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.

Previous Showing 41-50 of 50 results.

A167332 Totally multiplicative sequence with a(p) = 2*(4*p-1) = 8*p-2 for prime p.

Original entry on oeis.org

1, 14, 22, 196, 38, 308, 54, 2744, 484, 532, 86, 4312, 102, 756, 836, 38416, 134, 6776, 150, 7448, 1188, 1204, 182, 60368, 1444, 1428, 10648, 10584, 230, 11704, 246, 537824, 1892, 1876, 2052, 94864, 294, 2100, 2244, 104272, 326, 16632, 342, 16856
Offset: 1

Views

Author

Jaroslav Krizek, Nov 01 2009

Keywords

Crossrefs

Programs

  • Maple
    f:=n -> mul((8*t[1]-2)^t[2],t=ifactors(n)[2]):
    map(f, [$1..100]); # Robert Israel, Jun 06 2016
  • Mathematica
    a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((4*fi[[All, 1]] - 1)^fi[[All, 2]])); Table[a[n]*2^PrimeOmega[n], {n, 1, 100}] (* G. C. Greubel, Jun 06 2016 *)
    f[p_, e_] := (8*p-2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 18 2023 *)
  • PARI
    a(n) = {my(f=factor(n)); for (k=1, #f~, f[k,1] = 8*f[k,1]-2;); factorback(f);} \\ Michel Marcus, Jun 06 2016

Formula

Multiplicative with a(p^e) = (2*(4*p-1))^e. If n = Product p(k)^e(k) then a(n) = Product (2*(4*p(k)-1))^e(k).
a(n) = A061142(n) * A166653(n) = 2^bigomega(n) * A166653(n) = 2^A001222(n) * A166653(n).

A167333 Totally multiplicative sequence with a(p) = 2*(5p-1) = 10p-2 for prime p.

Original entry on oeis.org

1, 18, 28, 324, 48, 504, 68, 5832, 784, 864, 108, 9072, 128, 1224, 1344, 104976, 168, 14112, 188, 15552, 1904, 1944, 228, 163296, 2304, 2304, 21952, 22032, 288, 24192, 308, 1889568, 3024, 3024, 3264, 254016, 368, 3384, 3584, 279936, 408, 34272, 428
Offset: 1

Views

Author

Jaroslav Krizek, Nov 01 2009

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((5*fi[[All, 1]] - 1)^fi[[All, 2]])); Table[a[n]*2^PrimeOmega[n], {n, 1, 100}] (* G. C. Greubel, Jun 06 2016 *)
    f[p_, e_] := (10*p-2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 18 2023 *)
  • PARI
    a(n) = {my(f=factor(n)); for (k=1, #f~, f[k,1] = 10*f[k,1]-2;); factorback(f);} \\ Michel Marcus, Jun 06 2016

Formula

Multiplicative with a(p^e) = (2*(5p-1))^e. If n = Product p(k)^e(k) then a(n) = Product (2*(5*p(k)-1))^e(k).
a(n) = A061142(n) * A166654(n) = 2^bigomega(n) * A166654(n) = 2^A001222(n) * A166654(n).

A167334 Totally multiplicative sequence with a(p) = 2*(2p+1) = 4p+2 for prime p.

Original entry on oeis.org

1, 10, 14, 100, 22, 140, 30, 1000, 196, 220, 46, 1400, 54, 300, 308, 10000, 70, 1960, 78, 2200, 420, 460, 94, 14000, 484, 540, 2744, 3000, 118, 3080, 126, 100000, 644, 700, 660, 19600, 150, 780, 756, 22000, 166, 4200, 174, 4600, 4312, 940, 190, 140000, 900
Offset: 1

Views

Author

Jaroslav Krizek, Nov 01 2009

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((2*fi[[All, 1]] + 1)^fi[[All, 2]])); Table[a[n]*2^PrimeOmega[n], {n, 1, 100}] (* G. C. Greubel, Jun 06 2016 *)
    f[p_, e_] := (4*p+2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 19 2023 *)

Formula

Multiplicative with a(p^e) = (2*(2p+1))^e. If n = Product p(k)^e(k) then a(n) = Product (2*(2*p(k)+1))^e(k).
a(n) = A061142(n) * A166660(n) = 2^bigomega(n) * A166660(n) = 2^A001222(n) * A166660(n).

A167335 Totally multiplicative sequence with a(p) = 2*(3p+1) = 6p+2 for prime p.

Original entry on oeis.org

1, 14, 20, 196, 32, 280, 44, 2744, 400, 448, 68, 3920, 80, 616, 640, 38416, 104, 5600, 116, 6272, 880, 952, 140, 54880, 1024, 1120, 8000, 8624, 176, 8960, 188, 537824, 1360, 1456, 1408, 78400, 224, 1624, 1600, 87808, 248, 12320, 260, 13328, 12800
Offset: 1

Views

Author

Jaroslav Krizek, Nov 01 2009

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((3*fi[[All, 1]] + 1)^fi[[All, 2]])); Table[a[n]*2^PrimeOmega[n], {n, 1, 100}] (* G. C. Greubel, Jun 06 2016 *)
    f[p_, e_] := (6*p+2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 19 2023 *)

Formula

Multiplicative with a(p^e) = (2*(3p+1))^e. If n = Product p(k)^e(k) then a(n) = Product (2*(3*p(k)+1))^e(k).
a(n) = A061142(n) * A166661(n) = 2^bigomega(n) * A166661(n) = 2^A001222(n) * A166661(n).

