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 1-10 of 67 results. Next

A370900 Partial sums of the powerfree part function (A055231).

Original entry on oeis.org

1, 3, 6, 7, 12, 18, 25, 26, 27, 37, 48, 51, 64, 78, 93, 94, 111, 113, 132, 137, 158, 180, 203, 206, 207, 233, 234, 241, 270, 300, 331, 332, 365, 399, 434, 435, 472, 510, 549, 554, 595, 637, 680, 691, 696, 742, 789, 792, 793, 795, 846, 859, 912, 914, 969, 976, 1033
Offset: 1

Views

Author

Amiram Eldar, Mar 05 2024

Keywords

References

  • Steven R. Finch, Mathematical Constants II, Cambridge University Press, 2018, p. 52.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, p, 1]; pfp[n_] := Times @@ f @@@ FactorInteger[n]; pfp[1] = 1; Accumulate[Array[pfp[#] &, 100]]
  • PARI
    pfp(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, f[i, 1], 1));}
    lista(kmax) = {my(s = 0); for(k = 1, kmax, s += pfp(k); print1(s, ", "))};

Formula

a(n) = Sum_{k=1..n} A055231(k).
a(n) = c * n^2 / 2 + O(R(n)), where c = Product_{p prime} (1 - (p^2+p-1)/(p^3*(p+1))) = 0.649606699337... (A191622), R(n) = x^(3/2) * exp(-c_1 * log(n)^(3/5) / log(log(n))^(1/5)) unconditionally, or x^(7/5) * exp(c_2 * log(n) / log(log(n))) assuming the Riemann hypothesis, and c_1 and c_2 are positive constants (Tóth, 2017).

A370901 Partial alternating sums of the powerfree part function (A055231).

Original entry on oeis.org

1, -1, 2, 1, 6, 0, 7, 6, 7, -3, 8, 5, 18, 4, 19, 18, 35, 33, 52, 47, 68, 46, 69, 66, 67, 41, 42, 35, 64, 34, 65, 64, 97, 63, 98, 97, 134, 96, 135, 130, 171, 129, 172, 161, 166, 120, 167, 164, 165, 163, 214, 201, 254, 252, 307, 300, 357, 299, 358, 343, 404, 342
Offset: 1

Views

Author

Amiram Eldar, Mar 05 2024

Keywords

Crossrefs

Similar sequences: A068762, A068773, A307704, A357817, A362028.

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, p, 1]; pfp[n_] := Times @@ f @@@ FactorInteger[n]; pfp[1] = 1; Accumulate[Array[(-1)^(# + 1) * pfp[#] &, 100]]
  • PARI
    pfp(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, f[i, 1], 1));}
    lista(kmax) = {my(s = 0); for(k = 1, kmax, s += (-1)^(k+1) * pfp(k); print1(s, ", "))};

Formula

a(n) = Sum_{k=1..n} (-1)^(k+1) * A055231(k).
a(n) = (5/38) * c * n^2 + O(R(n)), where c = Product_{p prime} (1 - (p^2+p-1)/(p^3*(p+1))) = 0.649606... (A191622), R(n) = x^(3/2) * exp(-c_1 * log(n)^(3/5) / log(log(n))^(1/5)) unconditionally, or x^(7/5) * exp(c_2 * log(n) / log(log(n))) assuming the Riemann hypothesis, and c_1 and c_2 are positive constants (Tóth, 2017).

A328014 Numbers whose powerful part (A057521) is larger than their powerfree part (A055231).

Original entry on oeis.org

4, 8, 9, 12, 16, 18, 24, 25, 27, 32, 36, 40, 45, 48, 49, 50, 54, 56, 63, 64, 72, 75, 80, 81, 96, 98, 100, 108, 112, 121, 125, 128, 135, 144, 147, 150, 160, 162, 169, 175, 176, 180, 189, 192, 196, 200, 208, 216, 224, 225, 240, 242, 243, 245, 250, 252, 256, 270
Offset: 1

Views

Author

Amiram Eldar, Oct 01 2019

Keywords

Comments

Differs from A122145(n) at n >= 25.
Cloutier et al. showed that the number of terms of this sequence below x is D0 * x^(3/4) + O(x^(2/3)*log(x)), where D0 is a constant given in A328015.

