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 1-3 of 3 results.

A017173 a(n) = 9*n + 1.

Original entry on oeis.org

1, 10, 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109, 118, 127, 136, 145, 154, 163, 172, 181, 190, 199, 208, 217, 226, 235, 244, 253, 262, 271, 280, 289, 298, 307, 316, 325, 334, 343, 352, 361, 370, 379, 388, 397, 406, 415, 424, 433, 442, 451, 460, 469, 478
Offset: 0

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Author

Keywords

Comments

Also all the numbers with digital root 1; A010888(a(n)) = 1. - Rick L. Shepherd, Jan 12 2009
A116371(a(n)) = A156144(a(n)); positions where records occur in A156144: A156145(n+1) = A156144(a(n)). - Reinhard Zumkeller, Feb 05 2009
If A=[A147296] 9*n^2+2*n (n>0, 11, 40, 87, ...); Y=[A010701] 3 (3, 3, 3, ...); X=[A017173] 9*n+1 (n>0, 10, 19, 28, ...), we have, for all terms, Pell's equation X^2 - A*Y^2 = 1. Example: 10^2 - 11*3^2 = 1; 19^2 - 40*3^2 = 1; 28^2 - 87*3^2 = 1. - Vincenzo Librandi, Aug 01 2010

Crossrefs

Cf. A093644 ((9,1) Pascal, column m=1).
Numbers with digital root m: this sequence (m=1), A017185 (m=2), A017197 (m=3), A017209 (m=4), A017221 (m=5), A017233 (m=6), A017245 (m=7), A017257 (m=8), A008591 (m=9).

Programs

Formula

G.f.: (1 + 8*x)/(1 - x)^2.
a(n) = 2*a(n-1) - a(n-2) with a(0)=1, a(1)=10. - Vincenzo Librandi, Aug 01 2010
E.g.f.: exp(x)*(1 + 9*x). - Stefano Spezia, Apr 20 2023
a(n) = A016777(3*n). - Elmo R. Oliveira, Apr 12 2025

A185019 a(n) = n*(14*n-3).

Original entry on oeis.org

0, 11, 50, 117, 212, 335, 486, 665, 872, 1107, 1370, 1661, 1980, 2327, 2702, 3105, 3536, 3995, 4482, 4997, 5540, 6111, 6710, 7337, 7992, 8675, 9386, 10125, 10892, 11687, 12510, 13361, 14240, 15147, 16082, 17045, 18036, 19055, 20102, 21177, 22280, 23411, 24570
Offset: 0

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Author

Bruno Berselli, Oct 14 2011 - based on remarks and sequences by Omar E. Pol

Keywords

Crossrefs

Cf. A195020 (vertices of the numerical spiral in link).

Programs

  • Magma
    [n*(14*n-3): n in [0..42]];
    
  • Magma
    I:=[0,11,50]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Mathematica
    CoefficientList[Series[x (11 + 17 x)/(1 - x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)
  • PARI
    for(n=0, 42, print1(n*(14*n-3)", "));
    

Formula

G.f.: x*(11+17*x)/(1-x)^3.
a(n) = A195025(-n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. - Wesley Ivan Hurt, Dec 18 2020
From Elmo R. Oliveira, Dec 30 2024: (Start)
E.g.f.: exp(x)*x*(11 + 14*x).
a(n) = n + A195023(n). (End)

A334116 a(n) is the least number k greater than n such that the square roots of both k and n have continuous fractions with the same period p and, if p > 1, the same periodic terms except for the last term.

Original entry on oeis.org

1, 5, 8, 4, 10, 12, 32, 15, 9, 17, 40, 20, 74, 33, 24, 16, 26, 39, 1880, 30, 112, 660, 96, 35, 25, 37, 104, 299, 338, 42, 77600, 75, 60, 78, 48, 36, 50, 84, 68, 87, 130, 56, 288968, 468, 350, 3242817, 192, 63, 49, 65, 200, 2726, 1042, 1628, 180, 72, 308, 425, 5880, 95
Offset: 1

Views

Author

Gerhard Kirchner, Apr 14 2020

Keywords

Comments

Note that a(n)=n if n is a square. The square root of a squarefree integer n has a continued fraction of the form [e(0);[e(1),...,e(p)]] with e(p)=2e(0) and e(i)=e(p-i) for 0 < i < p, see reference. The symmetric part [e(1),...,e(p-1)] of the continued fraction [m;[e(1),...,e(p-1), 2m]] will be called the pattern of n. 2 has the empty pattern (sqrt(2)=[1,[2]]), 3 has the pattern [1] (sqrt(3)=[1,[1,2]]) and so on. In this sense, the description of the sequence can be simplified as "Least number greater than n with the same pattern".
It can be can proved (see link) that integers with the same pattern are terms of a quadratic sequence.
An ambiguity has to be fixed: sqrt(2)=[1,[2]] = [1,[2,2]] = [1,[2,2,2]] and so on. We define that the shortest pattern is correct, here it is empty. Comment on the third subsequence (2),6,12,... below: The second term 6 has the pattern [2], but the first term 2 in brackets has the "wrong" pattern, after fixing the ambiguity.

