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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A350900 Triangle read by rows: T(n, k) = Sum_{i=1..n} gcd(i,n) / gcd(gcd(i,k),n) for 1 <= k <= n.

Original entry on oeis.org

1, 3, 2, 5, 5, 3, 8, 5, 8, 4, 9, 9, 9, 9, 5, 15, 10, 9, 10, 15, 6, 13, 13, 13, 13, 13, 13, 7, 20, 12, 20, 9, 20, 12, 20, 8, 21, 21, 11, 21, 21, 11, 21, 21, 9, 27, 18, 27, 18, 15, 18, 27, 18, 27, 10, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 11, 40, 25, 24, 20, 40, 15, 40, 20, 24, 25, 40, 12
Offset: 1

Views

Author

Werner Schulte, Jan 21 2022

Keywords

Comments

Subtriangle (triangle without main diagonal) is symmetrical.
Conjecture: Let f be an arbitrary arithmetic function. Define for n > 0 the sequence a(f; n) = Sum_{i=1..n, k=1..n} f(gcd(i,n)/gcd(gcd(i,k),n)); a(f; n) equals Dirichlet convolution of f(n)*A000010(n) and A057660(n); if f is multiplicative, then a(f; n) is multiplicative; row sums of this triangle use f(n) = n (see formula section).

Examples

			The triangle T(n, k) for 1 <= k <= n starts:
n \k :   1   2   3   4   5   6   7   8   9  10  11  12
======================================================
   1 :   1
   2 :   3   2
   3 :   5   5   3
   4 :   8   5   8   4
   5 :   9   9   9   9   5
   6 :  15  10   9  10  15   6
   7 :  13  13  13  13  13  13   7
   8 :  20  12  20   9  20  12  20   8
   9 :  21  21  11  21  21  11  21  21   9
  10 :  27  18  27  18  15  18  27  18  27  10
  11 :  21  21  21  21  21  21  21  21  21  21  11
  12 :  40  25  24  20  40  15  40  20  24  25  40  12
  etc.
		

Crossrefs

Row sums gives A373059.

Programs

  • PARI
    T(n, k) = sum(i=1, n, gcd(i,n) / gcd(gcd(i,k),n));
    row(n) = vector(n, k, T(n,k)); \\ Michel Marcus, Jan 22 2022

Formula

T(n, 1) = A018804(n); T(n, n) = n.
T(n, k) = T(n, n-k) for 1 <= k < n.
Conjecture: Row sums equal Dirichlet convolution of A002618 and A057660.

A365076 Number of length-n binary words x such that the infinite word xxxx... is balanced.

Original entry on oeis.org

2, 4, 8, 12, 22, 22, 44, 44, 62, 64, 112, 78, 158, 130, 148, 172, 274, 184, 344, 232, 302, 334, 508, 302, 522, 472, 548, 474, 814, 442, 932, 684, 778, 820, 904, 672, 1334, 1030, 1100, 904, 1642, 904, 1808, 1222, 1282, 1522, 2164, 1198, 2102, 1564, 1912, 1728
Offset: 1

Views

Author

Jeffrey Shallit, Aug 20 2023

Keywords

Comments

A binary word w is "balanced" if for all lengths and all blocks b of the same length appearing in it, the number of 1's in b can take only two different values. For example, 00111 is not balanced because 00 has no 1's, 01 has one, and 11 has two.

Examples

			For n = 4, the 12 such words are 0000, 0001, 0010, 0100, 0101, 0111 and their bitwise binary complements.
		

Crossrefs

Programs

  • Python
    from math import gcd
    def A365076(n): return sum(n//gcd(n,k) for k in range(1,n+1))+1 # Chai Wah Wu, Aug 24 2023
    
  • Python
    from math import prod
    from sympy import factorint
    def A365076(n): return 1+prod((p**((e<<1)+1)+1)//(p+1) for p,e in factorint(n).items()) # Chai Wah Wu, Aug 05 2024

Formula

a(n) = 2*A057661(n).
a(n) = A057660(n) + 1.

