A099777 -id:A099777 - cp's OEIS frontend

A320773 Numbers (excluding squares) whose square root has a continued fraction with a period < 3.

2, 3, 5, 6, 8, 10, 11, 12, 15, 17, 18, 20, 24, 26, 27, 30, 35, 37, 38, 39, 40, 42, 48, 50, 51, 56, 63, 65, 66, 68, 72, 80, 82, 83, 84, 87, 90, 99, 101, 102, 104, 105, 110, 120, 122, 123, 132, 143, 145, 146, 147, 148, 150, 152, 156, 168, 170, 171, 182, 195, 197, 198, 200

Offset: 1

Paul Weisenhorn, Oct 21 2018

nonn

The Heron sequence of every number a(n) has the following relationship: numerator(h(k))^2 - a(n)*denominator(h(k))^2 = 1 for k > 1.
The Heron sequence of every number a(n) has the following relationship with the continued fraction f(s) convergent to sqrt(a(n)): h(k) = f(2^k-1).
From Gerhard Kirchner, Jan 17 2020: (Start)
Numbers k = m^2 + r with m > 0 and 0 < r <= 2m such that r is a divisor of 2m.
Continued fraction: k = [m; 2m/r, 2m, 2m/r, 2m, ...].
The number of terms that are between m^2 and (m+1)^2  is equal to the number of divisors of 2m, which is A099777(m).
Proof see link. The Maxima code below demonstrates the divisor property. Note that there is no divisor of 2m between m and 2m.
(End)

			The continued fraction of sqrt(6) = 2 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + 1/(2 + 1/(4 + ...)))))) = [2; 2, 4, 2, 4, 2, 4, ...] has repeating portion (2, 4) with period 2, so 6 is a term.

Gerhard Kirchner, Divisor property of a(n)

Digits:=40: nr:=0:
for z from 2 to 200 do
  test:=true: c:=sqrt(z):
  if (c=floor(c)) the test:=false: end if:
  while (test=true) do
    b[0]:=floor(c):
    r[0]:=c:
    for k from 1 to 2 do
      r[k]:=evalf(1/(r[k-1]-b[k-1])):
      b[k]:=floor(r[k]):
    end do:
    if (b[1]=2*b[0]) or (b[2]=2*b[0]) then nr:=nr+1: a[nr]:=z: printf("%4d",z): end if:
    test:=false:
  end do:
end do:

Select[Range[200], !IntegerQ[Sqrt[#]] && Length@ContinuedFraction[Sqrt[#]][[-1]]<3 &] (* Amiram Eldar, Nov 01 2018 *)

block([n: 2, m: 0, r: 0, k: 0, kmax: 10,v: ""],
  while kGerhard Kirchner, Jan 17 2020 */

Edited by Jon E. Schoenfield, Oct 19 2019

A372714 a(n) = tau(3n-1) = A000005(3n-1).

Original entry on oeis.org

2, 2, 4, 2, 4, 2, 6, 2, 4, 2, 6, 4, 4, 2, 6, 2, 6, 2, 8, 2, 4, 4, 6, 2, 4, 4, 10, 2, 4, 2, 6, 4, 6, 2, 8, 2, 8, 2, 6, 4, 4, 4, 8, 2, 4, 2, 12, 4, 4, 2, 8, 4, 4, 4, 6, 2, 8, 2, 10, 2, 8, 4, 6, 2, 4, 2, 12, 4, 4, 4, 6, 4, 4, 4, 12, 2, 8, 2, 6, 2, 6, 6, 8, 2, 4, 2

Offset: 1

Vaclav Kotesovec, May 11 2024

nonn

Cf. A372713, A372715.

Cf. A000005, A099777, A099774.

Table[DivisorSigma[0, 3*n-1], {n, 1, 150}]

Sum_{k=1..n} a(k) ~ 2*n * (log(n) + 2*gamma - 1 + 2*log(3)) / 3, where gamma is the Euler-Mascheroni constant A001620.

A372715 a(n) = tau(3n-2) = A000005(3n-2).

Original entry on oeis.org

1, 3, 2, 4, 2, 5, 2, 4, 3, 6, 2, 4, 2, 8, 2, 4, 3, 6, 4, 4, 2, 7, 2, 8, 2, 6, 2, 4, 4, 8, 4, 4, 2, 9, 2, 4, 2, 10, 4, 4, 3, 6, 2, 8, 4, 8, 2, 4, 4, 6, 2, 8, 2, 12, 2, 4, 3, 6, 6, 4, 2, 8, 4, 8, 2, 9, 2, 4, 4, 10, 2, 4, 4, 12, 2, 4, 2, 8, 4, 8, 2, 6, 4, 8, 4, 9

Offset: 1

Vaclav Kotesovec, May 11 2024

nonn

Cf. A372713, A372714.

