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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A157290 Decimal expansion of 105/Pi^4.

Original entry on oeis.org

1, 0, 7, 7, 9, 2, 8, 1, 3, 6, 7, 4, 1, 8, 5, 5, 1, 9, 4, 8, 6, 1, 0, 4, 2, 2, 4, 3, 0, 4, 7, 4, 6, 2, 8, 8, 4, 8, 9, 1, 9, 1, 9, 1, 9, 4, 6, 3, 2, 0, 1, 7, 5, 8, 5, 4, 0, 7, 6, 4, 3, 7, 2, 4, 5, 5, 7, 2, 3, 4, 5, 8, 0, 9, 3, 2, 9, 5, 1, 6, 2, 6, 1, 5, 2, 6, 0, 0, 1, 0, 2, 6, 0, 0, 5, 5, 0, 1, 5, 0, 9, 0, 8, 4, 8
Offset: 1

Views

Author

R. J. Mathar, Feb 26 2009

Keywords

Comments

The Product_{p = primes = A000040} (1+1/p^4), the quartic analog to A082020.

Examples

			1.077928136741855194... = (1+1/2^4)*(1+1/3^4)*(1+1/5^4)*(1+1/7^4)*...
		

References

  • Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 230.

Crossrefs

Programs

Formula

Equals A013662/A013666 = Product_{i>=1} (1+1/A030514(i)).
Equals Sum_{k>=1} 1/A005117(k)^4 = 1 + Sum_{k>=1} 1/A113849(k). - Amiram Eldar, May 22 2020
Equals 1/A347329. - Hugo Pfoertner, Jul 01 2024

A344705 a(n) = n + A001615(n) - sigma(n), where A001615 is the Dedekind psi-function, and sigma(n) gives the sum of divisors of n; difference between psi and the sum of proper divisors.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 5, 8, 10, 11, 8, 13, 14, 15, 9, 17, 15, 19, 14, 21, 22, 23, 12, 24, 26, 23, 20, 29, 30, 31, 17, 33, 34, 35, 17, 37, 38, 39, 22, 41, 42, 43, 32, 39, 46, 47, 20, 48, 47, 51, 38, 53, 42, 55, 32, 57, 58, 59, 36, 61, 62, 55, 33, 65, 66, 67, 50, 69, 70, 71, 21, 73, 74, 71, 56, 77, 78, 79, 38, 68, 82
Offset: 1

Views

Author

Antti Karttunen, May 28 2021

Keywords

Comments

First negative term occurs as a(1440) = -18.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(e+1) - 1)/(p-1); a[1] = 1; a[n_] := Module[{fct = FactorInteger[n]}, n * (Times @@ (1 + 1/fct[[;; , 1]]) + 1) - Times @@ f @@@ fct]; Array[a, 100] (* Amiram Eldar, Dec 08 2023 *)
  • PARI
    A001615(n) = (n * sumdivmult(n, d, issquarefree(d)/d));
    A344705(n) = ((n + A001615(n)) - sigma(n));

Formula

a(n) = A001615(n) - A001065(n) = n - A244963(n) = n + A001615(n) - sigma(n).
a(n) = A033879(n) + A306927(n).
a(n) = n + A344753(n) - 2*A001065(n).
Sum_{k=1..n} a(k) = c * n^2 / 2 + O(n*log(n)), where c = 15/Pi^2 + 1 - Pi^2/6 = 0.874883... . - Amiram Eldar, Dec 08 2023

A157291 Decimal expansion of Zeta(5)/Zeta(10).

Original entry on oeis.org

1, 0, 3, 5, 8, 9, 7, 4, 7, 7, 2, 7, 7, 5, 0, 0, 2, 2, 4, 3, 9, 4, 4, 9, 8, 5, 8, 7, 4, 5, 6, 0, 9, 5, 6, 8, 4, 2, 4, 7, 8, 8, 4, 2, 5, 6, 0, 7, 6, 8, 9, 4, 8, 0, 8, 2, 2, 4, 6, 6, 5, 4, 2, 3, 7, 4, 4, 6, 6, 9, 2, 5, 6, 1, 2, 4, 0, 3, 3, 7, 4, 1, 8, 9, 3, 2, 1, 5, 9, 8, 8, 3, 9, 3, 9, 0, 6, 8, 0, 1, 1, 4, 6, 3, 0
Offset: 1

Views

Author

R. J. Mathar, Feb 26 2009

Keywords

Comments

The product_{p = primes = A000040} (1+1/p^5), the fifth-power analog to A082020.

