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 1-7 of 7 results.

A353960 a(1)=1. Thereafter, if a(n) has occurred k (>1) times in a(j), 1 <= j <= n then a(n+1) = k*a(n). If a(n) is a first occurrence, a(n+1) = A078709(a(n)).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 1, 5, 2, 4, 8, 2, 6, 1, 6, 12, 2, 8, 16, 3, 6, 18, 3, 9, 3, 12, 24, 3, 15, 3, 18, 36, 4, 12, 36, 72, 6, 24, 48, 4, 16, 32, 5, 10, 2, 10, 20, 3, 21, 5, 15, 30, 3, 24, 72, 144, 9, 18, 54, 6, 30, 60, 5, 20, 40, 5, 25, 8, 24, 96, 8, 32, 64, 9, 27, 6, 36, 108, 9
Offset: 1

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Author

David James Sycamore, May 12 2022

Keywords

Comments

A nonprime term arises either following a multiple (>1) of a prior term, or following a novel term m such that A078709(m) is a nonprime number. A prime term arises either as a multiple of 1 (first occurrences of 2,3,5), or following a novel term m such that A078709(m) is a prime. Let C(m) be the cardinality of m in the sequence (starting 6,5,10,5,8,8,6,13,8,6,8,9,...).
Conjecture: C(m) = A125057(m) + k, where k is the number of divisors d of m such that 1 < d <= C(m/d). The cardinality of m due to occurrences of novel terms is A125057(m), whereas the cardinality due to repeat terms is counted by k, and all of the latter terms occur prior to any of the former. (Note comments in A125057, A126888, A125056, concerning definitions of those sequences.)

Examples

			a(8)=1, the 5th occurrence of 1, so the next term a(9)=5.
a(43)=32, a first occurrence, so a(44)=A078709(32)=5.
a(33)=36 (consequence of two occurrences of 18), and 36 has not occurred before, so a(34)=A078709(36)=4.
C(12)=9 because A125057(12)=6, and there are 3 occurrences of 12 which are consequent to repeat terms (2 terms 6, 4 terms 3, and 3 terms 4).
Alternatively, 12 has 5 divisors >1, of which 3 (d=2,3,4) satisfy d <= C(12/d), therefore k=3 and C(12) = A125057(12) + 3 = 6 + 3 = 9.
		

Crossrefs

Programs

Formula

For p prime = 2,3,5, C(p) = A125057(p) + 1; for p > 5, C(p) = A125057(p).

Extensions

More terms from Michel Marcus, May 16 2022

A126888 a(n) is the smallest positive integer such that floor(a(n)/d(a(n))) = n, or -1 if no such number exists, where d(m) is the number of positive divisors of m.

Original entry on oeis.org

1, 5, 7, 28, 11, 13, 44, 17, 19, 63, 23, 51, 55, 29, 31, 49, 69, 37, 77, 41, 43, 91, 47, 147, 153, 53, 111, 115, 59, 61, 125, 129, 67, 207, 71, 73, 296, 155, 79, 121, 83, 680, 261, 89, 183, 185, 284, 97, 399, 101, 103, 209, 107, 109, 221, 113, 459, 235, 237, 363, 247, 249
Offset: 1

Views

Author

Leroy Quet, Dec 30 2006

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{k = 1},While[Floor[k/Length[Divisors[k]]] != n, k++ ];k];Table[f[n], {n, 62}] (* Ray Chandler, Jan 04 2007 *)

Extensions

Extended by Ray Chandler, Jan 04 2007.
Added escape clause to definition at the suggestion of Hugo van der Sanden. - N. J. A. Sloane, Jul 01 2022

A125056 a(n) is the largest positive integer such that floor(a(n)/d(a(n))) = n, where d(m) is the number of positive divisors of m.

Original entry on oeis.org

6, 12, 30, 48, 60, 72, 120, 96, 144, 180, 140, 240, 216, 252, 360, 336, 420, 224, 312, 480, 504, 540, 378, 720, 600, 840, 660, 672, 352, 364, 756, 780, 1080, 960, 1260, 864, 594, 924, 936, 1440, 1320, 1680, 1050, 1056, 1092, 1120, 1512, 1560
Offset: 1

Views

Author

Hugo van der Sanden, Jan 09 2007

Keywords

Comments

We know the sequence is well-defined given the limit x/d(x) > 0.5*sqrt(x) from comments in A036763.
Does every positive integer n equal floor(m/d(m)) for some m?

Crossrefs

Programs

  • Mathematica
    t = Table[ Floor[ n / DivisorSigma[0, n]], {n, 10^5}]; f[n_] := Max@ Flatten@ Position[t, n]; Array[f, 51] (* Robert G. Wilson v, Jan 12 2007 *)

A125057 a(n) is the number of positive integers m such that floor(m/d(m)) = n, where d(m) is the number of positive divisors of m.

