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 1-4 of 4 results.

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

A318529 a(n) begins the first run of at least n consecutive numbers with same number of exponential divisors.

Original entry on oeis.org

1, 1, 1, 242, 3624, 22020, 671346, 8870024, 49250144, 463239475, 1407472722, 82462576220, 82462576220, 5907695879319
Offset: 1

Views

Author

Amiram Eldar, Aug 28 2018

Keywords

Comments

From David A. Corneth, Aug 28 2018: (Start)
For 4 <= n <= 10, a(n) has two exponential divisors. Most numbers have 1 or 2 exponential divisors.
For n > 3, a(n) isn't squarefree. (End)
For n >= 2^(k+1), A049419(a(n)) must be divisible by A051548(k), because for 1 <= j <= k at least one of a(n),...,a(n)+n-1 has 2-adic order j. - Robert Israel, Sep 07 2018

Examples

			a(4) = 242 since the number of exponential divisors of 242, 243, 244, and 245 is 2, and this is the first run of 4 consecutive numbers.
		

Crossrefs

Programs

  • Mathematica
    edivnum[1] = 1; edivnum [p_?PrimeQ] = 1; edivnum [p_?PrimeQ, e_] := DivisorSigma[ 0, e ]; edivnum [n_] := Times @@ (edivnum [#[[1]], #[[2]]] & ) /@ FactorInteger[ n ]; Seq[n_,q_] := Map[edivnum, Range[n, n + q - 1]]; findConsec[q_, nmin_, nmax_] := Module[{}, s = Seq[1, q]; n = q + 1; found = False; Do[ If[ CountDistinct[s] == 1, found = True; Break[] ]; s = Rest[AppendTo[s, edivnum[n] ]]; n++, {k, nmin, nmax}]; If[found, n - q, 0]]; seq = {1}; nmax = 100000000; Do[n1 = Last[seq]; s1 = findConsec[m, n1, nmax]; If[s1 == 0, Break[]]; AppendTo[ seq, s1 ], {m, 2, 7}]; seq (* after Jean-François Alcover in A049419 *)

Extensions

a(11)-a(13) from Giovanni Resta, Aug 28 2018
a(14) from Giovanni Resta, Sep 07 2018

A349262 a(n) is the start of the least run of exactly n consecutive numbers with the same value of A349258.

Original entry on oeis.org

1, 14, 20, 2, 91, 6850, 2302, 141, 56014, 184171, 2800171, 27805034, 35297611, 8313366182, 1791416073, 3618621410
Offset: 1

Views

Author

Amiram Eldar, Nov 12 2021

Keywords

Comments

a(17) > 10^11, if it exists.

Examples

			a(2) = 14 since A349258(14) = A349258(15) = 2, but A349258(13) != 2 and A349258(16) != 2.
		

Crossrefs

Cf. A349258.
Similar sequences: A006558, A045983, A048932, A067813, A077657, A318166.

Programs

  • Mathematica
    f[p_, e_] := 2^DigitCount[e, 2, 1] - 1; d[1] = 0; d[n_] := Plus @@ f @@@ FactorInteger[n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], dprev = 0, n = 2, c = 1, k = 1}, s[[1]] = 1; While[k < len && n < nmax, d1 = d[n]; If[d1 == dprev, c++, If[c > 0 && c <= len && s[[c]] == 0, k++; s[[c]] = n - c]; c = 1]; n++; dprev = d1]; TakeWhile[s, # > 0 &]]; seq[8, 10^4]

A349305 a(n) is the start of the least run of exactly n consecutive numbers with the same number of nonunitary divisors.

Original entry on oeis.org

4, 10, 1, 19940, 54584, 204323, 2789143044, 27092041443
Offset: 1

Views

Author

Amiram Eldar, Nov 14 2021

Keywords

Comments

a(9) > 10^11, if it exists.

Examples

			a(2) = 10 since A048105(10) = A048105(11) = 0, and A048105(9) != 0 and A048105(12) != 0.
		

Crossrefs

Similar sequences: A006558, A045983, A048932, A067813, A077657, A318166.

Programs

  • Mathematica
    d[n_] := DivisorSigma[0, n] - 2^PrimeNu[n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], dprev = -1, n = 1, c = 0, k = 0}, While[k < len && n < nmax, d1 = d[n]; If[d1 == dprev, c++, If[c > 0 && c <= len && s[[c]] == 0, k++; s[[c]] = n - c]; c = 1]; n++; dprev = d1]; TakeWhile[s, # > 0 &]]; seq[6, 10^6]
Showing 1-4 of 4 results.