Examples

			1.035897477277500224... = (1+1/2^5)*(1+1/3^5)*(1+1/5^5)*(1+1/7^5)*...
		

Crossrefs

Programs

  • Maple
    evalf(Zeta(5)/Zeta(10)) ;
  • Mathematica
    RealDigits[Zeta[5]/Zeta[10],10,120][[1]] (* Harvey P. Dale, Apr 06 2013 *)

Formula

Equals A013663/A013668 = Product_{i>=1} (1+1/A050997(i)).
Equals Sum_{k>=1} 1/A005117(k)^5 = 1 + Sum_{k>=1} 1/A113850(k). - Amiram Eldar, May 22 2020
Equals 93555 * zeta(5) / Pi^10. - Vaclav Kotesovec, May 22 2020

A289320 a(n) = A289310(n)^2 + A289311(n)^2.

Original entry on oeis.org

1, 5, 10, 25, 26, 50, 50, 125, 100, 130, 122, 250, 170, 250, 260, 625, 290, 500, 362, 650, 500, 610, 530, 1250, 676, 850, 1000, 1250, 842, 1300, 962, 3125, 1220, 1450, 1300, 2500, 1370, 1810, 1700, 3250, 1682, 2500, 1850, 3050, 2600, 2650, 2210, 6250, 2500
Offset: 1

Views

Author

Rémy Sigrist, Jul 02 2017

Keywords

Comments

This sequence is totally multiplicative.
a(n) > n^2 for any n > 1.
If n is a square, then a(n) is a square.
If a(n) and a(m) are squares, then a(n*m) is a square.
a(n) is also a square for nonsquares n = 42, 168, 246, 287, 378, 672, 984, 1050, 1148, 1434, 1512, 1673, 2058, 2214, 2583, 2688, ...

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^2 + 1)^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Nov 13 2022 *)
  • PARI
    a(n) = my (f=factor(n)); return (prod(i=1, #f~, (1 + f[i,1]^2) ^ f[i,2]))
    
  • Python
    from sympy import factorint
    from operator import mul
    from functools import reduce
    def a(n): return 1 if n==1 else reduce(mul, [(1 + p**2)**k for p, k in factorint(n).items()])
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Aug 03 2017

Formula

Totally multiplicative, with a(p^k) = (1 + p^2)^k for any prime p and k > 0.
Sum_{k=1..n} a(k) ~ c * n^3, where c = 2/(Pi^2 * Product_{p prime} (1 - 1/p^2 - 1/p^3 - 1/p^4)) = 0.4778963213... . - Amiram Eldar, Nov 13 2022
Sum_{n>=1} 1/a(n) = 15/Pi^2 (A082020). - Amiram Eldar, Dec 15 2022

A375073 Numbers whose prime factorization exponents include at least one 2, at least one 3 and no other exponents.

Original entry on oeis.org

72, 108, 200, 392, 500, 675, 968, 1125, 1323, 1352, 1372, 1800, 2312, 2700, 2888, 3087, 3267, 3528, 4232, 4500, 4563, 5292, 5324, 5400, 6125, 6728, 7688, 7803, 8575, 8712, 8788, 9000, 9747, 9800, 10584, 10952, 11979, 12168, 12348, 13068, 13448, 13500, 14283, 14792
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2024

Keywords

Comments

Numbers k such that the set of distinct prime factorization exponents of k (row k of A136568) is {2, 3}.
Number k such that A051904(k) = 2 and A051903(k) = 3.

Crossrefs

Equals A338325 \ (A062503 UNION A062838).
Subsequence of A001694 and A046100.
A143610 is a subsequence.

Programs

  • Mathematica
    Select[Range[15000], Union[FactorInteger[#][[;; , 2]]] == {2, 3} &]
  • PARI
    is(k) = Set(factor(k)[,2]) == [2, 3];

Formula

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/p^2 + 1/p^3) - 15/Pi^2 - zeta(3)/zeta(6) + 1 = A330595 - A082020 - A157289 + 1 = 0.047550294197921818806... .

A082017 First row of square array T(n,k) with T(1,1) = 1 where antidiagonals are filled alternating upwards and downwards with the smallest number not already used such that the n-th antidiagonal sum is a multiple of n.

