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 1-5 of 5 results.

A114962 a(n) = n^2 + 14.

Original entry on oeis.org

14, 15, 18, 23, 30, 39, 50, 63, 78, 95, 114, 135, 158, 183, 210, 239, 270, 303, 338, 375, 414, 455, 498, 543, 590, 639, 690, 743, 798, 855, 914, 975, 1038, 1103, 1170, 1239, 1310, 1383, 1458, 1535, 1614, 1695, 1778, 1863, 1950, 2039, 2130, 2223, 2318, 2415, 2514
Offset: 0

Views

Author

Cino Hilliard, Feb 21 2006

Keywords

Comments

Old name was: "Numbers of the form x^2 + 14".
x^2 + 14 != y^n for all x,y and n > 1.

Crossrefs

Cf. A155136, n^2 - 28; A000290, n^2; A114948, n^2 + 10.
Cf. sequences of the type n^2 + k: A002522 (k=1), A059100 (k=2), A117950 (k=3), A087475 (k=4), A117951 (k=5), A114949 (k=6), A117619 (k=7), A189833 (k=8), A189834 (k=9), A114948 (k=10), A189836 (k=11), A241748 (k=12), A241749 (k=13), this sequence (k=14), A241750 (k=15), A241751 (k=16), A241847 (k=17), A241848 (k=18), A241849 (k=19), A241850 (k=20), A241851 (k=21), A114963 (k=22), A241889 (k=23), A241890 (k=24), A114964 (k=30).

Programs

Formula

G.f.: (14-27*x+15*x^2)/(1-x)^3. - Colin Barker, Jan 11 2012
From Amiram Eldar, Nov 02 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(14)*Pi*coth(sqrt(14)*Pi))/28.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(14)*Pi*cosech(sqrt(14)*Pi))/28. (End)
From Elmo R. Oliveira, Nov 29 2024: (Start)
E.g.f.: exp(x)*(14 + x + x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

Extensions

Added 14 from Vincenzo Librandi, Apr 30 2014
Definition changed by Bruno Berselli, Mar 13 2015
Offset corrected by Amiram Eldar, Nov 02 2020

A293859 Prime factors of numbers of the form k^2 + 10.

Original entry on oeis.org

2, 5, 7, 11, 13, 19, 23, 37, 41, 47, 53, 59, 89, 103, 127, 131, 139, 157, 167, 173, 179, 197, 211, 223, 241, 251, 263, 277, 281, 293, 317, 331, 367, 373, 379, 383, 397, 401, 409, 419, 449, 463, 487, 491, 499, 503, 521, 557, 569, 571, 601, 607, 613, 619, 641
Offset: 1

Views

Author

J. Lowell, Oct 17 2017

Keywords

Comments

Primes p such that Legendre(-10,p) = 0 or 1. - N. J. A. Sloane, Dec 26 2017
Question: Is there a comment of the form "a prime number is in this sequence if and only if it is congruent to (list of appropriate values) mod n" for this sequence?
From Robert Israel, Nov 19 2017: (Start)
Prime p > 5 is in the sequence iff -10 is a quadratic residue mod p.
Thus p is either in the intersection of A002144 and A038879 or in neither of them.
Primes == 1, 2, 5, 7, 9, 11, 13, 19, 23, or 37 (mod 40). (End)

Examples

			7 is in the sequence because 2^2 + 10 = 14 is 2 times 7.
19 is in the sequence because 3^2 + 10 = 19.
		

Crossrefs

Programs

  • Maple
    select(isprime, [seq(seq(i*40+j, j = [1, 2, 5, 7, 9, 11, 13, 19, 23, 37]), i=0..40)]); # Robert Israel, Nov 19 2017
    # Load the Maple program HH given in A296920. Then run HH(-10, 200); This produces A155488, A296925, A293859. - N. J. A. Sloane, Dec 26 2017
  • Mathematica
    Select[Prime@ Range@ 120, {} != FindInstance[# x == n^2 + 10 && n >= 0 && x > 0, {n, x}, Integers, 1] &] (* Giovanni Resta, Oct 19 2017 *)

Extensions

More terms from Giovanni Resta, Oct 19 2017

A114965 a(n) = n^2 + 34.

Original entry on oeis.org

34, 35, 38, 43, 50, 59, 70, 83, 98, 115, 134, 155, 178, 203, 230, 259, 290, 323, 358, 395, 434, 475, 518, 563, 610, 659, 710, 763, 818, 875, 934, 995, 1058, 1123, 1190, 1259, 1330, 1403, 1478, 1555, 1634, 1715, 1798, 1883, 1970, 2059, 2150, 2243, 2338, 2435
Offset: 0

Views

Author

Cino Hilliard, Feb 21 2006

Keywords

Comments

Conjecture: n^2 + 34 != x^k for all n,x and k > 1.
The conjecture is true: See Cohn. - James Rayman, Feb 14 2023

Crossrefs

Programs

  • Mathematica
    34+Range[50]^2  (* Harvey P. Dale, Jan 28 2011 *)
  • PARI
    a(n)=n^2+34