Cf. A000005, A099777, A099774.

Table[DivisorSigma[0, 3*n-2], {n, 1, 150}]

Sum_{k=1..n} a(k) ~ 2*n * (log(n) + 2*gamma - 1 + 2*log(3)) / 3, where gamma is the Euler-Mascheroni constant A001620.

A337566 a(n) is the number of possible decompositions of the polynomial n * (x + x^2 + ... + x^q), where q > 1, into a sum of k polynomials, not necessarily all different; each of these polynomials is to be of the form b_1 * x + b_2 * x^2 + ... + b_q * x^q where each b_i is one of the numbers 1, 2, 3, ..., q and no two b_i are equal.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 1, 2, 3, 3, 1, 5, 1, 3, 5, 3, 1, 6, 1, 5, 5, 3, 1, 7, 3, 3, 5, 5, 1, 9, 1, 4, 5, 3, 5, 9, 1, 3, 5, 7, 1, 9, 1, 5, 9, 3, 1, 9, 3, 6, 5, 5, 1, 9, 5, 7, 5, 3, 1, 13, 1, 3, 9, 5, 5, 9, 1, 5, 5, 9, 1, 12, 1, 3, 9, 5, 5, 9, 1, 9, 7, 3, 1, 13, 5, 3, 5, 7, 1, 15, 5, 5

Offset: 1

Bernard Schott, Sep 01 2020

nonn

Inspired by the 6th problem of the 13th British Mathematical Olympiad in 1977 (see the link BMO) where the problem asked to find for n = 26 all the values of q for which this decomposition is possible (see 2nd example).
As mentioned by Tony Gardiner in his book (see reference), "the wording" of this problem "is very strange". Letter n in Olympiad exercise becomes q in the Name.
If a solution is the sum of k polynomials of degree q, then, the relation between (n,k,q) is: k*(q+1) = 2*n with q > 1 (as in the problem) and q < n (because one proves there is no solution when q >= n); then, a(n) is the number of pairs (k,q) that are solutions of this last relation.

			For n = 3, the only solution, that corresponds to q = 2 and k = 2, is:
    3 * (x + x^2) = (x + 2x^2) + (2x + x^2).
For n = 26 as in the British Olympiad problem, a(26) = 3, and these three possible decompositions are:
   for k = 2, q = 25:
     26 * (x +   x^2 +   x^3 + ... +   x^24 +   x^25) =
          (x +  2x^2 +  3x^3 + ... + 24x^24 + 25x^25) +
        (25x + 24x^2 + 23x^3 + ... +  2x^24 +   x^25);
   for k = 4, q = 12:
     26 * (x +   x^2 +   x^3 + ... +   x^11 +   x^12) =
          (x +  2x^2 +  3x^3 + ... + 11x^11 + 12x^12) +
        (12x + 11x^2 + 10x^3 + ... +  2x^11 +   x^12) +
          (x +  2x^2 +  3x^3 + ... + 11x^11 + 12x^12) +
        (12x + 11x^2 + 10x^3 + ... +  2x^11 +   x^12);
   for  k = 13, q = 3:
      26 *  (x +  x^2 +  x^3) =
       4 *  (x + 2x^2 + 3x^3) +
       4 * (2x + 3x^2 +  x^3) +
       3 * (3x +  x^2 + 2x^3) +
           (2x +  x^2 + 3x^3) +
           (3x + 2x^2 +  x^3).

A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Problem 6 pp. 212-213 (1977).

Antti Karttunen, Table of n, a(n) for n = 1..20000
British Mathematical Olympiad 1977, Problem 6.
Index to sequences related to Olympiads.

Cf. A065091, A099777.

with(numtheory):
Data:= 0, seq(tau(2*n)-3, n=2..150);

MapAt[# + 1 &, Array[DivisorSigma[0, 2 #] - 3 &, 92], 1] (* Michael De Vlieger, Dec 12 2021 *)

a(n) = if (n==1, 0, numdiv(2*n)-3); \\ Michel Marcus, Sep 06 2020

a(1) = 0 then, for n >= 2, a(n) = tau(2n) - 3 = A099777(n) - 3.
a(n) = 1 iff n = 4 or n = p odd prime (A065091).
a(n) = p-3, p odd prime > 3 iff n = 2^(p-2).

A349693 Dirichlet convolution of the ruler function (A001511) with itself.

