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 1-3 of 3 results.

A374671 Positive numbers k such that k! and (k+1)! have an equal number of infinitary divisors.

Original entry on oeis.org

8, 19, 23, 44, 45, 57, 67, 76, 80, 83, 84, 85, 105, 107, 116, 120, 123, 140, 141, 146, 158, 161, 165, 174, 177, 187, 201, 208, 214, 225, 235, 239, 241, 243, 244, 246, 247, 263, 269, 272, 277, 284, 297, 309, 315, 321, 322, 325, 337, 339, 341, 342, 344, 360, 363
Offset: 1

Views

Author

Amiram Eldar, Jul 16 2024

Keywords

Comments

Positive numbers such that k! and (k+1)! have an equal number of Fermi-Dirac factors (A064547).
Positive numbers k such that A037445(k!) = A037445((k+1)!).
Positive numbers k such that A064547(k!) = A064547((k+1)!).
Positive numbers k such that A177329(k) = A177329(k+1).

Examples

			8 is a term since A037445(8!) = A037445(9!) = 64.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := s[n] = Module[{e = FactorInteger[n!][[;; , 2]]}, Sum[DigitCount[e[[k]], 2, 1], {k, 1, Length[e]}]]; Select[Range[2, 400], s[#] == s[# + 1] &]
  • PARI
    s(n) = {my(e = factor(n!)[, 2]); sum(k=1, #e, hammingweight(e[k]));}
    lista(kmax) = {my(s1 = s(1), s2); for(k = 2, kmax, s2 = s(k); if(s1 == s2, print1(k-1, ", ")); s1 = s2);}

A374672 Numbers k such that k! has more infinitary divisors than (k+1)!.

Original entry on oeis.org

5, 9, 17, 27, 33, 34, 35, 43, 48, 51, 53, 59, 65, 68, 69, 75, 77, 87, 91, 97, 98, 99, 103, 115, 119, 125, 129, 134, 135, 139, 147, 149, 151, 155, 163, 164, 171, 179, 183, 189, 194, 195, 197, 199, 203, 211, 215, 221, 227, 229, 230, 231, 237, 245, 249, 257, 259
Offset: 1

Views

Author

Amiram Eldar, Jul 16 2024

Keywords

Comments

Numbers k such that k! has more Fermi-Dirac factors (A064547) than (k+1)!.
Numbers k such that A037445(k!) > A037445((k+1)!).
Numbers k such that A064547(k!) > A064547((k+1)!).
Numbers k such that A177329(k) > A177329(k+1).

Examples

			5 is a term since A037445(5!) = 16 > A037445(6!) = 8.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := s[n] = Module[{e = FactorInteger[n!][[;; , 2]]}, Sum[DigitCount[e[[k]], 2, 1], {k, 1, Length[e]}]]; Select[Range[2, 300], s[#] > s[# + 1] &]
  • PARI
    s(n) = {my(e = factor(n!)[, 2]); sum(k=1, #e, hammingweight(e[k]));}
    lista(kmax) = {my(s1 = s(1), s2); for(k = 2, kmax, s2 = s(k); if(s1 > s2, print1(k-1, ", ")); s1 = s2);}

A374673 a(n) is the start of the least run of exactly n consecutive positive numbers with an equal value of A177329, or -1 if no such run exists.

Original entry on oeis.org

2, 8, 44, 83, 4475, 75093, 164903, 59480, 1342805
Offset: 1

Views

Author

Amiram Eldar, Jul 16 2024

Keywords

Comments

For n > 1, a(n)! is the start of the least run of successive factorials of positive numbers (i.e., ignoring 0!) with an equal number of infinitary divisors (A037445).
a(9) > 320000, if it exists.

Examples

			  n |   a(n) | A177329(k), k = a(n), a(n)+1, ..., a(n)+n-1
  --|--------|------------------------------------------------
  1 |      2 | A177329(2) = 1
  2 |      8 | A177329(8) = A177329(9) = 6
  3 |     44 | A177329(44) = A177329(45) = A177329(46) = 21
  4 |     83 | A177329(83) = ... = A177329(86) = 35
  5 |   4475 | A177329(4475) = ... A177329(4479) = 923
  6 |  75093 | A177329(75093) = ... = A177329(75098) = 10857
  7 | 164903 | A177329(164903) = ... = A177329(164909) = 22038
  8 |  59480 | A177329(59480) = ... = A177329(59487) = 8814
		

Crossrefs

Programs

  • Mathematica
    s[n_] := Module[{e = FactorInteger[n!][[;; , 2]]}, Sum[DigitCount[e[[k]], 2, 1], {k, 1, Length[e]}]]; seq[len_] := Module[{v = Table[0, {len}], w = {0}, c = 0, k = 3, m, s1}, While[c < len, s1 = s[k]; m = Length[w]; If[s1 == w[[m]], AppendTo[w, s1], If[m <= len && v[[m]] == 0, v[[m]] = k-m; c++]; w = {s1}]; k++]; v]; seq[5]
  • PARI
    s(n) = {my(e = factor(n!)[, 2]); sum(k=1, #e, hammingweight(e[k]));}
    lista(len) = {my(v = vector(len), w = [0], c = 0, k = 3, m, s1); while(c < len, s1 = s(k); m = #w; if(s1 == w[m], w = concat(w, s1), if(m < = len && v[m] == 0, v[m] = k-m; c++); w = [s1]); k++); v;}
    
  • Python
    from itertools import count
    from collections import Counter
    from sympy import factorint
    def A374673(n):
        if n==1: return 2
        c, a, l = Counter(), 0, 0
        for m in count(2):
            c += Counter(factorint(m))
            b = sum(map(int.bit_count,c.values()))
            if b==a:
                l += 1
            else:
                if l==n-1:
                    return m-n
                l = 0
            a = b # Chai Wah Wu, Jul 18 2024

Extensions

a(9) from Chai Wah Wu, Jul 18 2024
Showing 1-3 of 3 results.