A344590 Number of divisors d of n for which A011772(d) = A011772(n), where A011772(n) is the smallest number m such that m(m+1)/2 is divisible by n.
1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 3, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 3, 1, 1, 1, 1, 2
Offset: 1
Keywords
Examples
36 has 9 divisors: 1, 2, 3, 4, 6, 9, 12, 18, 36. When A011772 is applied to them, one obtains values [1, 3, 2, 7, 3, 8, 8, 8, 8], thus there are four divisors that obtain the maximal value 8 obtained at 36 itself, therefore a(36) = 4.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
Crossrefs
Programs
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Mathematica
A011772[n_] := Module[{m = 1}, While[Not[IntegerQ[m(m+1)/(2n)]], m++]; m]; a[n_] := With[{m = A011772[n]}, Count[Divisors[n], d_ /; A011772[d] == m]]; Array[a, 100] (* Jean-François Alcover, Jun 12 2021 *)
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PARI
A011772(n) = { if(n==1, return(1)); my(f=factor(if(n%2, n, 2*n)), step=vecmax(vector(#f~, i, f[i, 1]^f[i, 2]))); forstep(m=step, 2*n, step, if(m*(m-1)/2%n==0, return(m-1)); if(m*(m+1)/2%n==0, return(m))); }; \\ From A011772 A344590(n) = { my(x=A011772(n)); sumdiv(n,d,A011772(d)==x); };
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Python
from itertools import combinations from functools import reduce from operator import mul from sympy import factorint, divisors from sympy.ntheory.modular import crt def A011772(n): plist = [p**q for p, q in factorint(2*n).items()] if len(plist) == 1: return n-1 if plist[0] % 2 else 2*n-1 return min(min(crt([m,2*n//m],[0,-1])[0],crt([2*n//m,m],[0,-1])[0]) for m in (reduce(mul, d) for l in range(1,len(plist)//2+1) for d in combinations(plist,l))) def A344590(n): m = A011772(n) return sum(1 for d in divisors(n) if A011772(d) == m) # Chai Wah Wu, Jun 02 2021
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