Examples

			12 is in the sequence since A057521(12) = 4 > A055231(12) = 3.
		

Crossrefs

Programs

  • Mathematica
    funp[p_, e_] := If[e > 1, p^e, 1]; pow[n_] := Times @@ (funp @@@ FactorInteger[n]); aQ[n_] := pow[n] > n/pow[n]; Select[Range[1000], aQ]
  • PARI
    pful(f) = prod(i=1, #f~, if(f[i, 2]>1, f[i, 1]^f[i, 2], 1)); \\ A057521
    pfree(f) = for (k=1, #f~, if (f[k, 2] > 1, f[k, 2] = 0); ); factorback(f); \\ A055231
    isok(n) = my(f=factor(n)); pful(f) > pfree(f); \\ Michel Marcus, Oct 02 2019

A368825 a(1) = 1; for n > 1, a(n) = A055231(a(n-1) + n), where A055231(k) is the powerfree part of k.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 7, 15, 3, 13, 3, 15, 7, 21, 1, 17, 34, 13, 1, 21, 42, 1, 3, 1, 26, 13, 5, 33, 62, 23, 2, 34, 67, 101, 17, 53, 10, 3, 42, 82, 123, 165, 13, 57, 102, 37, 21, 69, 118, 21, 1, 53, 106, 5, 15, 71, 1, 59, 118, 178, 239, 301, 91, 155, 55, 1, 17, 85, 154, 7, 78, 6, 79, 17, 23, 11
Offset: 1

Views

Author

Scott R. Shannon, Jan 07 2024

Keywords

Comments

The sequence is conjectured to contain all the squarefree numbers; see A005117.
The fixed points begin 1, 7, 5759, 35435, 58403, 62051, 182363, 261763, ... , although it is likely there are infinitely more.
See A368827 for the indices where a(n) = 1.

Examples

			a(9) = 3 as a(8)+9 = 15+9 = 24 = 2*2*2*3, and the powerfree part of 24 is 3.
		

Crossrefs

Cf. A368827, A368823 (multiplication), A055231, A005117, A124010.

A379579 Numerators of the partial sums of the reciprocals of the powerfree part function (A055231).

Original entry on oeis.org

1, 3, 11, 17, 91, 16, 117, 152, 187, 381, 4261, 13553, 178499, 90322, 30441, 35446, 607587, 1300259, 24875091, 25521737, 77027101, 38733998, 895731799, 932913944, 1044460379, 2097501253, 2320594123, 2352464533, 68444564327, 11443370128, 355822756173, 389249504528
Offset: 1

Views

Author

Amiram Eldar, Dec 26 2024

Keywords

Examples

			Fractions begin with 1, 3/2, 11/6, 17/6, 91/30, 16/5, 117/35, 152/35, 187/35, 381/70, 4261/770, 13553/2310, ...
		

References

  • D. Suryanarayana and P. Subrahmanyam, The maximal k-full divisor of an integer, Indian J. Pure Appl. Math., Vol. 12, No. 2 (1981), pp. 175-190.

Crossrefs

Cf. A055231, A328013, A370900, A370901, A379580 (denominators), A379581.

Programs

  • Mathematica
    f[p_, e_] := If[e==1, p, 1]; powfree[n_] := Times @@ f @@@ FactorInteger[n]; Numerator[Accumulate[Table[1/powfree[n], {n, 1, 50}]]]
  • PARI
    powfree(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] == 1, f[i, 1], 1)); }
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += 1 / powfree(k); print1(numerator(s), ", "))};

Formula

a(n) = numerator(Sum_{k=1..n} 1/A055231(k)).
a(n)/A379580(n) = A * n^(1/2) + B * n^(1/3) + O(n^(1/5)), where A = A328013, and B = (zeta(2/3)/zeta(2)) * Product_{p prime} (1 + (p^(1/3)-1)/(p*(p^(2/3)-p^(1/3)+1))) = -2.59305556147555965163... .

