A054861 -id:A054861 - cp's OEIS frontend

A180017 Difference of sums of digits of n in ternary and in binary.

0, 0, 1, -1, 1, 1, 0, 0, 3, -1, 0, 0, 0, 0, 1, -1, 3, 3, 0, 0, 2, 0, 1, 1, 2, 2, 3, -3, -1, -1, -2, -2, 3, 1, 2, 2, 0, 0, 1, -1, 2, 2, 1, 1, 3, -1, 0, 0, 2, 2, 3, 1, 3, 3, -2, -2, 1, -1, 0, 0, 0, 0, 1, -3, 3, 3, 2, 2, 4, 2, 3, 3, 2, 2, 3, 1, 3, 3, 2, 2, 6, -2, -1, -1, -1, -1, 0, -2, 1, 1, -2, -2, 0

Offset: 0

Reinhard Zumkeller, Aug 06 2010

This sequence is positive on average, since 1/log(3) > 1/log(4). Do all integers appear infinitely often? - Charles R Greathouse IV, Feb 07 2013

			For n = 7 = 21_3 = 111_2, a(n) = (2+1) - (1+1+1) = 0.
For n = 8 = 22_3 = 1000_2, a(n) = (2+2) - (1+0+0+0) = 3.
For n = 9 = 100_3 = 1001_2, a(n) = (1+0+0) - (1+0+0+1) = -1.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A180018, A180019, A007088, A007089.

Table[Total[IntegerDigits[n,3]]-Total[IntegerDigits[n,2]],{n,0,100}] (* Harvey P. Dale, Dec 08 2015 *)

a(n) = sumdigits(n,3) - sumdigits(n,2); \\ Michel Marcus, Nov 12 2023

a(n) = A053735(n) - A000120(n);
a(A037301(n)) = 0;
a(A000244(n)) = 1 - A000120(A000244(n));
a(A000079(n)) = A053735(A000079(n)) - 1;
a(A024023(n)) = 2*n - A000120(A024023(n)); a(A000225(n)) = A053735(A000225(n)) - n.
a(n) = A011371(n) - 2*A054861(n). - Henry Bottomley, Feb 16 2024

A064459 a(n) = Sum_{k>=1} floor(n/12^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8

Offset: 0

Robert G. Wilson v, Oct 03 2001

Original incorrect name was: "Highest power of 12 dividing n!": that sequence is A090619. If p is prime, Legendre's formula says the highest power of p dividing n! is Sum_{k>=1} floor(n/p^k), but of course 12 is not prime. - Robert Israel, Mar 23 2018

Harry J. Smith, Table of n, a(n) for n=0..1000
Wikipedia, Legendre's formula.

Cf. A011371, A054861, A090619.

List([0..110],n->Sum([1..n],k-?Int(n/(12^k)))); # Muniru A Asiru, Mar 24 2018

f:= proc(n) add(floor(n/12^k), k=1..floor(log[12](n))) end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Mar 23 2018

Table[t = 0; p = 12; While[s = Floor[n/p]; t = t + s; s > 0, p *= 12]; t, {n, 0, 100} ]
Join[{0},Accumulate[Table[If[Divisible[n,12],1,0],{n,110}]]] (* Harvey P. Dale, Feb 14 2016 *)

{ for (n=0, 1000, a=0; p=12; while (s = n\p, a+=s; p*=12); write("b064459.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 15 2009

a(n) = floor[n/12] + floor[n/144] + floor[n/1728] + floor[n/20736] + ....

Corrected by Robert Israel, Mar 23 2018

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.

A261237 Number of steps needed when starting from (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k) to encounter the first number whose base-3 representation begins with a digit other than 2.

Original entry on oeis.org

1, 1, 2, 5, 13, 34, 92, 251, 687, 1885, 5184, 14292, 39557, 110094, 308351, 868716, 2458964, 6984467, 19890809, 56775186, 162427605, 465816503, 1339163192, 3858600035, 11138726760, 32199805820

Offset: 0

Antti Karttunen, Aug 16 2015

a(n) = How many numbers whose base-3 representation begins with digit "2" are encountered before (3^n)-1 is reached when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k by k - (sum of digits in base-3 representation of k). Note that (3^n)-1 (in base-3: "222...", with digit "2" repeated n times) is not included in the count, although the starting point (3^(n+1))-1 is.

