A379258 a(n) is the number of iterations of the Euler phi function needed to reach 1 starting at the n-th 3-smooth number.

1, 2, 3, 3, 3, 4, 4, 4, 5, 4, 5, 5, 6, 5, 6, 5, 7, 6, 6, 7, 6, 8, 7, 6, 8, 7, 7, 9, 8, 7, 9, 8, 7, 10, 9, 8, 8, 10, 9, 8, 11, 10, 9, 8, 11, 10, 9, 12, 9, 11, 10, 9, 12, 11, 10, 13, 9, 12, 11, 10, 13, 10, 12, 11, 14, 10, 13, 12, 11, 14, 10, 13, 12, 15, 11, 14, 11

Offset

1,2

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

a(n) = A049108(A003586(n)).

a(n) = valuation(A003586(n), 2) + valuation(A003586(n), 3) + 1 + [valuation(A003586(n), 2) == 0] for n > 1, where [] is the Iverson bracket.

a(n) = A022328(n) + A022329(n) + 1 + [n is in A022330], for n > 1.

a(A022330(n)) = n + 2 for n >= 1.

a(A022331(n)) = n + 1 for n >= 0.

a(A202821(n)) = 2*n + 1, for n >= 0.

Example

a(6) = 4 because the 6th 3-smooth number is A003586(6) = 8, and 4 iterations of phi are needed to reach 1: 8 -> 4 -> 2 -> 1.

Mathematica

f[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3]}, e2 + e3 + 1 + Boole[e2 == 0]]; f[1] = 1; With[{max = 3*10^4}, f /@ Sort[Flatten[Table[2^i*3^j, {i, 0, Log2[max]}, {j, 0, Log[3, max/2^i]}]]]]

Prog

(PARI) list(lim) = {my(e2, e3); print1(1, ", "); for(k = 2, lim, e2 = valuation(k, 2); e3 = valuation(k, 3); if(k == (1 << e2) * 3^e3, print1(e2 + e3 + 1 + (e2 == 0), ", "))); }

Crossrefs

Cf. A000010, A003586, A049108, A086420, A022328, A022329, A022330, A022331, A202821.

Sequence in context: A013940 A029129 A087842 * A201206 A101776 A189721

Adjacent sequences: A379255 A379256 A379257 * A379259 A379260 A379261

Keyword

nonn,easy

Author

Amiram Eldar, Dec 19 2024

Status

approved