Original entry on oeis.org

1, 4, 2, 10, 2, 8, 2, 20, 3, 8, 2, 20, 2, 8, 4, 35, 2, 12, 2, 20, 4, 8, 2, 40, 3, 8, 4, 20, 2, 16, 2, 56, 4, 8, 4, 30, 2, 8, 4, 40, 2, 16, 2, 20, 6, 8, 2, 70, 3, 12, 4, 20, 2, 16, 4, 40, 4, 8, 2, 40, 2, 8, 6, 84, 4, 16, 2, 20, 4, 16, 2, 60, 2, 8, 6, 20, 4, 16, 2, 70

Offset: 1

Ilya Gutkovskiy, Nov 25 2021

Dirichlet convolution of A000005 with A104117. - Ridouane Oudra, Jul 23 2025

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A000217, A001511, A115364, A129628.

Cf. A000005, A104117, A007814, A085058, A090739, A099777, A372784, A001227, A000292,

a:= n-> (f-> add(f(d)*f(n/d), d=numtheory[divisors](n)))(k-> padic[ordp](2*k, 2)):
seq(a(n), n=1..80);  # Alois P. Heinz, Nov 25 2021

Table[Sum[IntegerExponent[2 d, 2] IntegerExponent[2 n/d, 2], {d, Divisors[n]}], {n, 1, 80}]
f[p_, e_] := If[p == 2, Binomial[e + 3, 3], e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 80] (* Amiram Eldar, Nov 25 2021 *)

A001511(n) = (1+valuation(n,2));
A349693(n) = sumdiv(n,d,A001511(n/d)*A001511(d)); \\ Antti Karttunen, Nov 25 2021

from sympy import divisor_count
def A349693(n): return divisor_count(n)*(m:=(n&-n).bit_length()+1)*(m+1)//6 # Chai Wah Wu, Jul 13 2022

Dirichlet g.f.: zeta(s)^2 * 4^s / (2^s-1)^2.
a(n) = Sum_{d|n} A001511(d) * A001511(n/d).
a(n) = Sum_{d|n} A000217(A001511(d)).
Multiplicative with a(p^e) = binomial(e+3,3) if p = 2 and e+1 otherwise. - Amiram Eldar, Nov 25 2021
Sum_{k=1..n} a(k) ~ 4*n*(log(n) - 1 + 2*gamma - 2*log(2)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Nov 26 2021
From Ridouane Oudra, Jul 23 2025: (Start)
a(n) = Sum_{i=0..A007814(n)} (i+1)*tau(n/2^i).
a(n) = Sum_{d|n} A115364(d).
a(n) = (1/6)*A090739(n)*A085058(n-1)*A000005(n).
a(n) = (1/6)*A001511(n)*A090739(n)*A099777(n).
a(n) = (1/3)*A115364(n)*A372784(n).
a(n) = A001227(n)*A000292(A001511(n)).
a(2*n+1) = tau(2*n+1).
a(2^k*(2*n+1)) = binomial(k+3, 3)*tau(2*n+1), for k, n >= 0. (End)

A355750 Sum of the divisors of 2n minus the number of divisors of 2n.

Original entry on oeis.org

1, 4, 8, 11, 14, 22, 20, 26, 33, 36, 32, 52, 38, 50, 64, 57, 50, 82, 56, 82, 88, 78, 68, 114, 87, 92, 112, 112, 86, 156, 92, 120, 136, 120, 136, 183, 110, 134, 160, 176, 122, 212, 128, 172, 222, 162, 140, 240, 165, 208, 208, 202, 158, 268, 208, 238, 232, 204, 176, 344, 182

Offset: 1

Wesley Ivan Hurt, Jul 15 2022

Consider the partitions of 2n into 2 parts (s,t), where s <= t. a(n) gives the sum of all the quotients t/s such that t/s is an integer. (See example.)

			a(7) = 20; the partitions of 2*7 = 14 into two parts (s,t) where s <= t are: (1,13), (2,12), (3,11), (4,10), (5,9), (6,8), and (7,7). The sum of the quotients t/s such that each t/s is an integer is then: 13/1 + 12/2 + 7/7 = 13 + 6 + 1 = 20.

Cf. A000005 (tau), A000203 (sigma), A062731, A099777.

Bisection of A065608.

Table[DivisorSigma[1, 2 n] - DivisorSigma[0, 2 n], {n, 80}]

a(n) = my(f=factor(2*n)); sigma(f) - numdiv(f); \\ Michel Marcus, Jul 16 2022

a(n) = sigma(2n) - tau(2n).
a(n) = Sum_{d|2n} (2n-d)/d.
a(n) = A065608(2n) = A000203(2n) - A000005(2n).
a(n) = A062731(n) - A099777(n).
a(n) = Sum_{k=1..n} m*c(m), where m=(2n-k)/k and c(m)=1-ceiling(m)+floor(m).

