n contain Durfee squares in the range of 1 <= s <= floor(sqrt(n)) (with perfect square n also including k = 0), the distinct Durfee square excesses must be the differences n - s^2 for 1 <= s <= floor(sqrt(n)).
We borrow the term "square excess" from A053186(n), which is simply the difference n - floor(sqrt(n)).
Row n of this sequence contains distinct Durfee square excesses among all integer partitions of n (see example below).

Examples

			Table begins:
   1:  0;
   2:  1;
   3:  2;
   4:  0,  3;
   5:  1,  4;
   6:  2,  5;
   7:  3,  6;
   8:  4,  7;
   9:  0,  5,  8;
  10:  1,  6,  9;
  11:  2,  7, 10;
  12:  3,  8, 11;
  13:  4,  9, 12;
  14:  5, 10, 13;
  15:  6, 11, 14;
  16:  0,  7, 12, 15;
  ...
For n = 4, the partitions are {4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}. The partition {2, 2} has Durfee square s = 2; for all partitions except {2, 2}, we have Durfee square with s = 1. Therefore we have two unique solutions to n - s^2 for n = 4, i.e., {0, 3}, so row 4 contains these values.
		

Crossrefs

Programs

  • Mathematica
    Array[# - Reverse@ Range[Sqrt@ #]^2 &, 625] // Flatten

Formula

Let s = floor(sqrt(n));
T(n,1) = A053186(n) = n - s;
T(n,k) = T(n,1) + partial sums of 2(s - k + 1) + 1 for 2 <= k <= s + 1.
A000196(n) = Length of row n.
A022554(n) = Sum of row n.
Last term in row n = T(n, A000196(n)) = n - 1.
Previous Showing 11-15 of 15 results.