Formula

From Elmo R. Oliveira, Jan 25 2025: (Start)
G.f.: (34 - 67*x + 35*x^2)/(1 - x)^3.
E.g.f.: (34 + x + x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

Extensions

Edited by Charles R Greathouse IV, Aug 09 2010
a(0) = 34 prepended by Elmo R. Oliveira, Jan 26 2025

A214870 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places counterclockwise. T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 9, 8, 13, 17, 16, 6, 14, 21, 26, 25, 11, 12, 22, 31, 37, 36, 18, 15, 20, 32, 43, 50, 49, 27, 24, 23, 30, 44, 57, 65, 64, 38, 35, 19, 33, 42, 58, 73, 82, 81, 51, 48, 28, 29, 45, 56, 74, 91, 101, 100, 66, 63, 39, 34, 41, 59, 72, 92, 111
Offset: 1

Views

Author

Boris Putievskiy, Mar 11 2013

Keywords

Comments

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1).
Enumeration table T(n,k) layer by layer. The order of the list:
T(1,1)=1;
T(1,2), T(2,1), T(2,2);
. . .
T(1,n), T(3,n), ... T(n,3), T(n,1); T(n,2), T(n,4), ... T(4,n), T(2,n);
. . .

Examples

			The start of the sequence as table:
   1   2   5  10  17  26 ...
   3   4   9  16  25  36 ...
   7   8   6  11  18  27 ...
  13  14  12  15  24  35 ...
  21  22  20  23  19  28 ...
  31  32  30  33  29  34 ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   5,  4,  7;
  10,  9,  8, 13;
  17, 16,  6, 14, 21;
  26, 25, 11, 12, 22, 31;
  ...
		

Crossrefs

Programs

  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    if i > j:
       result=i*i-i+(j%2)*(2-(j+1)/2)+((j+1)%2)*(j/2+1)
    else:
       result=j*j-2*(i%2)*j + (i%2)*((i+1)/2+1) + ((i+1)%2)*(-i/2+1)

Formula

As table
T(n,k) = k*k-2*(n mod 2)*k+(n mod 2)*((n+1)/2+1)+((n+1) mod 2)*(-n/2+1), if n<=k;
T(n,k) = n*n-n+(k mod 2)*(2-(k+1)/2)+((k+1) mod 2)*(k/2+1), if n>k.
As linear sequence
a(n) = j*j-2*(i mod 2)*j+(i mod 2)*((i+1)/2+1)+((i+1) mod 2)*(-i/2+1), if i<=j;
a(n) = i*i-i+(j mod 2)*(2-(j+1)/2)+((j+1) mod 2)*(j/2+1), if i>j; where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).

A214871 Natural numbers placed in table T(n,k) layer by layer. The order of placement - T(n,n), T(1,n), T(n,1), T(2,n), T(n,2),...T(n-1,n), T(n,n-1). Table T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 3, 4, 6, 2, 7, 11, 8, 9, 12, 18, 13, 5, 14, 19, 27, 20, 15, 16, 21, 28, 38, 29, 22, 10, 23, 30, 39, 51, 40, 31, 24, 25, 32, 41, 52, 66, 53, 42, 33, 17, 34, 43, 54, 67, 83, 68, 55, 44, 35, 36, 45, 56, 69, 84, 102, 85, 70, 57, 46, 26, 47, 58, 71, 86, 103, 123
Offset: 1

Views

Author

Boris Putievskiy, Mar 11 2013

Keywords

Comments

a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1).
Enumeration table T(n,k) layer by layer. The order of the list:
T(1,1)=1;
T(2,2), T(1,2), T(2,1);
. . .
T(n,n), T(1,n), T(n,1), T(2,n), T(n,2),...T(n-1,n), T(n,n-1);
. . .

Examples

			The start of the sequence as table:
  1....3...6..11..18..27...
  4....2...8..13..20..29...
  7....9...5..15..22..31...
  12..14..16..10..24..33...
  19..21..23..25..17..35...
  28..30..32..34..36..26...
  . . .
The start of the sequence as triangle array read by rows:
  1;
  3,4;
  6,2,7;
  11,8,9,12;
  18,13,5,14,19;
  27,20,15,16,21,28;
  . . .
		

Crossrefs

Cf. A060734, A060736, A185725, A213921, A213922; table T(n,k) contains: in rows A059100, A087475, A114949, A189833, A114948, A114962; in columns A117950, A117951, A117619, A189834, A189836; the main diagonal is A002522.

Programs

  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    if i == j:
       result=(i-1)**2+1
    if i > j:
       result=(i-1)**2+2*j+1
    if i < j:
       result=(j-1)**2+2*i

Formula

As table
T(n,k) = (n-1)^2+1, if n=k;
T(n,k) = (n-1)^2+2*k+1, if n>k;
T(n,k) = (k-1)^2+2*n, if n
As linear sequence
a(n) = (i-1)^2+1, if i=j;
a(n) = (i-1)^2+2*j+1, if i>j;
a(n) = (j-1)^2+2*i, if i>j; where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).
Showing 1-5 of 5 results.