Examples

			1) p=1: f(1)=2, f(2)=a(2)=5, f(3)=a(5)=10, f(4)=a(10)=17,..
sqrt(2)=[1,[2]], sqrt(5)=[2,[4]], sqrt(10)=[3,[6]], sqrt(17)=[4,[8]],..
2) p=2: f(1)=3, f(2)=a(3)=8, f(3)=a(8)=15, f(4)=a(15)=24,..
sqrt(3)=[1,[1,2]], sqrt(8)=[2,[1,4]], sqrt(15)=[3,[1,6]], sqrt(24)=[4,[1,8]],..
3) p=3: f(1)=41, f(2)=a(41)=130, f(3)=a(130)=269,..
sqrt(41)=[6,[2,2,12]], sqrt(130)=[11,[2,2,121]], sqrt(269)=[16,[2,2,256]],..
4) p=4: f(1)=33, f(2)=a(33)=60, f(3)=a(60)=95,..
sqrt(33)=[5,[1,2,1,10]], sqrt(60)=[7,[1,2,1,49]], sqrt(95)=[9,[1,2,1,81]],..
Several subsequences f(k) with f(k+1)=a(f(k)).
k>1 if first term in brackets, k>0 otherwise.
First terms  Period  Formula           Example
1) 2,5,10,17   1  A002522(k)=k^2+1           1
2) 3,8,15,24   2  A005563(k)=(k+1)^2-1       2
3)(2),6,12     2  A002378(k)=k*(k+1)
4) 7,32,75     4  A013656(k)=k*(9*k-2)
5) 11,40,87    2  A147296(k)=k*(9*k+2)
6) 13,74,185   5  A154357(k)=25*k^2-14*k+2
7) (3),14,33   4  A033991(k)=k*(4*k-1)       4
8) (5),18,39   2  A007742(k)=k*(4*k+1)
9) 21,112,275  6  A157265(k)=36*k^2-17*k+2
10)23,96,219   4  A154376(k)=25*k^2-2*k
11)27,104,231  2  A154377(k)=25*k^2+2*k
12)28,299,858  4  A156711(k)=144*k^2-161*k+45
13)29,338,985  5  A156640(k)=169*k^2+140*k+29
14)(8),34,78   4  A154516(k)=9*k^2-k
15)(10),38,84  2  A154517(k)=9*k^2+k
16)(2),41,130  3  A154355(k)=25*k^2-36*k+13  3
17)47,192,435  4  A157362(k)=49*k^2-2*k
		

References

  • Kenneth H. Rosen, Elementary number theory and its applications, Addison-Wesley, 3rd ed. 1993, page 428.

Crossrefs

Programs

  • Maxima
    block([nmax: 100],
    /*saves the first nmax terms in the current directory*/
    algebraic: true, local(coeff), showtime: true,
    fl: openw(sconcat("terms",nmax, ".txt")),
    coeff(w,m):=
      block(a: m, p: 0, s: w, vv:[],
       while a<2*m do
        (p: p+1, s: ratsimp(1/(s-floor(s))), a: floor(s),
         if a<2*m then vv: append(vv, [a])),
       j: floor((p-1)/2),
       if mod(p,2)=0 then v: [1,0,vv[j+1]] else v: [0,1,1],
       for i from j thru 1 step(-1) do
        (h: vv[i], u: [v[1]+h*v[3], v[3], 2*h*v[1]+v[2]+h^2*v[3]], v: u),
       return(v)),
       for n from 1 thru nmax do
        (w: sqrt(n), m: floor(w),
         if w=m then  b: n else
          (v: coeff(w,m),  x: v[1], y: v[2], z: v[3], q: mod(z,2),
           if q=0 then (z: z/2, y: y/2) else x: 2*x,
           fr: (x*m+y)/z, m: m+z, fr: fr+x, b: m^2+fr),
          printf( fl, "~d, ", b)),
          close(fl));
    
  • Python
    from sympy import floor, S, sqrt
    def coeff(w,m):
        a, p, s, vv = m, 0, w, []
        while a < 2*m:
            p += 1
            s = S.One/(s-floor(s))
            a = floor(s)
            if a < 2*m:
                vv.append(a)
        j = (p-1)//2
        v = [0,1,1] if p % 2 else [1, 0, vv[j]]
        for i in range(j-1,-1,-1):
            h = vv[i]
            v = [v[0]+h*v[2], v[2], 2*h*v[0]+v[1]+h**2*v[2]]
        return v
    def A334116(n):
        w = sqrt(n)
        m = floor(w)
        if w == m:
            return n
        else:
            x, y, z = coeff(w,m)
            if z % 2:
                x *= 2
            else:
                z //= 2
                y //= 2
            return (m+z)**2+x+(x*m+y)//z # Chai Wah Wu, Sep 30 2021, after Maxima code
Showing 1-3 of 3 results.