A334767 a(n) = Product_{k=1..n} d(2*k), where d() is the number of divisors function A000005.

Original entry on oeis.org

2, 6, 24, 96, 384, 2304, 9216, 46080, 276480, 1658880, 6635520, 53084160, 212336640, 1274019840, 10192158720, 61152952320, 244611809280, 2201506283520, 8806025134080, 70448201072640, 563585608581120, 3381513651486720, 13526054605946880

Offset: 1

Ctibor O. Zizka, May 10 2020

nonn

			a(4) = d(2)*d(4)*d(6)*d(8) = 2*3*4*4 = 96.

Cf. A000005, A066843, A099777.

Rest @ FoldList[Times, 1, DivisorSigma[0, Range[2, 40, 2]]] (* Amiram Eldar, May 10 2020 *)

a(n) = prod(k=1, n, numdiv(2*k)); \\ Michel Marcus, May 10 2020

A348554 Irregular triangle read by rows: row n gives the divisors d of 2n with 1 < d < 2n, for n >= 2.

Original entry on oeis.org

2, 2, 3, 2, 4, 2, 5, 2, 3, 4, 6, 2, 7, 2, 4, 8, 2, 3, 6, 9, 2, 4, 5, 10, 2, 11, 2, 3, 4, 6, 8, 12, 2, 13, 2, 4, 7, 14, 2, 3, 5, 6, 10, 15, 2, 4, 8, 16, 2, 17, 2, 3, 4, 6, 9, 12, 18, 2, 19, 2, 4, 5, 8, 10, 20, 2, 3, 6, 7, 14, 21, 2, 4, 11, 22, 2, 23, 2, 3, 4, 6, 8, 12, 16, 24, 2, 5, 10, 25

Offset: 2

Wolfdieter Lang, Oct 22 2021

This gives the rows 2*n of A137510, for n >= 2.
The length of row n is A069930(n) = tau(2*n) - 2 = A099777(n) - 2.
The sum of row n is A346880(n) = A062731(n) - (2*n + 1).

			The irregular triangle T(n, k) begins:
n, 2*n / k 1  2  3  4  5  6  7 ...
----------------------------------
2,   4:    2
3,   6:    2  3
4,   8:    2  4
5,  10:    2  5
6   12:    2  3  4 6
7,  14:    2  7
8,  16:    2  4  8
9,  18:    2  3  6  9
10, 20:    2  4  5 10
11, 22:    2 11
12, 24:    2  3  4  6  8 12
13, 26:    2 13
14, 28:    2  4  7 14
15, 30:    2  3  5  6 10 15
16, 32:    2  4  8  1
17, 34:    2 17
18, 36:    2  3  4  6  9 12 18
19, 38:    2 19
20, 40:    2  4  5  8 10 20
...

Cf. A069930, A099777, A137510, A346880.

Flatten@Table[Select[Divisors[2n],1<#<2n&],{n,2,25}] (* Giorgos Kalogeropoulos, Oct 22 2021 *)

row(n) = select(x->((x>1) && (x<2*n)), divisors(2*n)); \\ Michel Marcus, Oct 23 2021

T(n, k) = A137510(2*n, k), for n >= 2 and k = 1, 2, ..., A069930(n).

A350159 Number of subgroups of the dicyclic group Dic_n.

Original entry on oeis.org

3, 6, 8, 11, 10, 18, 12, 20, 19, 24, 16, 36, 18, 30, 32, 37, 22, 48, 24, 50, 40, 42, 28, 70, 37, 48, 48, 64, 34, 84, 36, 70, 56, 60, 56, 103, 42, 66, 64, 100, 46, 108, 48, 92, 90, 78, 52, 136, 63, 102, 80, 106, 58, 132, 80, 130, 88, 96, 64, 184, 66, 102, 116

Offset: 1

Firdous Ahmad Mala, Dec 17 2021

nonn

			a(2) = A000005(4) + A000203(2) = 3+3 = 6.
Given the fact that Dic_2 is isomorphic to the quaternion group Q_8, the subgroups of Dic_2 are isomorphic to the subgroups of Q_8 which are {1}, {1,-1}, {1,i,-1,-i}, {1,j,-1,-j}, {1,k,-1,-k} and Q_8.