A379580 Denominators of the partial sums of the reciprocals of the powerfree part function (A055231).

Original entry on oeis.org

1, 2, 6, 6, 30, 5, 35, 35, 35, 70, 770, 2310, 30030, 15015, 5005, 5005, 85085, 170170, 3233230, 3233230, 9699690, 4849845, 111546435, 111546435, 111546435, 223092870, 223092870, 223092870, 6469693230, 1078282205, 33426748355, 33426748355, 9116385915, 18232771830
Offset: 1

Views

Author

Amiram Eldar, Dec 26 2024

Keywords

References

  • D. Suryanarayana and P. Subrahmanyam, The maximal k-full divisor of an integer, Indian J. Pure Appl. Math., Vol. 12, No. 2 (1981), pp. 175-190.

Crossrefs

Cf. A055231, A370900, A370901, A379579 (numerators), A379582.

Programs

  • Mathematica
    f[p_, e_] := If[e==1, p, 1]; powfree[n_] := Times @@ f @@@ FactorInteger[n]; Denominator[Accumulate[Table[1/powfree[n], {n, 1, 50}]]]
  • PARI
    powfree(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] == 1, f[i, 1], 1)); }
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += 1 / powfree(k); print1(denominator(s), ", "))};

Formula

a(n) = denominator(Sum_{k=1..n} 1/A055231(k)).

A379581 Numerators of the partial alternating sums of the reciprocals of the powerfree part function (A055231).

Original entry on oeis.org

1, 1, 5, -1, 1, -2, 1, -104, 1, -19, 1, -769, -7687, -4916, -261, -1262, -20453, -57923, -1066503, -5979161, -17475593, -8958244, -201189767, -79457304, -42275159, -87410483, -13046193, -23669663, -612055937, -1025912126, -28568429291, -128848674356, -125809879051
Offset: 1

Views

Author

Amiram Eldar, Dec 26 2024

Keywords

Examples

			Fractions begin with 1, 1/2, 5/6, -1/6, 1/30, -2/15, 1/105, -104/105, 1/105, -19/210, 1/2310, -769/2310, ...
		

Crossrefs

Cf. A055231, A328013, A370900, A370901, A379579, A379582 (denominators).

Programs

  • Mathematica
    f[p_, e_] := If[e==1, p, 1]; powfree[n_] := Times @@ f @@@ FactorInteger[n]; Numerator[Accumulate[Table[(-1)^(n+1)/powfree[n], {n, 1, 50}]]]
  • PARI
    powfree(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] == 1, f[i, 1], 1)); }
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += (-1)^(k+1) / powfree(k); print1(numerator(s), ", "))};

Formula

a(n) = numerator(Sum_{k=1..n} (-1)^(k+1)/A055231(k)).
a(n)/A379582(n) = A * n^(1/2) + B * n^(1/3) + O(n^(1/5)), where A = ((9-12*sqrt(2))/23) * A328013, and B = ((2^(5/3) - 3*2^(1/3) - 1)/(2^(5/3) - 2^(1/3) + 1)) * (zeta(2/3)/zeta(2)) * Product_{p prime} (1 + (p^(1/3)-1)/(p*(p^(2/3)-p^(1/3)+1))) = 1.42776088919948241359... .

A379582 Denominators of the partial alternating sums of the reciprocals of the powerfree part function (A055231).

Original entry on oeis.org

1, 2, 6, 6, 30, 15, 105, 105, 105, 210, 2310, 2310, 30030, 15015, 1001, 1001, 17017, 34034, 646646, 3233230, 9699690, 4849845, 111546435, 37182145, 37182145, 74364290, 74364290, 74364290, 2156564410, 3234846615, 100280245065, 100280245065, 100280245065, 200560490130
Offset: 1

Views

Author

Amiram Eldar, Dec 26 2024

Keywords

Crossrefs

Cf. A055231, A370900, A370901, A379580, A379581 (numerators).