A292302 Expansion of (1 - x)*Sum_{k>=1} k*phi(k)*x^k/(1 - x^k), where phi() is the Euler totient function (A000010).

Original entry on oeis.org

1, 2, 4, 4, 10, 0, 22, 0, 18, 2, 48, -34, 80, -28, 18, 24, 102, -90, 160, -112, 70, 32, 174, -206, 220, -50, 76, -74, 340, -372, 490, -248, 94, 42, 84, -232, 662, -304, 70, -196, 738, -738, 904, -586, 60, 240, 642, -966, 904, -538, 348, -184, 1030, -1116, 690, -482, 552, 38, 984, -1806
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 14 2017

Keywords

Comments

First differences of A057660.

Crossrefs

Programs

  • Mathematica
    nmax = 60; Rest[CoefficientList[Series[(1 - x) Sum[EulerPhi[k] k x^k /(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]]

Formula

G.f.: (1 - x)*Sum_{k>=1} k*phi(k)*x^k/(1 - x^k).

A307705 Expansion of Product_{k>=1} 1/(1 - x^k)^(k-phi(k)), where phi() is the Euler totient function (A000010).

Original entry on oeis.org

1, 0, 1, 1, 3, 2, 8, 5, 16, 15, 34, 30, 75, 66, 144, 150, 285, 292, 566, 585, 1062, 1170, 1988, 2205, 3729, 4159, 6755, 7785, 12214, 14147, 21957, 25560, 38709, 45839, 67884, 80747, 118332, 141244, 203614, 245330, 348396, 420971, 592439, 717659, 998248, 1215439, 1672544, 2040210, 2786687
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 22 2019

Keywords

Comments

Euler transform of A051953.

Crossrefs

Programs

  • Mathematica
    nmax = 48; CoefficientList[Series[Product[1/(1 - x^k)^(k - EulerPhi[k]), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 48; CoefficientList[Series[Exp[Sum[(DivisorSigma[2, k] - DivisorSigma[2, k^2]/DivisorSigma[1, k^2]) x^k/k, {k, 1, nmax}]], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d^2 - EulerPhi[d^2], {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 48}]

Formula

G.f.: exp(Sum_{k>=1} (sigma_2(k) - sigma_2(k^2)/sigma_1(k^2)) * x^k/k).
G.f.: exp(Sum_{k>=1} ( Sum_{d|k} cototient(d^2) ) * x^k/k).
a(n) ~ exp(3*((Pi^2 - 6)*Zeta(3))^(1/3) * n^(2/3) / (2*Pi)^(2/3) + 1/4) * ((Pi^2 - 6)*Zeta(3))^(1/4) / (A^3 * 2^(1/12) * 3^(1/2) * Pi^(5/6) * n^(3/4)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, May 06 2019

A317480 Numbers m such that Sum_{k=1..m} 1/gcd(k,m) is an integer.

Original entry on oeis.org

1, 614341, 618233, 1854699, 11746427, 26584019, 35239281, 79752057, 85393399, 118082503, 345592247, 354247509, 505096361, 802597537, 1036776741, 1062742527, 1515289083, 2149579159, 2243567557, 3695178641, 5077547629, 5250772527, 6566252693, 6730702671
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2018

Keywords

Comments

Also numbers n such that n | A057660(n).
a(1)-a(21) are taken from De Koninck's book.

References

  • Jean-Marie De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 265.

Crossrefs

Cf. A057660.

Programs

  • Mathematica
    fun[p_, e_]:=(p^(2e+1)+1)/(p+1); aQ[n_] := Divisible[Times @@ (fun @@@ FactorInteger[n]), n]; Select[Range[10000000], aQ]
  • PARI
    xi(n) = {f = factor(n); for (i=1, #f~, p = f[i, 1]; e = f[i, 2]; f[i, 1] = (p^(2*e+1)+1)/(p+1); f[i, 2] = 1; ); factorback(f); } for(n=1, 1e7, if(xi(n) % n == 0, print(n)))
Previous Showing 61-65 of 65 results.