Hayder Baqer Shelash and A. R. Ashrafi, The Number of Subgroups of a Given Type in Certain Finite Groups, Iranian Journal of Mathematical Sciences and Informatics, Vol. 16, No. 2 (2021), pp. 73-87.
Wikipedia, Dicyclic group

Cf. A000005, A000203, A099777, A345628.

a[n_] := DivisorSigma[0, 2*n] + DivisorSigma[1, n]; Array[a, 50] (* Amiram Eldar, Dec 17 2021 *)

a(n) = numdiv(2*n) + sigma(n); \\ Michel Marcus, Dec 18 2021

a(n) = A000005(2n) + A000203(n) = A099777(n) + A000203(n).

A372793 Sequence related to the asymptotic expansion of Sum_{k=1..n} tau(m*k).

Original entry on oeis.org

1, 2, 3, 16, 5, 864, 7, 4096, 729, 64000, 11, 6879707136, 13, 2809856, 61509375, 4294967296, 17, 812479653347328, 19, 26843545600000000, 26795786661, 2791309312, 23, 4019988717840603673710821376, 9765625, 73719087104, 7625597484987, 25962355635465062711296, 29

Offset: 1

Vaclav Kotesovec, May 13 2024

nonn

For m>=1, Sum_{k=1..n} tau(m*k) = A018804(m) * n * log(n) + O(n).

If p is prime, then Sum_{k=1..n} tau(p*k) ~ (2*p - 1) * n * (log(n) - 1 + 2*gamma)/p + n*log(p)/p, where gamma is the Euler-Mascheroni constant A001620.

			Sum_{k=1..n} tau(4*k) ~ (8*n*(log(n) + 2*gamma - 1) + n*4*log(2)) / 4, a(4) = exp(4*log(2)) = 16.
Sum_{k=1..n} tau(6*k) ~ (15*n*(log(n) + 2*gamma - 1) + n*(5*log(2) + 3*log(3))) / 6, a(6) = exp(5*log(2) + 3*log(3)) = 864.
Sum_{k=1..n} tau(8*k) ~ (20*n*(log(n) + 2*gamma - 1) + n*12*log(2)) / 8, a(8) = exp(12*log(2)) = 4096.
Sum_{k=1..n} tau(9*k) ~ (21*n*(log(n) + 2*gamma - 1) + n*6*log(3)) / 9, a(9) = exp(6*log(3)) = 729.
Sum_{k=1..n} tau(10*k) ~ (27*n*(log(n) + 2*gamma - 1) + n*(9*log(2) + 3*log(5))) / 10, a(10) = exp(9*log(2) + 3*log(5)) = 64000.
Sum_{k=1..n} tau(12*k) ~ (40*n*(log(n) + 2*gamma - 1) + n*(20*log(2) + 8*log(3))) / 12, a(12) = exp(20*log(2) + 8*log(3)) = 6879707136.

Cf. A000005 (m=1), A099777 (m=2), A372713 (m=3), A372784 (m=4), A372785 (m=5), A372786 (m=6), A372787 (m=7), A372788 (m=8), A372789 (m=9), A372790 (m=10), A372791 (m=11), A372792 (m=12).

Cf. A018804, A193679.

Sum_{k=1..n} tau(m*k) ~ A018804(m) * n * (log(n) - 1 + 2*gamma)/m + n*log(a(m))/m.
a(m) = exp(limit_{n->oo} (m * (Sum_{k=1..n} tau(m*k)) - A018804(m)*n*(log(n) - 1 + 2*gamma))/n).
If p is prime, then a(p) = p.
If p is prime, then a(p^k) = p^(k*p^(k-1)).
If p and q are distinct primes, then a(p*q) = p^(2*q-1) * q^(2*p-1).

cp's OEIS Frontend

A320773 Numbers (excluding squares) whose square root has a continued fraction with a period < 3.

Views

Author

Keywords

Comments

Examples

Links

Programs

Maple

Mathematica

Maxima

Extensions

A372714 a(n) = tau(3*n-1) = A000005(3*n-1).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Formula

A372715 a(n) = tau(3*n-2) = A000005(3*n-2).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Formula

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A349693 Dirichlet convolution of the ruler function (A001511) with itself.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A355750 Sum of the divisors of 2n minus the number of divisors of 2n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Formula

A334767 a(n) = Product_{k=1..n} d(2*k), where d() is the number of divisors function A000005.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

PARI

A348554 Irregular triangle read by rows: row n gives the divisors d of 2*n with 1 < d < 2*n, for n >= 2.

Views

Author

Keywords

Comments

A372714 a(n) = tau(3n-1) = A000005(3n-1).

A372715 a(n) = tau(3n-2) = A000005(3n-2).

A348554 Irregular triangle read by rows: row n gives the divisors d of 2n with 1 < d < 2n, for n >= 2.