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.

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A374484 Index of A006899(n) in A003586.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 12, 13, 17, 19, 22, 27, 28, 34, 37, 41, 48, 49, 56, 62, 65, 74, 77, 84, 93, 95, 106, 111, 118, 130, 131, 143, 152, 157, 171, 175, 186, 199, 202, 218, 225, 235, 252, 253, 271, 281, 290, 309, 312, 329, 344, 350, 371, 378, 393, 413, 416, 439
Offset: 1

Views

Author

Chai Wah Wu, Sep 16 2024

Keywords

Comments

Index of powers of 2 and 3 in 3-smooth numbers.

Examples

			A006899(10) = 64 which is the 17th term of A003586, therefore a(10) = 17.
		

Crossrefs

Disjoint union of A022330 and A022331.

Programs

  • Mathematica
    seq[lim_] := Position[Times @@ IntegerExponent[#, {2, 3}] & /@ Sort[Flatten[ Table[2^i*3^j, {i, 0, Log2[lim]}, {j, 0, Log[3, lim/2^i]}] ]], 0] // Flatten; seq[10^11] (* Amiram Eldar, Sep 18 2024 *)
  • Python
    from sympy import integer_log
    def A374484(n): return sum(((1<
    				

Formula

A003586(a(n)) = A006899(n).
a(n) ~ c * n^2, where c = log(2)*log(3)/(2*(log(2) + log(3))^2) = 0.118598856384648... - Vaclav Kotesovec and Amiram Eldar, Sep 19 2024

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.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Dec 19 2024

Keywords

Examples

			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.
		

Crossrefs

Programs

  • 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]}]]]]
  • 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), ", ")));}

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.
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