A090616 -id:A090616 - cp's OEIS frontend

A319316 Numbers k such that A090616(k) < A054861(k).

3, 9, 15, 27, 28, 29, 30, 31, 39, 45, 54, 55, 57, 63, 81, 82, 83, 84, 85, 87, 90, 91, 93, 94, 95, 99, 108, 109, 110, 111, 117, 118, 119, 123, 126, 127, 135, 162, 163, 165, 171, 174, 175, 183, 189, 190, 191, 207, 219, 243, 244, 245, 246, 247, 248, 249, 250, 251

Offset: 1

Jianing Song, Sep 17 2018

nonn

Numbers k such that the highest power of 12 dividing n! is determined by the highest power of 4 dividing n!.
Note that A054861 and A090616 are both asymptotic to a(n) = n/2 + O(log(n)), nevertheless, it seems that the number of k such that A090616(k) is bigger predominates. Conjecture: the ratio of k <= N such that A090616(k) > A054861(k) tends to 1 as N tends to infinity, while the ratio of k <= N such that A090616(k) < A054861(k) and A090616(k) = A054861(k) both tend to 0.
Number of k in range [0, N] such that A090616(k) =, < or > A054861(k):
..N....A090616(k) = A054861(k)...A090616(k) < A054861(k)...A090616(k) > A054861(k)
10^2...............38........................26........................37
10^3..............344.......................228.......................429
10^4.............2703......................2227......................5071
10^5............23003.....................19892.....................57106
10^6...........203478....................185152....................611371
10^7..........1762288...................1726062...................6511651

			The highest power of 3 dividing 9! is 3^4, while the highest power of 4 dividing 9! is 4^3, so 9 is a term, and the highest power of 12 dividing 9! is 12^3.
The highest power of 3 dividing 15! is 3^6, while the highest power of 4 dividing 15! is 4^5, so 15 is a term, and the highest power of 12 dividing 15! is 12^5.

Jianing Song, Table of n, a(n) for n = 1..2227 (all terms <= 10000)

Cf. A217445 (k such that A090616(k) = A054861(k)), A319317 (k such that A090616(k) > A054861(k)).

isA319316(n)=(n-vecsum(digits(n, 2)))\2<(n-vecsum(digits(n, 3)))\2

A319317 Numbers k such that A090616(k) > A054861(k).

Original entry on oeis.org

8, 16, 17, 20, 24, 25, 26, 32, 34, 35, 40, 41, 44, 48, 49, 50, 52, 53, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 96, 97, 98, 104, 106, 107, 116, 128, 129, 130, 131, 132, 133, 134, 136, 137, 140, 142, 143, 144, 145, 146, 148, 149, 150, 151

Offset: 1

Jianing Song, Sep 17 2018

nonn

Numbers k such that the highest power of 12 dividing n! is determined by the highest power of 3 dividing n!.
Note that A054861 and A090616 are both asymptotic to a(n) = n/2 + O(log(n)), nevertheless, it seems that the number of k such that A090616(k) is bigger predominates. Conjecture: the ratio of k <= N such that A090616(k) >A054861(k) tends to 1 as N tends to infinity, while the ratio of k <= N such thatA090616(k) < A054861(k) and A090616(k) = A054861(k) both tend to 0.
Number of k in range [0, N] such that A090616(k) =, < or > A054861(k):
  N    A090616(k) = A054861(k)   A090616(k) < A054861(k)   A090616(k) > A054861(k)
10^2               38                        26                        37
10^3              344                       228                       429
10^4             2703                      2227                      5071
10^5            23003                     19892                     57106
10^6           203478                    185152                    611371
10^7          1762288                   1726062                   6511651

			The highest power of 3 dividing 8! is 3^2, while the highest power of 4 dividing 8! is 4^3, so 8 is a term, and the highest power of 12 dividing 8! is 12^2.
The highest power of 3 dividing 16! is 3^6, while the highest power of 4 dividing 16! is 4^7, so 16 is a term, and the highest power of 12 dividing 16! is 12^6.