A167336 Totally multiplicative sequence with a(p) = 2*(4p+1) = 8p+2 for prime p.

Original entry on oeis.org

1, 18, 26, 324, 42, 468, 58, 5832, 676, 756, 90, 8424, 106, 1044, 1092, 104976, 138, 12168, 154, 13608, 1508, 1620, 186, 151632, 1764, 1908, 17576, 18792, 234, 19656, 250, 1889568, 2340, 2484, 2436, 219024, 298, 2772, 2756, 244944, 330, 27144, 346
Offset: 1

Views

Author

Jaroslav Krizek, Nov 01 2009

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((4*fi[[All, 1]] + 1)^fi[[All, 2]])); Table[a[n]*2^PrimeOmega[n], {n, 1, 100}] (* G. C. Greubel, Jun 06 2016 *)
    f[p_, e_] := (8*p+2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 19 2023 *)

Formula

Multiplicative with a(p^e) = (2*(4p+1))^e. If n = Product p(k)^e(k) then a(n) = Product (2*(4*p(k)+1))^e(k).
a(n) = A061142(n) * A166662(n) = 2^bigomega(n) * A166662(n) = 2^A001222(n) * A166662(n).

A167337 Totally multiplicative sequence with a(p) = 2*(5p+1) = 10p+2 for prime p.

Original entry on oeis.org

1, 22, 32, 484, 52, 704, 72, 10648, 1024, 1144, 112, 15488, 132, 1584, 1664, 234256, 172, 22528, 192, 25168, 2304, 2464, 232, 340736, 2704, 2904, 32768, 34848, 292, 36608, 312, 5153632, 3584, 3784, 3744, 495616, 372, 4224, 4224, 553696, 412, 50688
Offset: 1

Views

Author

Jaroslav Krizek, Nov 01 2009

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((5*fi[[All, 1]] + 1)^fi[[All, 2]])); Table[a[n]*2^PrimeOmega[n], {n, 1, 100}] (* G. C. Greubel, Jun 06 2016 *)
    f[p_, e_] := (10*p+2)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 19 2023 *)

Formula

Multiplicative with a(p^e) = (2*(5p+1))^e. If n = Product p(k)^e(k) then a(n) = Product (2*(5*p(k)+1))^e(k).
a(n) = A061142(n) * A166663(n) = 2^bigomega(n) * A166663(n) = 2^A001222(n) * A166663(n).

A328854 Dirichlet g.f.: Product_{p prime} 1 / (1 - 2 * p^(-s))^2.

Original entry on oeis.org

1, 4, 4, 12, 4, 16, 4, 32, 12, 16, 4, 48, 4, 16, 16, 80, 4, 48, 4, 48, 16, 16, 4, 128, 12, 16, 32, 48, 4, 64, 4, 192, 16, 16, 16, 144, 4, 16, 16, 128, 4, 64, 4, 48, 48, 16, 4, 320, 12, 48, 16, 48, 4, 128, 16, 128, 16, 16, 4, 192, 4, 16, 48, 448, 16, 64, 4, 48, 16, 64, 4, 384, 4, 16, 48
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 28 2019

Keywords

Comments

Dirichlet convolution of A061142 with itself.

Crossrefs

Programs

  • Mathematica
    Table[2^PrimeOmega[n] DivisorSigma[0, n], {n, 1, 75}]
    f[p_, e_] := (e + 1)*2^e; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 02 2020 *)
  • PARI
    a(n) = numdiv(n)*2^bigomega(n); \\ Michel Marcus, Dec 02 2020
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1 - 2*X)^2)[n], ", ")) \\ Vaclav Kotesovec, Aug 22 2021

Formula

If n = Product (p_j^k_j) then a(n) = Product (2^k_j * (k_j + 1)).
a(n) = 2^bigomega(n) * tau(n), where bigomega = A001222 and tau = A000005.

A353172 a(n) is the least k > 1 such that Omega(n) = Omega(n mod k), where Omega = A001222.

Original entry on oeis.org

2, 3, 4, 5, 3, 7, 4, 9, 5, 6, 3, 13, 5, 5, 9, 17, 3, 10, 4, 12, 11, 6, 3, 25, 7, 10, 15, 10, 3, 11, 4, 33, 9, 5, 13, 20, 5, 8, 5, 24, 3, 15, 4, 9, 25, 6, 3, 49, 5, 14, 9, 11, 3, 19, 7, 20, 12, 6, 3, 22, 7, 7, 11, 65, 11, 18, 4, 10, 5, 25, 3, 40, 5, 5, 19, 16
Offset: 1

Views

Author

Thomas Scheuerle, Apr 28 2022

Keywords

Comments

It appears that a(m) = m*k/p if m = p*2^n ... . Are these formulas related to some well-known sequence of rational numbers?