Original entry on oeis.org

1, 2, 7, 6, 18, 15, 28, 29, 49, 45, 75, 66, 94, 91, 130, 120, 155, 153, 201, 190, 232, 231, 288, 276, 325, 326, 391, 378, 462, 435, 508, 496, 589, 561, 641, 630, 732, 703, 790, 780, 891, 861, 955, 946, 1066, 1035, 1136, 1128, 1257, 1225, 1333, 1326, 1464, 1431
Offset: 1

Views

Author

Amarnath Murthy, Apr 05 2003

Keywords

Examples

			T(n,k) begins:
1,   2,  7,  6, 18, 15, ...
4,   5,  8, 14, 16, 27, ...
3,   9, 12, 17, 26, 31, ...
13, 11, 19, 25, 32, 42, ...
10, 20, 24, 33, 41, 50, ...
21, 23, 34, 39, 51, 60, ...
		

Crossrefs

Extensions

Edited and more terms from Alois P. Heinz, Oct 26 2011

A082018 First column of square array T(n,k) with T(1,1) = 1 where antidiagonals are filled alternating upwards and downwards with the smallest number not already used such that the n-th antidiagonal sum is a multiple of n.

Original entry on oeis.org

1, 4, 3, 13, 10, 21, 22, 40, 36, 64, 55, 81, 78, 115, 105, 138, 136, 182, 171, 211, 210, 265, 253, 300, 301, 364, 351, 433, 406, 477, 465, 556, 528, 606, 595, 695, 666, 751, 741, 850, 820, 912, 903, 1021, 990, 1089, 1081, 1208, 1176, 1282, 1275, 1411, 1378
Offset: 1

Views

Author

Amarnath Murthy, Apr 05 2003

Keywords

Comments

This is the boustrophedon method of filling an array. Sums of antidiagonals of T are in A074132. Sums of antidiagonals of T divided by number of antidiagonals are in A074133. Diagonal of T is in A082019.

Examples

			T(n,k) begins:
1,   2,  7,  6, 18, 15, ...
4,   5,  8, 14, 16, 27, ...
3,   9, 12, 17, 26, 31, ...
13, 11, 19, 25, 32, 42, ...
10, 20, 24, 33, 41, 50, ...
21, 23, 34, 39, 51, 60, ...
		

Crossrefs

Extensions

Edited and more terms from Alois P. Heinz, Oct 26 2011

A295657 Multiplicative with a(p^e) = p^floor((e-1)/2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Antti Karttunen, Nov 28 2017

Keywords

Crossrefs

Programs

  • Mathematica
    Array[Apply[Times, FactorInteger[#] /. {p_, e_} /; p > 0 :> p^Floor[(e - 1)/2]] &, 105] (* Michael De Vlieger, Nov 28 2017 *)
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^floor((f[i,2]-1)/2));} \\ Amiram Eldar, Nov 30 2022

Formula

a(1) = 1; for n > 1, a(n) = A020639(n)^A004526(A067029(n)-1) * a(A028234(n)).
a(n) = A000188(A003557(n)).
a(n) = 1 iff A212793(n) = 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 15/Pi^2 = 1.519817... (A082020). - Amiram Eldar, Nov 30 2022

A361810 a(n) is the sum of divisors of n that are both infinitary and exponential.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 10, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 30, 25, 26, 30, 28, 29, 30, 31, 34, 33, 34, 35, 36, 37, 38, 39, 50, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 60, 55, 70, 57, 58, 59, 60, 61, 62, 63, 68, 65, 66, 67, 68
Offset: 1

Views

Author

Amiram Eldar, Mar 25 2023

Keywords

Comments

The number of these divisors is A359411(n).
The indices of records of a(n)/n are the primorials (A002110) cubed, i.e., 1 and the terms of A115964.

Examples

			a(8) = 10 since 8 has 2 divisors that are both infinitary and exponential, 2 and 8, and 2 + 8 = 10.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := DivisorSum[e, p^# &, BitOr[#, e] == e &]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    s(p,e) = sumdiv(e, d, p^d*(bitor(d, e) == e));
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, s(f[i, 1], f[i, 2])); }

Formula

Multiplicative with a(p^e) = Sum_{d|e, bitor(d, e) == e} p^d.
a(n) >= n, with equality if and only if n is in A138302.
limsup_{n->oo} a(n)/n = Product_{p prime} (1 + 1/p^2) = 15/Pi^2 (A082020).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((1 - 1/p)*(1 + Sum_{e>=1} Sum_{d|e, bitor(d, e) == e} p^(d-2*e))) = 0.51015879911178031024... .