Jianing Song, Table of n, a(n) for n = 1..5071 (all terms <= 10000)

Cf. A217445 (k such that A090616(k) = A054861(k)), A319316 (k such that A090616(k) < A054861(k)).

isA319317(n)=(n-vecsum(digits(n, 2)))\2>(n-vecsum(digits(n, 3)))\2

A090622 Square array read by antidiagonals of highest power of k dividing n! (with n,k>1).

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 0, 1, 3, 0, 0, 1, 1, 4, 0, 1, 0, 1, 2, 4, 0, 0, 1, 1, 2, 2, 7, 0, 0, 0, 1, 1, 2, 2, 7, 0, 0, 1, 0, 2, 1, 3, 4, 8, 0, 0, 0, 1, 0, 2, 1, 3, 4, 8, 0, 0, 0, 0, 1, 1, 2, 1, 4, 4, 10, 0, 0, 0, 1, 1, 1, 1, 4, 2, 4, 5, 10, 0, 0, 1, 0, 1, 1, 2, 1, 4, 2, 5, 5, 11, 0, 0, 0, 1, 0, 1, 1, 2, 1, 4, 2, 5, 5, 11

Offset: 2

Henry Bottomley, Dec 06 2003

			Square array starts:
1, 0, 0, 0, 0, 0, 0, ...
1, 1, 0, 0, 1, 0, 0, ...
3, 1, 1, 0, 1, 0, 1, ...
3, 1, 1, 1, 1, 0, 1, ...
4, 2, 2, 1, 2, 0, 1, ...
4, 2, 2, 1, 2, 1, 1, ...
7, 2, 3, 1, 2, 1, 2, ...

Alois P. Heinz, Antidiagonals n = 2..142, flattened

Columns include A011371, A054861, A090616, A027868, A054861, A054896, A090617, A090618, A027868, A064458, A090619, A090620, A054896, A027868, A090621.
Cf. A090623, A090624, A115627.
Diagonal gives A011776. - Alois P. Heinz, Oct 04 2012

f:= proc(n, p) local c, k; c, k:= 0, p;
       while n>=k do c:= c+iquo(n, k); k:= k*p od; c
    end:
T:= (n, k)-> min(seq(iquo(f(n, i[1]), i[2]), i=ifactors(k)[2])):
seq(seq(T(n, 2+d-n), n=2..d), d=2..20);  # Alois P. Heinz, Oct 04 2012

f[n_, p_] := Module[{c = 0, k = p}, While[n >= k , c = c + Quotient[n, k]; k = k*p ]; c ]; t[n_, k_] := Min[ Table[ Quotient[f[n, i[[1]]], i[[2]]], {i, FactorInteger[k]}]]; Table[ Table[t[n, 2 + d - n], {n, 2, d}], {d, 2, 20}] // Flatten (* Jean-François Alcover, Oct 03 2013, translated from Alois P. Heinz's Maple program *)

For k=p prime: T(n,p) = [n/p] + [n/p^2] + [n/p^3] + .... For k = p^m a prime power: T(n,p^m) = [T(n,p)/m]. For k = b*c with b and c coprime: T(n,a*b) = min(T(n,a), T(n,b)). T(n,k) is close to, but below, n/A090624(k).

A054893 a(n) = Sum_{j > 0} floor(n/4^j).

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24

Offset: 0

Henry Bottomley, May 23 2000

Different from highest power of 4 dividing n! (see A090616).

			  a(10^0) = 0.
  a(10^1) = 2.
  a(10^2) = 32.
  a(10^3) = 330.
  a(10^4) = 3331.
  a(10^5) = 33330.
  a(10^6) = 333330.
  a(10^7) = 3333329.
  a(10^8) = 33333328.
  a(10^9) = 333333326.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A053737, A235127 (first differences).