			For n=0, we start from 3^(0+1) - 1 = 2 (also "2" in base-3), and subtract 2 to get 0, which doesn't begin with 2, thus a(0) = 1.
For n=1, we start from 3^(1+1) - 1 = 8 ("22" in base-3), and subtract 2*2 = 4 to get 4 ("11" in base-3) which doesn't begin with 2, thus a(1) = 1.
For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract first 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), so in two steps we have reached the first such number that does not begin with "2" in base-3, thus a(2) = 2.

Cf. A000244, A054861, A122586, A261230, A261234, A261236.

/* Use the C-program given in A261234. */

Flatten@ Table[FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &]] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

a(n)=my(k=3^(n+1)-1,t=2*3^n,s); while(k>=t, k-=sumdigits(k,3); s++); s \\ Charles R Greathouse IV, Aug 21 2015

(definec (A261237 n) (let loop ((k (- (A000244 (+ 1 n)) 1)) (s 0)) (if (< (A122586 k) 2) s (loop (* 2 (A054861 k)) (+ 1 s)))))

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016

A375933 The second-largest exponent in the prime factorization of n, or 0 if it does not exist.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0

Offset: 1

Amiram Eldar, Sep 03 2024

First differs from A363127 at n = 60, and from A363131 at n = 72.

The position of the first occurrence of k = 1, 2, ..., is A167747(k+1) = 2*6^k.

			12 = 2^2 * 3^1 has 2 exponents in its prime factorization: 1 and 2. 2 is the largest and 1 is the second-largest. Therefore a(12) = 1.

Cf. A051903, A054861, A072774, A167747, A375932, A375934.

Cf. A363127, A363131.

a[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, Max[0, Max[Select[e, # < Max[e] &]]]]; Array[a, 100]

a(n) = if(n == 1, 0, my(e = factor(n)[,2]); e = select(x -> x < vecmax(e), e); if(#e == 0, 0, vecmax(e)));

a(n) = A051903(A375932(n)).
a(n) = 0 if and only if n is a power of a squarefree number (A072774).
a(n) = 1 if and only if n is in A375934.
a(n) <= A051903(n), with equality if and only if n = 1.
a(n!) = A054861(n) for n != 3.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{i >= 1} i * d(i) = 0.42745228287872473252..., where d(i) = Sum_{j >= i+1} d_2(i, j) and d_2(i, j) = Product_{p prime} (1 - 1/p^(i+1) + 1/p^j - 1/p^(j+1)) - Product_{p prime} (1 - 1/p^(i+1)) + [i > 1] * (Product_{p prime} (1 - 1/p^i) - Product_{p prime} (1 - 1/p^i + 1/p^j - 1/p^(j+1))), and [] is the Iverson bracket.

A054900 a(n) = Sum_{j >= 1} floor(n/16^j).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 0

Henry Bottomley, May 23 2000

nonn

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.

Cf. A053836, A054897, A054899, A067080, A098844, A132032.

m:=16;
function a(n) // a = A054900, m = 16
  if n eq 0 then return 0;
  else return a(Floor(n/m)) + Floor(n/m);
  end if; end function;
[a(n): n in [0..127]]; // G. C. Greubel, Apr 28 2023

a[n_, m_]:= If[n==0, 0, a[Floor[n/m], m] +Floor[n/m]];
Table[a[n, 16], {n,0,127}] (* G. C. Greubel, Apr 28 2023 *)

m=16 # a = A054900
def a(n): return 0 if (n==0) else a(n//m) + (n//m)
[a(n) for n in range(128)] # G. C. Greubel, Apr 28 2023

a(n) = (n - A053836(n))/15.
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence:
a(n) = a(floor(n/16)) + floor(n/16).
a(16*n) = a(n) + n.
a(n*16^m) = a(n) + n*(16^m - 1)/15.
a(k*16^m) = k*(16^m - 1)/15, for 0 <= k < 16, m>=0.
Asymptotic behavior:
a(n) = n/15 + O(log(n)).
a(n+1) - a(n) = O(log(n)) (this follows from the inequalities below).
a(n) <= (n-1)/15; equality holds for powers of 16.
a(n) >= (n-15)/15 - floor(log_16(n)); equality holds for n = 16^m - 1, m > 0.
Limits:
lim inf (n/15 - a(n)) = 1/15, for n --> oo.
lim sup (n/15 - log_16(n) - a(n)) = 0, for n --> oo.
lim sup (a(n+1) - a(n) - log_16(n)) = 0, for n --> oo.
Series:
G.f.: (1/(1-x))*Sum_{k > 0} x^(16^k)/(1-x^(16^k)). (End)