A378769 Intersection of A375055 and A376936.

Original entry on oeis.org

5400, 9000, 10584, 10800, 13500, 16200, 18000, 21168, 21600, 24696, 26136, 27000, 31752, 32400, 36000, 36504, 37044, 40500, 42336, 43200, 45000, 48600, 49000, 49392, 52272, 54000, 62424, 63504, 64800, 67500, 68600, 72000, 73008, 74088, 77976, 78408, 81000, 84672
Offset: 1

Views

Author

Michael De Vlieger, Dec 13 2024

Keywords

Comments

Let omega = A001221, bigomega = A001222, rad = A007947.
Powerful numbers k with bigomega(k) > omega(k) > 2 that are divisible by two distinct prime cubes p^3 and q^3.
Numbers k such that there exists (d, k/d), d | k, such that d neither divides nor is coprime to k/d and vice versa in the following 3 ways:
Type A: rad(d) does not divide d/k and rad(d/k) does not divide d
Type B: rad(d) divides d/k but rad(d/k) does not divide d
Type C: rad(d) | d/k and rad(d/k) | d, hence rad(d) = rad(d/k) = rad(k), a kind of coreful divisor pair.
Since (d, d/k) are noncoprime and do not divide one another, both must be composite, thus k is also composite.
In addition the following kinds of divisor pairs are also seen:
Type D: (d, k/d) such that d | k/d but there exists a factor Q | k/d that does not divide d. Then omega(d) < omega(k/d) = omega(k).
Type E: Nontrivial unitary divisor pairs (d, k/d) such that gcd(d, k/d) = 1, d > 1, k/d > 1. Let prime power factor p^m | k be such that m is maximized. Then set d = p^m and it is clear that for any k in A024619, there exists at least 1 nontrivial unitary divisor pair.
A378767 = { k : omega(k) > 1, p^3 | k for some prime p }, and
A376936 = { k : rad(k)^2 | k, p^3 | k and q^3 | k for distinct primes p, q }.
Therefore, we need only take intersection of A375055 and A376936.

Examples

			Table of the first 12 terms of this sequence, showing examples of types A, B, and C described in Comments.
   n     a(n)  Factors of a(n)    Type A      Type B      Type C
  ----------------------------------------------------------------
   1    5400   2^3 * 3^3 * 5^2    24 * 225    4 * 1350    60 * 90
   2    9000   2^3 * 3^2 * 5^3    18 * 500    4 * 2250    60 * 150
   3   10584   2^3 * 3^3 * 7^2    24 * 441    4 * 2646    84 * 126
   4   10800   2^4 * 3^3 * 5^2    48 * 225    8 * 1350    90 * 120
   5   13500   2^2 * 3^3 * 5^3    12 * 1125   9 * 1500    90 * 150
   6   16200   2^3 * 3^4 * 5^2    24 * 675    4 * 4050    60 * 270
   7   18000   2^4 * 3^2 * 5^3    18 * 1000   8 * 2250   120 * 150
   8   21168   2^4 * 3^3 * 7^2    48 * 441    8 * 2646   126 * 168
   9   21600   2^5 * 3^3 * 5^2    50 * 432    8 * 2700    90 * 240
  10   24696   2^3 * 3^2 * 7^3    18 * 1372   4 * 6174    84 * 294
  11   26136   2^3 * 3^3 * 11^2   24 * 1089   4 * 6534   132 * 198
  12   27000   2^3 * 3^3 * 5^3    24 * 1125   4 * 6750    60 * 450
		

Crossrefs

Programs

  • Mathematica
    s = Union@ Select[Flatten@ Table[a^2*b^3, {b, Surd[#, 3]}, {a, Sqrt[#/b^3]}] &[2^16],
      Length@ Select[FactorInteger[#][[All, -1]], # > 2 &] >= 2 &];
    Select[s, PrimeOmega[#] > PrimeNu[#] > 2 &]

Formula

Intersection of A375055, A376936, and A378767.
This sequence is { k : rad(k)^2 | k, bigomega(k) > omega(k) > 2, p^3 | k and q^3 | k for distinct primes p, q }.
Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - (15/Pi^2) * (1 + Sum_{prime} 1/((p-1)*(p^2+1))) - ((Sum_{p prime} (1/(p^2*(p-1))))^2 - Sum_{p prime} (1/(p^4*(p-1)^2)))/2 = 0.0025524144364532126894... . - Amiram Eldar, Dec 21 2024
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