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A344759
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a(n) = n divided by the smallest divisor 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.
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6
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1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 1, 7, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 3, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 5, 1, 3, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 1, 2, 3, 1, 1, 6, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 3, 1, 2, 1, 4, 1, 1, 1, 1, 3
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OFFSET
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1,6
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COMMENTS
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It seems that A006516 gives the positions of records after its initial zero.
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LINKS
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FORMULA
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PROG
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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
(Python 3.8+)
from itertools import combinations
from math import prod
from sympy import factorint, divisors
from sympy.ntheory.modular import crt
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 (prod(d) for l in range(1, len(plist)//2+1) for d in combinations(plist, l)))
for d in divisors(n):
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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