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.

Showing 1-4 of 4 results.

A372401 Position of 210^n among 7-smooth numbers A002473.

Original entry on oeis.org

1, 68, 547, 2119, 5817, 13008, 25412, 45078, 74409, 116147, 173379, 249532, 348375, 474018, 630922, 823885, 1058051, 1338898, 1672260, 2064302, 2521535, 3050825, 3659361, 4354687, 5144682, 6037582, 7041946, 8166692, 9421074, 10814695, 12357491, 14059744, 15932086, 17985473
Offset: 0

Views

Author

Michael De Vlieger, Jun 03 2024

Keywords

Comments

Also position of 210^(n+1) in A147571.

Crossrefs

Cf. A002110, A002473, A022330, A147571, A202821, A372400, A372402.

Programs

  • Mathematica
    Table[
  Sum[Floor@ Log[7, 210^n/(2^i*3^j*5^k)] + 1,
    {i, 0, Log[2, 210^n]},
    {j, 0, Log[3, 210^n/2^i]},
    {k, 0, Log[5, 210^n/(2^i*3^j)]}],
  {n, 0, 12}]
  • Python
    import heapq
from itertools import islice
from sympy import primerange
def A372401gen(p=7): # generator for p-smooth terms
    v, oldv, psmooth_primes, = 1, 0, list(primerange(1, p+1))
    h = [(1, [0]*len(psmooth_primes))]
    idx = {psmooth_primes[i]:i for i in range(len(psmooth_primes))}
    loc = 0
    while True:
        v, e = heapq.heappop(h)
        if v != oldv:
            loc += 1
            if len(set(e)) == 1:
                yield loc
            oldv = v
            for p in psmooth_primes:
                vp, ep = v*p, e[:]
                ep[idx[p]] += 1
                heapq.heappush(h, (v*p, ep))
print(list(islice(A372401gen(), 15))) # Michael S. Branicky, Jun 05 2024
  • Python
    from sympy import integer_log
def A372401(n):
    c, x = 0, 210**n
    for i in range(integer_log(x,7)[0]+1):
        for j in range(integer_log(m:=x//7**i,5)[0]+1):
            for k in range(integer_log(r:=m//5**j,3)[0]+1):
                c += (r//3**k).bit_length()
    return c # Chai Wah Wu, Sep 16 2024

Formula

a(n) ~ c * n^4, where c = log(210)^4/(24*log(2)*log(3)*log(5)*log(7)) = 14.282278766622... - Vaclav Kotesovec and Amiram Eldar, Sep 22 2024

A372400 Position of 30^n among 5-smooth numbers A051037.

Original entry on oeis.org

1, 18, 83, 228, 486, 888, 1466, 2255, 3283, 4583, 6189, 8134, 10445, 13158, 16305, 19916, 24027, 28667, 33870, 39665, 46086, 53166, 60937, 69429, 78675, 88709, 99561, 111263, 123849, 137347, 151793, 167219, 183658, 201139, 219695, 239359, 260165, 282141, 305320
Offset: 0

Views

Author

Michael De Vlieger, Jun 03 2024

Keywords

Comments

Also position of 30^(n+1) in A143207.

Links

Crossrefs

Cf. A002110, A051037, A143207, A202821, A372401, A372402.

Programs

  • Mathematica
    Table[Sum[Floor@ Log[5, 30^n/(2^i*3^j)] + 1, {i, 0, Log[2, 30^n]}, {j, 0, Log[3, 30^n/2^i]}], {n, 0, 38}]
  • PARI
    a(n)=my(t=30^n,u=5*t); sum(a=0,logint(t,5), u\=5; sum(b=0,logint(u,3), logint(u\3^b,2)+1)) \\ Charles R Greathouse IV, Sep 18 2024
  • Python
    # uses imports/function in A372401
print(list(islice(A372401gen(p=5), 40))) # Michael S. Branicky, Jun 05 2024
  • Python
    from sympy import integer_log
def A372400(n):
    c, x = 0, 30**n
    for i in range(integer_log(x,5)[0]+1):
        for j in range(integer_log(y:=x//5**i,3)[0]+1):
            c += (y//3**j).bit_length()
    return c # Chai Wah Wu, Sep 16 2024

Formula

a(n) = k*n^3 + (3k/2)*n^2 + O(n) where k = (log 30)^3/(6 log 2 log 3 log 5) = 5.35057081984.... - Charles R Greathouse IV, Sep 19 2024

A372744 If the n-th 3-smooth number, A003586(n), equals 2^i * 3^j for some i, j >= 0, then the a(n)-th 3-smooth number, A003586(a(n)), equals 2^j * 3^i.

Original entry on oeis.org

1, 3, 2, 7, 5, 12, 4, 10, 19, 8, 16, 6, 27, 14, 24, 11, 37, 21, 9, 33, 18, 49, 30, 15, 44, 26, 13, 62, 40, 23, 57, 36, 20, 77, 52, 32, 17, 71, 47, 29, 93, 66, 43, 25, 87, 60, 39, 111, 22, 81, 55, 35, 104, 75, 51, 131, 31, 98, 69, 46, 123, 28, 91, 64, 152, 42
Offset: 1

Views

Author

Rémy Sigrist, May 12 2024

Keywords

Comments

This sequence is a self-inverse permutation of the positive integers with infinitely many fixed points (A202821).

Examples

			A003586(8) = 12 = 2^2 * 3^1, A003586(10) = 18 = 2^1 * 3^2, so a(8) = 10 and.

Links

Crossrefs

Cf. A003586, A022328, A022329, A202821 (fixed points).

Programs

  • PARI
    \\ See Links section.

Formula

A022328(a(n)) = A022329(n).
A022329(a(n)) = A022328(n).
a(n) = n iff n belongs to A202821.
sign(a(n) - n) = sign(A022328(n) - A022329(n)).

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.

Links

Crossrefs

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

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.
Showing 1-4 of 4 results.

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