Cf. A011371, A027868, A054861, A054899, A067080, A090616, A098844, A132028.

function A054893(n)
  if n eq 0 then return n;
  else return A054893(Floor(n/4)) + Floor(n/4);
  end if; return A054893;
end function;
[A054893(n): n in [0..103]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=4; While[s=Floor[n/p]; t=t+s; s>0, p *= 4]; t, {n,0,100}]
Table[Total[Floor/@(n/NestList[4#&,4,6])],{n,0,80}] (* Harvey P. Dale, Jun 12 2022 *)

a(n) = (n - sumdigits(n,4))/3; \\ Kevin Ryde, Jan 08 2024

def A054893(n):
    if (n==0): return 0
    else: return A054893(n//4) + (n//4)
[A054893(n) for n in range(104)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/4) + floor(n/16) + floor(n/64) + floor(n/256) + ...
a(n) = (n - A053737(n))/3.
From Hieronymus Fischer, Sep 15 2007: (Start)
a(n) = a(floor(n/4)) + floor(n/4).
a(4*n) = a(n) + n.
a(n*4^m) = a(n) + n*(4^m-1)/3.
a(k*4^m) = k*(4^m-1)/3, for 0 <= k < 4, m >= 0.
Asymptotic behavior:
a(n) = n/3 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/3; equality holds true for powers of 4.
a(n) >= (n-3)/3 - floor(log_4(n)); equality holds true for n = 4^m - 1, m>0. lim inf (n/3 - a(n)) = 1/3, for n-->oo.
lim sup (n/3 - log_4(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_4(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(4^k)/(1-x^(4^k)). (End)
Partial sums of A235127. - R. J. Mathar, Jul 08 2021

Edited by Hieronymus Fischer, Sep 15 2007

Examples added by Hieronymus Fischer, Jun 06 2012

A090617 Exponent of highest power of 8 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

			a(8)=2 since 8! = 40320 = 8^2 * 630.

Cf. A011371, A054861, A090616, A027868, A054896.

Table[IntegerExponent[n!,8],{n,0,80}] (* Harvey P. Dale, Mar 21 2013 *)

a(n) = valuation(n!, 8); \\ Michel Marcus, Jul 10 2022

a(n) = A090622(n, 8) = floor(A011371(n)/3) = floor((floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ...)/3).

a(n) = A244413(n!) . - R. J. Mathar, Jul 08 2021

A090618 Highest power of 9 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 11, 13, 13, 13, 13, 13, 13, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 20, 20, 20, 20

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

			a(9)=2 since 9!=362880=9^2*4480.

Cf. A011371, A054861, A090616, A027868, A054896, A090617.

IntegerExponent[Range[0,90]!,9] (* Harvey P. Dale, Jun 07 2016 *)

a(n) =A090622(n, 9) =[A054861(n)/2] =[([n/3]+[n/9]+[n/27]+[n/81]+...)/2].

A090619 Highest power of 12 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, 5, 5, 6, 6, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 12, 12, 13, 13, 14, 15, 15, 15, 17, 17, 17, 17, 18, 18, 19, 19, 19, 20, 21, 21, 22, 22, 22, 23, 23, 23, 25, 25, 26, 26, 27, 27, 28, 28, 28, 28, 30, 30, 31, 31, 31, 32, 32, 32, 34, 34, 34, 35, 35

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

Most sequences of the form "highest power of k dividing n!" essentially depend on one of the primes or prime powers dividing k. But in this case, the sequences with k=3 (A054861) and k=4 (A090616) are both close to n/2 and vary in which one is lower for different values of n.

a(2^n) = A090616(2^n) and a(3^n-1) = A090616(3^n-1) while a(2^n-1) = A054861(2^n-1) and a(3^n) = A054861(3^n). - Robert Israel, Mar 25 2018

			a(6)=2 since 6!=720=12^2*5.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A011371, A054861, A090616, A027868, A054896, A090617, A090618.

f2:= n -> n - convert(convert(n,base,2),`+`):
f3:= n -> (n - convert(convert(n,base,3),`+`))/2:
f:= n -> min(f3(n), floor(f2(n)/2)):
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Mar 23 2018

Table[IntegerExponent[n!, 12], {n, 0, 100}] (* Jean-François Alcover, Mar 26 2018 *)

a(n) = valuation(n!, 12); \\ Michel Marcus, Mar 24 2018

a(n) =A090622(n, 12) =min(A054861(n), A090616(n)). Close to n/2, indeed for n>3: n/2-log3(n+1) <= a(n) < n/2.

