A384004
a(n) = smallest k such that A010846(k) = n.
Original entry on oeis.org
1, 2, 4, 8, 6, 10, 22, 12, 44, 18, 24, 50, 98, 36, 48, 54, 224, 30, 42, 70, 108, 66, 78, 162, 102, 60, 138, 84, 174, 260, 132, 90, 126, 228, 354, 120, 234, 168, 350, 306, 150, 516, 408, 180, 252, 552, 696, 294, 240, 336, 612, 378, 270, 1416, 300, 702, 1332, 360
Offset: 1
Table of n, a(n) for n=1..10, showing row a(n) of A162306, replacing lpf(a(n)) with p, and A119288(a(n)) with q. Note: A010846(n) is the length of row n of A162306.
n a(n) row n of A162306
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1: 1 {1}
2: 2 {1, p}
3: 4 {1, p, p^2}
4: 8 {1, p, p^2, p^3}
5: 6 {1, p, q, p^2, p*q}
6: 10 {1, p, p^2, q, p^3, p*q}
7: 22 {1, p, p^2, p^3, q, p^4, p*q}
8: 12 {1, p, q, p^2, p*q, p^3, q^2, p^2*q}
9: 44 {1, p, p^2, p^3, q, p^4, p*q, p^5, p^2*q}
10: 18 {1, p, q, p^2, p*q, p^3, q^2, p^2*q, p^4, p*q^2}
- Michael De Vlieger, Table of n, a(n) for n = 1..4647
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..4647, showing primes in large red, proper prime powers in large gold, composite primorials in large bright green, other squarefree composites in small green, and numbers neither squarefree nor prime powers in blue or magenta, with magenta signifying powerful numbers that are not prime powers.
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(* First, load the theta program from the algorithms linked in A369609, then: *)
nn = 2310; t[_] := 0; u = 1; Do[(If[t[#] == 0, t[#] = n]; If[# == u, While[t[u] != 0, u++]]) &[theta[n]], {n, nn}]; Array[t, u - 1]
A301893
Numbers m that set records for the ratio A010846(m)/A000005(m).
Original entry on oeis.org
1, 6, 10, 18, 22, 30, 42, 66, 78, 102, 114, 138, 150, 174, 210, 330, 390, 510, 570, 690, 870, 1110, 1230, 1290, 1410, 1590, 1770, 1830, 2010, 2130, 2190, 2310, 2730, 3570, 3990, 4830, 6090, 6510, 7770, 8610, 9030, 9870, 11130, 12390, 12810, 14070, 14910, 15330
Offset: 1
The number 1 sets a record as it is the first term; the ratio A010846(1)/A000005(1) = 1. Since 2 <= m <= 5 have omega(m) = 1, they too have ratio = 1 and do not appear.
6 is the next term since those numbers 1 <= k <= 6 that divide some nonnegative integer power of 6 are {1, 2, 3, 4, 6}; of these, 4 are divisors, thus we have the ratio 5/4. This exceeds 1, so 6 follows 1 in the sequence. The numbers 7 <= m <= 9 have omega(m) = 1.
10 appears next since the regular m of 10 are {1, 2, 4, 5, 8, 10}; of these 4 divide 10, thus we have ratio 6/4 which exceeds that of 6, so 10 follows 6.
12 is not in the sequence since the regular m of 12 are {1, 2, 3, 4, 6, 8, 9, 12} and 6 of these divide 12, giving us the ratio 8/6 which is less than the 6/4 of 10.
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With[{s = Table[Count[Range@ n, _?(PowerMod[n, Floor@ Log2@ n, #] == 0 &)]/DivisorSigma[0, n], {n, 3000}]}, Map[FirstPosition[s, #][[1]] &, Union@ FoldList[Max, s]]]
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a000005(n) = if(n==0, 0, numdiv(n)) \\ after Michael Somos in A000005
a010846(n) = sum(k=1, n, if(gcd(n, k)-1, 0, moebius(k)*(n\k))) \\ after Benoit Cloitre in A010846
r=0; for(m=1, oo, if(a010846(m)/a000005(m) > r, print1(m, ", "); r=a010846(m)/a000005(m))) \\ Felix Fröhlich, Mar 30 2018
A322164
Numbers n > 1 such that phi(n) <= phi(k) + phi(n-k) for all 1 <= k <= n-1, where phi(n) is the Euler totient function (A000010).
Original entry on oeis.org
2, 3, 4, 6, 10, 12, 18, 24, 30, 42, 60, 84, 90, 120, 150, 180, 210, 330, 390, 420, 630, 840, 1050, 1260, 1470, 1680, 1890, 2100, 2310, 2730, 3570, 3990, 4620, 5460, 6930, 8190, 9240, 10920, 11550, 13650, 13860, 16170, 18480, 20790, 23100, 25410, 27720, 30030
Offset: 1
6 is in the sequence since phi(k) + phi(6-k) = 5, 3, 4, 3, 5 for k = 1 to 5 are all larger than phi(6) = 2.
- J. Sandor and B. Crstici, Handbook of Number Theory II, Springer Verlag, 2004, Chapter 3.3, p. 224.
- C. A. Nicol, Problem E2590, The American Mathematical Monthly, Vol. 83, No. 4 (1976), p. 284, solution by Lorraine L. Foster, Vol. 84, No. 8 (1977), p. 654-655.
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aQ[n_] := Module[{e=EulerPhi[n]}, LengthWhile[Range[1,n-1], EulerPhi[n-#] + EulerPhi[#] >= e &] == n-1]; Select[Range[2, 10000], aQ]
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isok(n) = {if (n == 1, return(0)); my(t = eulerphi(n)); for (k=1, n-1, if (t > eulerphi(k) + eulerphi(n-k), return(0));); return (1);} \\ Michel Marcus, Nov 29 2018
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