Original entry on oeis.org

5, 4, 9, 3, 7, 5, 6, 11, 7, 4, 8, 6, 9, 5, 4, 16, 7, 4, 8, 7, 11, 5, 10, 7, 7, 8, 7, 12, 9, 6, 10, 8, 8, 8, 10, 6, 4, 7, 7, 15, 8, 4, 11, 11, 8, 12, 7, 11, 7, 9, 8, 8, 12, 14, 8, 12, 8, 8, 11, 5, 14, 7, 7, 9, 5, 8, 4, 13, 7, 8, 12, 10, 6, 9, 14, 11, 9, 8, 9, 12, 13, 8, 8, 9, 9, 10, 7, 11, 14, 3, 10
Offset: 1

Views

Author

Hugo van der Sanden, Jan 09 2007

Keywords

Comments

We know the sequence is well-defined given the limit x/d(x) > 0.5*sqrt(x) from comments in A036763.
Does every positive integer n equal floor(m/d(m)) for some m?
First occurrence of k>2: 4, 2, 1, 7,5 , 11, 3, 23, 8, 28, 68, 54, 40, 16, 251, 572, 141, ???, ???, ???, 529, ..., (630). - Robert G. Wilson v, Jan 11 2007

Crossrefs

Programs

  • Mathematica
    t = Table[Floor[n/DivisorSigma[0, n]], {n, 10^5}]; f[n_] := Length@Select[t, # == n &]; Array[f, 91] (* Robert G. Wilson v, Jan 11 2007 *)

Extensions

Edited by Robert G. Wilson v, Jan 11 2007

A334762 a(n) = ceiling (n / A000005(n)).

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 4, 2, 3, 3, 6, 2, 7, 4, 4, 4, 9, 3, 10, 4, 6, 6, 12, 3, 9, 7, 7, 5, 15, 4, 16, 6, 9, 9, 9, 4, 19, 10, 10, 5, 21, 6, 22, 8, 8, 12, 24, 5, 17, 9, 13, 9, 27, 7, 14, 7, 15, 15
Offset: 1

Views

Author

Ctibor O. Zizka, May 10 2020

Keywords

Comments

ceiling (n / A000005(n)) - floor (n / A000005(n)) = 0 for A033950(n).

Examples

			a(1) = ceiling (1 / 1) = 1;
a(5) = ceiling (5 / 2) = 3;
		

Crossrefs

Programs

  • Maple
    seq(ceil(n/numtheory:-tau(n)), n=1..100); # Robert Israel, May 13 2020
  • Mathematica
    a[n_] := Ceiling[n / DivisorSigma[0, n]]; Array[a, 60] (* Amiram Eldar, May 10 2020 *)
  • Python
    from sympy import divisor_count
    def A334762(n): return (a := divmod(n,divisor_count(n)))[0] + int((a[1] > 0) == True) # Chai Wah Wu, Jun 20 2022

Formula

a(n) = ceiling (n / A000005(n)).

A078710 Integer part of the subinterval sample variance in the partition of [0,n] by the divisors of n.

Original entry on oeis.org

0, 0, 0, 4, 1, 12, 2, 7, 3, 40, 4, 60, 9, 17, 8, 112, 9, 144, 13, 35, 27, 220, 14, 104, 41, 60, 25, 364, 23, 420, 34, 93, 75, 166, 30, 612, 97, 133, 45, 760, 47, 840, 67, 124, 147, 1012, 50, 500, 93, 235, 96, 1300, 78, 410, 91, 297, 243, 1624, 69, 1740, 281, 254, 130, 576
Offset: 2

Views

Author

Joseph L. Pe, Dec 19 2002

Keywords

Examples

			The divisors of 9 partition the closed interval [0,9] into subintervals [0,1), [1,3), [3,9], with lengths 1, 2, 6, respectively. The sample variance of these lengths has integer part = 7. Hence a(9) = 7. [Corrected by _Sean A. Irvine_, Jul 12 2025]
		

Crossrefs

Cf. A078709.

Programs

  • Mathematica
    f[n_] := Module[{d, l, a, i}, d = Divisors[n]; l = Length[d]; a = {1}; For[i = 1, i <= l - 1, i++, a = Append[a, d[[i + 1]] - d[[i]]]]; a]; Table[Floor[Variance[f[i]]], {i, 2, 100}]

A355140 n/d(n) rounded to the nearest integer, where d(n) is the number of divisors of n (A000005).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 2, 3, 3, 6, 2, 7, 4, 4, 3, 9, 3, 10, 3, 5, 6, 12, 3, 8, 7, 7, 5, 15, 4, 16, 5, 8, 9, 9, 4, 19, 10, 10, 5, 21, 5, 22, 7, 8, 12, 24, 5, 16, 8, 13, 9, 27, 7, 14, 7, 14, 15, 30, 5, 31, 16, 11, 9, 16, 8, 34, 11, 17, 9, 36, 6, 37, 19, 13, 13, 19
Offset: 1

Views

Author

Sameer Khan, Jun 20 2022

Keywords

Comments

In the ambiguous case, fractions are rounded up.

Examples

			a(1) = round (1 / 1) = 1;
a(4) = round (4 / 3) = 1;
a(5) = round (5 / 2) = 3;
		

Crossrefs

Cf. A000005, A078709 (floor), A334762 (ceiling), A090395 (numerators), A090387 (denominators).

Programs

  • Mathematica
    Table[Floor[n/DivisorSigma[0,n]+1/2],{n,100}] (* Harvey P. Dale, Dec 22 2022 *)
  • Python
    from sympy import divisor_count
    def A355140(n): return (2*n+(d:=divisor_count(n)))//(2*d) # Chai Wah Wu, Jun 20 2022

Formula

a(n) = round (n / A000005(n)).
Showing 1-7 of 7 results.