A089792 a(n) = n-(exponent of highest power of 3 dividing n!).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 6, 5, 6, 7, 7, 8, 9, 9, 10, 11, 10, 11, 12, 12, 13, 14, 14, 15, 16, 14, 15, 16, 16, 17, 18, 18, 19, 20, 19, 20, 21, 21, 22, 23, 23, 24, 25, 24, 25, 26, 26, 27, 28, 28, 29, 30, 28, 29, 30, 30, 31, 32, 32, 33, 34, 33

Offset: 0

Paul Boddington, Jan 09 2004

The exponent of the highest power of 3 dividing binomial(n,k) is given by a(k)+a(n-k)-a(n).

Cf. A007949, A054861, A125824.

Table[n-IntegerExponent[n!,3],{n,0,70}] (* Harvey P. Dale, Aug 09 2015 *)

vector(70, n, n--; n-valuation(n!, 3)) \\ Michel Marcus, Aug 19 2015

a(n) = a(n-1)+1-A007949(n).
a(n) = log(denominator(n!/3^n))/log(3); a(n) = log_3(A125824(n)). - Paul Barry, Apr 02 2007
a(n) = n - A054861(n).

A193260 G.f.: x+x^2 = Sum_{n>=1} x^n * ((1+x+x^2)^n - x^(2*n)) / (1+x+x^2)^a(n).

Original entry on oeis.org

1, 2, 5, 6, 7, 9, 10, 11, 15, 16, 17, 19, 20, 21, 23, 24, 25, 28, 29, 30, 32, 33, 34, 36, 37, 38, 43, 44, 45, 47, 48, 49, 51, 52, 53, 56, 57, 58, 60, 61, 62, 64, 65, 66, 69, 70, 71, 73, 74, 75, 77, 78, 79, 83, 84, 85, 87, 88, 89, 91, 92, 93, 96, 97, 98, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 114, 115, 117, 118, 119, 125, 126, 127, 129, 130, 131, 133, 134, 135

Offset: 1

Paul D. Hanna, Jul 19 2011

nonn

			G.f.: x+x^2 = x*((1+x+x^2) - x^2)/(1+x+x^2)  + x^2*((1+x+x^2)^2 - x^4)/(1+x+x^2)^2  + x^3*((1+x+x^2)^3 - x^6)/(1+x+x^2)^5  + x^4*((1+x+x^2)^4 - x^8)/(1+x+x^2)^6  + x^5*((1+x+x^2)^5 - x^10)/(1+x+x^2)^7  + x^6*((1+x+x^2)^6 - x^12)/(1+x+x^2)^9  + x^7*((1+x+x^2)^7 - x^14)/(1+x+x^2)^10  + x^8*((1+x+x^2)^8 - x^16)/(1+x+x^2)^11  + x^9*((1+x+x^2)^9 - x^18)/(1+x+x^2)^15 +...+ x^n*((1+x+x^2)^n - x^(2*n))/(1+x+x^2)^a(n) +...

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A193259, A054861.

Table[n+Floor[Log[3,n]]+IntegerExponent[n!,3],{n,90}] (* Harvey P. Dale, Oct 10 2012 *)

{a(n)=if(n<1,0,n + floor(log(n+1/2)/log(3)) + valuation(n!,3))}

{a(n)=if(n<1,0,if(n==1,1,polcoeff(sum(m=1,n+1,x^m*((1+x+x^2)^m-x^(2*m))/(1+x+x^2 +x^2*O(x^n))^if(m>=n,1,a(m)))+x^(n+1),n+1)))}

a(n) = n + floor(log_3(n)) + A054861(n) for n>=1, where A054861(n) = highest power of 3 dividing n!.

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

cp's OEIS Frontend

A180017 Difference of sums of digits of n in ternary and in binary.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A064459 a(n) = Sum_{k>=1} floor(n/12^k).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Formula

Extensions

A090618 Highest power of 9 dividing n!.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A090619 Highest power of 12 dividing n!.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A261237 Number of steps needed when starting from (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k) to encounter the first number whose base-3 representation begins with a digit other than 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

C

Mathematica

PARI

Scheme

Extensions

A375933 The second-largest exponent in the prime factorization of n, or 0 if it does not exist.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A054900 a(n) = Sum_{j >= 1} floor(n/16^j).

Views

Author

Keywords

Links

Crossrefs