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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A357310 a(n) is the number of j in the range 1 <= j <= n with the same maximal exponent in prime factorization as n.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 5, 1, 2, 6, 7, 3, 8, 9, 10, 1, 11, 4, 12, 5, 13, 14, 15, 2, 6, 16, 3, 7, 17, 18, 19, 1, 20, 21, 22, 8, 23, 24, 25, 4, 26, 27, 28, 9, 10, 29, 30, 2, 11, 12, 31, 13, 32, 5, 33, 6, 34, 35, 36, 14, 37, 38, 15, 1, 39, 40, 41, 16, 42, 43, 44, 7, 45, 46, 17, 18, 47, 48, 49, 3
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 23 2022

Keywords

Crossrefs

Cf. A000079 (positions of 1's), A051903, A058933, A289023.

Programs

  • Maple
    f:= proc(n) option remember; `if`(n=1, 0,
          max(map(i-> i[2], ifactors(n)[2])))
        end:
    b:= proc(n) option remember; `if`(n<1, 0, b(n-1)+x^f(n)) end:
    a:= n-> coeff(b(n), x, f(n)):
    seq(a(n), n=1..80);  # Alois P. Heinz, Sep 23 2022
  • Mathematica
    Table[Length[Select[Range[n], If[# == 1, 0, Max @@ Last /@ FactorInteger[#]] == If[n == 1, 0, Max @@ Last /@ FactorInteger[n]] &]], {n, 1, 80}]
    seq[max_] := Module[{e = Join[{0}, Table[Max @@ FactorInteger[n][[;; , 2]], {n, 2, max}]], c = Table[0, {max}]}, Do[c[[k]] = 1 + Count[e[[1 ;; k - 1]], e[[k]]], {k, 1, max}]; c]; seq[100] (* Amiram Eldar, Jan 05 2024 *)
  • PARI
    lista(nmax) = {my(e = vector(nmax, k, if(k==1, 0, vecmax(factor(k)[,2]))), c); for(k = 1, nmax, c =  1; for(j = 1, k-1, c += (e[j] == e[k])); print1(c, ", "));} \\ Amiram Eldar, Jan 05 2024

Formula

a(n) = |{j <= n : A051903(j) = A051903(n)}|.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1/zeta(2)^2 + Sum_{k>=3} (1/zeta(k+1) - 1/zeta(k))^2 = 0.43029326822775728041... . - Amiram Eldar, Jan 05 2024

A380653 Number of positive integers less than or equal to n that have the same sum of prime factors (with repetition) as n.

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Jan 29 2025

Keywords

Comments

Ordinal transform of A001414.

Crossrefs

Programs

  • Maple
    b:= n-> add(i[1]*i[2], i=ifactors(n)[2]):
    p:= proc() 0 end:
    a:= proc(n) option remember; local t;
          t:= b(n); p(t):= p(t)+1
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, Jan 30 2025
  • Mathematica
    sopfr[1] = 0; sopfr[n_] := Plus @@ Times @@@ FactorInteger@ n; Table[Length[Select[Range[n], sopfr[#] == sopfr[n] &]], {n, 1, 100}]
  • Python
    from sympy import factorint
    from collections import Counter
    from itertools import count, islice
    def agen(): # generator of terms
        sopfrcount = Counter()
        for n in count(1):
            key = sum(p*e for p, e in factorint(n).items())
            sopfrcount[key] += 1
            yield sopfrcount[key]
    print(list(islice(agen(), 100))) # Michael S. Branicky, Jan 30 2025

Formula

a(n) = |{j <= n : sopfr(j) = sopfr(n)}|.

A380049 Partial sums of A072203.

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 9, 12, 14, 15, 17, 20, 24, 27, 29, 30, 32, 35, 39, 44, 48, 51, 55, 58, 60, 61, 63, 66, 70, 75, 81, 88, 94, 99, 103, 106, 110, 113, 115, 116, 118, 121, 125, 130, 136, 141, 147, 154, 160, 167, 173, 180, 188, 195, 201, 206, 210, 213, 217, 220, 224, 227, 231, 234, 236
Offset: 1

Views

Author

Tsuyoshi Hanatate, Jan 10 2025

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Accumulate[Table[-LiouvilleLambda[n], {n, 2, 100}]]] (* Vaclav Kotesovec, Jan 15 2025 *)
  • PARI
    f(n) = 1 - sum(i=1, n, (-1)^bigomega(i)); \\ A072203
    a(n) = sum(k=1, n, f(k)); \\ Michel Marcus, Feb 06 2025

Formula

a(n) = Sum_{k=1..n} A072203(k).
Conjecture: The average value of a(n) is 2*n^(3/2)/(-3*zeta(1/2)). - Vaclav Kotesovec, Jan 15 2025
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