A217445 Numbers n such that n! has the same number of terminating zeros in bases 3 and 4.

Original entry on oeis.org

1, 2, 4, 5, 6, 7, 10, 11, 12, 13, 14, 18, 19, 21, 22, 23, 33, 36, 37, 38, 42, 43, 46, 47, 51, 56, 58, 59, 60, 61, 62, 75, 86, 88, 89, 92, 100, 101, 102, 103, 105, 112, 113, 114, 115, 120, 121, 122, 124, 125, 138, 139, 141, 147, 153, 159, 164, 166, 167, 168

Offset: 1

Tanya Khovanova, Oct 03 2012

The number of zeros of n! base 3 is approaching n/2 as n grows. Similarly, the number of zeros of n! base 4 is approaching n/2 as n grows. Consequently, this sequence is expected to have high density.
From Robert Israel, Jan 19 2017: (Start)
Numbers n such that A000120(n) + (n + A000120(n) mod 2) = A053735(n).
Since typically A000120(n) ~ log_2(n) while typically A053735(n) ~ log_3(n), the density of this sequence should go to 0, contrary to the previous comment. (End)
Comment from N. J. A. Sloane, Dec 06 2019: (Start)
Appears to be the same as the list of positive numbers n such that the last nonzero digit of n! in base 12 belongs to the set [1, 2, 5, 7, 10, 11].
The first footnote in Deshouillers et al. (2016) says: "if the last nonzero digit of n! in base 12 belongs to {1, 2, 5, 7, 10, 11} then |(digit-sum of n in base 3) - (digit-sum of n in base 2)| is <= 1; this seems to occur infinitely many times." Compare A096288. (End)

			6! is 222200 in base 3 and 23100 in base 4, both of them have 2 zeros at the end, so 6 is in the sequence.

Jean-Marc Deshouillers, Laurent Habsieger, Shanta Laishram, Bernard Landreau, Sums of the digits in bases 2 and 3, arXiv:1611.08180, 2016

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A054861 (base 3), A090616 (base 4), A090622, A096288.

s2:= n -> convert(convert(n,base,2),`+`):
s3:= n -> convert(convert(n,base,3),`+`):
select(n -> s2(n) + (n+s2(n) mod 2) = s3(n), [$1..1000]); # Robert Israel, Jan 19 2017

sntzQ[n_]:=Module[{f=n!},Last[Split[IntegerDigits[f,3]]]==Last[ Split[ IntegerDigits[ f,4]]]]; Select[Range[200],sntzQ] (* Harvey P. Dale, Jul 11 2020 *)

is(n)=my(L=log(n+1));sum(k=1,L\log(3),n\3^k)==sum(k=1,L\log(2),n>>k)\2 \\ Charles R Greathouse IV, Oct 04 2012

More terms from Alois P. Heinz, Oct 03 2012

A090621 Exponent of highest power of 16 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

			a(10)=2 since 10! = 3628800 = 16^2 * 14175.

Cf. A011371, A054861, A090616, A027868, A054896, A090617, A090618, A090619, A090620.

a(n) = valuation(n!, 16); \\ Michel Marcus, Jul 10 2022

a(n) = A090622(n, 16) = floor(A011371(n)/4) = floor(A090616(n)/2) = floor((floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ...)/4). Almost n/4.

cp's OEIS Frontend

A319316 Numbers k such that A090616(k) < A054861(k).

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A319317 Numbers k such that A090616(k) > A054861(k).

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A090622 Square array read by antidiagonals of highest power of k dividing n! (with n,k>1).

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A054893 a(n) = Sum_{j > 0} floor(n/4^j).

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A090617 Exponent of highest power of 8 dividing n!.

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A090618 Highest power of 9 dividing n!.

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A090619 Highest power of 12 dividing n!.

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A217445 Numbers n such that n! has the same number of terminating zeros in bases 3 and 4.