Examples

			a(10) = 6 because 10 = 5*2 and 10 mod 6 = 4 = 2*2.
		

Crossrefs

Programs

  • PARI
    a(n) = my(k=2); while(bigomega(n) != bigomega(max(n%k,1)), k++); k
    
  • Python
    from itertools import count
    from sympy.ntheory.factor_ import primeomega
    def A353172(n):
        a = primeomega(n)
        for k in count(2):
            if (m := n % k) > 0 and primeomega(m) == a:
                return k # Chai Wah Wu, Jun 20 2022

Formula

a(A029744(n)) = A029744(n) + 1.
a(A003627(n)) = 3.
a(A000040(n)) = A095925(n).
a(A077065(n)) = 6. For n > 2.
If a(n) = 10, then n mod 10 is in most cases 8 and seldom 6.
a(m) = m*3/5 if m = 5*2^n or m = 15. This formula is valid for all positive n because (5*2^n) mod (5*2^n)*(3/5) = 2^(n+1). If the sequence of solutions does not create powers of two in the modulo operation, it will be of finite length. See next two formulas:
a(m) = m*3/11 if m = 11, 22, 33 or 66.
a(m) = m*4/43 if m = 43*2^n for n < 4. This series of solutions terminates because of the next formula which replaces the powers of two:
a(m) = m*41/(43*2^4) if m = 43*2^4*2^n. This formula is valid for all positive n.
a(m) = m*5/9 if m = 9*2^n or m = 27 or 45. This formula is valid for all positive n.
For each k = a(p) if k < p and gcd(k, p) = 1 such a formula, of the form a(m) = m*k/p, if m = p*2^n ..., can be developed.

A375286 a(n) = f(1) + f(2) + ... + f(n), where f(n) = (-2)^Omega(n) = A165872(n).

Original entry on oeis.org

1, -1, -3, 1, -1, 3, 1, -7, -3, 1, -1, -9, -11, -7, -3, 13, 11, 3, 1, -7, -3, 1, -1, 15, 19, 23, 15, 7, 5, -3, -5, -37, -33, -29, -25, -9, -11, -7, -3, 13, 11, 3, 1, -7, -15, -11, -13, -45, -41, -49, -45, -53, -55, -39, -35, -19, -15, -11, -13, 3, 1, 5, -3, 61
Offset: 1

Views

Author

Keywords

Crossrefs

Partial sums of A165872.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<1, 0,
          a(n-1)+(-2)^numtheory[bigomega](n))
        end:
    seq(a(n), n=1..64);  # Alois P. Heinz, Apr 25 2025
  • PARI
    s=0; vector(60,n,s+=(-2)^bigomega(n))

Formula

Johnston, Leong, & Tudzi prove that |a(n)| < 2260n. Sun conjectures that |a(n)| < n for n >= 3078. Mossinghoff & Trudgian verify this to 2.5 * 10^14.
Because of powers of two, |a(n)| >= n/2 infinitely often.

A354273 Square array read by ascending antidiagonals: A(n,k) = k^Omega(n).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 4, 3, 4, 1, 1, 2, 9, 4, 5, 1, 1, 4, 3, 16, 5, 6, 1, 1, 2, 9, 4, 25, 6, 7, 1, 1, 8, 3, 16, 5, 36, 7, 8, 1, 1, 4, 27, 4, 25, 6, 49, 8, 9, 1, 1, 4, 9, 64, 5, 36, 7, 64, 9, 10, 1, 1, 2, 9, 16, 125, 6, 49, 8, 81, 10, 11, 1, 1, 8, 3, 16, 25, 216, 7, 64, 9, 100, 11, 12, 1
Offset: 1

Views

Author

Stefano Spezia, May 22 2022

Keywords

Examples

			Array begins:
    1, 1,  1,  1,   1,   1,   1,   1, ...
    1, 2,  3,  4,   5,   6,   7,   8, ...
    1, 2,  3,  4,   5,   6,   7,   8, ...
    1, 4,  9, 16,  25,  36,  49,  64, ...
    1, 2,  3,  4,   5,   6,   7,   8, ...
    1, 4,  9, 16,  25,  36,  49,  64, ...
    1, 2,  3,  4,   5,   6,   7,   8, ...
    1, 8, 27, 64, 125, 216, 343, 512, ...
    ...
		

Crossrefs

Cf. A000012 (n = 1 or k = 1), A061142 (k = 2), A165824 - A165871 (k = 3..50), A176029 (diagonal).

Programs

  • Mathematica
    A[n_,k_]:=k^PrimeOmega[n]; Flatten[Table[A[n-k+1,k],{n,13},{k,n}]]

Formula

A(n, k) = A051129(A001222(n), k).
The columns are totally multiplicative: A(i*j, k) = A(i, k)*A(j, k).
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