Programs

  • Mathematica
    f[p_, e_] := If[e==1, p, 1]; powfree[n_] := Times @@ f @@@ FactorInteger[n]; Denominator[Accumulate[Table[(-1)^(n+1)/powfree[n], {n, 1, 50}]]]
  • PARI
    powfree(n) = {my(f = factor(n)); prod(i=1, #f~, if(f[i, 2] == 1, f[i, 1], 1)); }
    list(nmax) = {my(s = 0); for(k = 1, nmax, s += (-1)^(k+1) / powfree(k); print1(denominator(s), ", "))};

Formula

a(n) = denominator(Sum_{k=1..n} (-1)^(k+1)/A055231(k)).

A140394 Numbers n, satisfying A055231(n+1) - A055231(n) = 1, and with n and n+1 not squarefree.

Original entry on oeis.org

49, 1681, 18490, 23762, 39325, 57121, 182182, 453962, 656914, 843637, 1431125, 1608574, 1609674, 1940449, 2328482, 2948406, 3203050, 3721549, 5606230, 6352825, 8984002, 10000165, 13502254, 19326874, 19740249, 21006589, 26623750, 35558770, 38067925, 46297822
Offset: 1

Views

Author

Michel Lagneau, Dec 19 2011

Keywords

Comments

There exists an infinite number of numbers that are divisible by a square and satisfy A055231(n+1) - A055231(n) = 1 because the Fermat-Pell equation 2x^2 - y^2 = 1 admits an infinite number of solutions.

Examples

			49 is in the sequence because A055231(50) - A055231(49) = A055231(2*5^2) - A055231(7^2) = 2 - 1 = 1;
18490 is in the sequence because A055231(18491) - A055231(18490) = A055231(11*41^2) -A055231(2*5*43^2)  = 11 - 10 = 1.
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 49, p. 18, Ellipses, Paris 2008.

Crossrefs

Programs

  • Maple
    isA013929 := proc(n)
        n>3 and not numtheory[issqrfree](n) ;
    end proc:
    isA140394 := proc(n)
        isA013929(n) and isA013929(n+1) and (A055231(n+1) -A055231(n) = 1)  ;
    end proc:
    for n from 1 do
        if isA140394(n) then
            print(n);
        end if;
    end do: # R. J. Mathar, Dec 23 2011
  • Mathematica
    rad[n_] := Times @@ First /@ FactorInteger[n]; pow[n_] := Denominator[n / rad[n]^2]; aQ[n_] := !SquareFreeQ[n] && !SquareFreeQ[n + 1] && pow[n + 1] - pow[n] == 1; Select[Range[10^6], aQ] (* Amiram Eldar, Oct 01 2019 *)

Extensions

a(24)-a(30) from Amiram Eldar, Oct 01 2019

A368823 a(1) = 1; for n > 1, a(n) = A055231(a(n-1) * n), where A055231(k) is the powerfree part of k.

Original entry on oeis.org

1, 2, 6, 3, 15, 10, 70, 35, 35, 14, 154, 231, 3003, 858, 1430, 715, 12155, 24310, 461890, 46189, 969969, 176358, 4056234, 676039, 676039, 104006, 104006, 7429, 215441, 6463230, 200360130, 100180065, 367326905, 43214930, 60500902, 30250451, 1119266687, 117817546, 4594884294, 11487210735
Offset: 1

Views

Author

Scott R. Shannon, Jan 07 2024

Keywords

Comments

Terms are squarefree. - Michael De Vlieger, Jan 07 2024.

Examples

			a(4) = 3 as a(3)*4 = 6*4 = 24 = 2*2*2*3, and the powerfree part of 24 is 3.
		

Crossrefs

Cf. A368825 (addition), A055231, A005117, A124010.

Programs

  • Mathematica
    f[x_] := Times @@ Map[#1^(#2 Boole[#2 == 1]) & @@ # &, FactorInteger[x]]; a[n_] := f[n* a[n - 1]]; a[1] = 1; Array[a, 120] (* Michael De Vlieger, Jan 07 2024 *)
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