A235127 -id:A235127 - cp's OEIS frontend

A309891 a(n) is the total number of trailing zeros in the representations of n over all bases b >= 2.

0, 1, 1, 3, 1, 3, 1, 5, 3, 3, 1, 6, 1, 3, 3, 8, 1, 6, 1, 6, 3, 3, 1, 9, 3, 3, 5, 6, 1, 7, 1, 10, 3, 3, 3, 11, 1, 3, 3, 9, 1, 7, 1, 6, 6, 3, 1, 13, 3, 6, 3, 6, 1, 9, 3, 9, 3, 3, 1, 12, 1, 3, 6, 14, 3, 7, 1, 6, 3, 7, 1, 15, 1, 3, 6, 6, 3, 7, 1, 13, 8, 3, 1, 12

Offset: 1

Rémy Sigrist, Aug 21 2019

a(n) depends only on the prime signature of n.

a(n) is the sum of the k-adic valuations of n for k >= 2. - Friedjof Tellkamp, Jan 25 2025

			For n = 12: 12 has 2 trailing zeros in base 2 (1100), 1 trailing zero in bases 3, 4, 6 and 12 (110, 30, 20, 10) and no trailing zero in other bases, hence a(12) = 1*2 + 4*1 = 6.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000
Index entries for sequences computed from exponents in factorization of n

Cf. A001221, A006218, A007814, A007949, A033093, A056239, A112765, A122841, A169594, A214411, A235127, A244413, A286561, A293514, A369180.
Cf. A111003.
First differences of A078632.

Table[DivisorSum[n, IntegerExponent[n, #] &, # > 1 &], {n, 84}] (* Jon Maiga, Aug 25 2019 *)

a(n) = sumdiv(n, d, if (d>1, valuation(n,d), 0))

a(n) = {if(n == 1, return(0)); my(f = factor(n)[, 2], res = 0, t = 2, of = f, nf = f >> 1, nd(v) = prod(i = 1, #v, v[i] + 1)); while(Set(of) != [0], res += (nd(of) - nd(nf)) * (t-1); of = nf; t++; nf = f \ t); res} \\ David A. Corneth, Aug 22 2019

a(n) = Sum_{d|n, d>1} A286561(n,d), where A286561 gives the d-valuation of n.
a(p) = 1 for any prime number p.
a(p^k) = A006218(k) for any k >= 0 and any prime number p.
a(n) = 2^A001221(n) - 1 for any squarefree number n.
a(n) = 3 for any semiprime number n.
a(m*n) >= a(m) + a(n).
a(n) >= A007814(n) + A007949(n) + A235127(n) + A112765(n) + A122841(n) + A214411(n) + A244413(n).
a(n) = A056239(A293514(n)). - Antti Karttunen, Aug 22 2019
a(n) <= A033093(n). - Michel Marcus, Aug 22 2019
a(n) = A169594(n) - 1. - Jon Maiga, Aug 25 2019
From Friedjof Tellkamp, Feb 27 2024: (Start)
G.f.: Sum_{k>=2, j>=1} x^(k^j)/(1-x^(k^j)).
Dirichlet g.f.: zeta(s) * Sum_{k>=1} (zeta(k*s) - 1).
Sum_{n>=1} a(n)/n^2 = Pi^2/8 (A111003). (End)

A217319 Numbers with binary representation ending in 4k+2 or 4k+3 zeros.

Original entry on oeis.org

4, 8, 12, 20, 24, 28, 36, 40, 44, 52, 56, 60, 64, 68, 72, 76, 84, 88, 92, 100, 104, 108, 116, 120, 124, 128, 132, 136, 140, 148, 152, 156, 164, 168, 172, 180, 184, 188, 192, 196, 200, 204, 212, 216, 220, 228, 232, 236, 244, 248, 252, 260, 264, 268, 276, 280

Offset: 1

Vladimir Shevelev, Mar 18 2013

Or numbers having infinitary divisor 4, or the same, having factor 4 in Fermi-Dirac representation as a product of distinct terms of A050376.
From Peter Munn, Aug 25 2020: (Start)
Compare the terms, as a set, with A145204\{0} (numbers having 3 as a Fermi-Dirac factor). The self-inverse function defined by A225546 maps the members of either one of these sets 1:1 onto the other set.
Numbers whose 4th-power-free part is divisible by 4.
(End)
Numbers k such that the exponent of the highest power of 4 dividing k, A235127(k), is odd. The asymptotic density of this sequence is 1/5. - Amiram Eldar, Sep 20 2020

Peter J. C. Moses, Table of n, a(n) for n = 1..10000

Cf. A050376, A171949, A235127.

Related to A145204\{0} via A225546.

isA007814 := proc(n)
    if modp( A007814(n),4) in {2,3} then
        true ;
    else
        false ;
    end if;
end proc:
for n from 1 to 1000 do
    if isA007814(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Nov 22 2023

okQ[n_] := (cnt = Count[ Split[ IntegerDigits[n, 2]] // Last, 0]; k0 = k /. ToRules@ Reduce[ (cnt == 2*k || cnt == 2*k+1), k, Integers]; OddQ[k0]); Select[ Range[312], okQ] (* Jean-François Alcover, Mar 18 2013 *)
Select[Map[# Boole[IntegerQ[(1/4 (1+#))]||IntegerQ[(1/4 (2+#))]&[Length[Last[Split[IntegerDigits[#,2]]]]]]&,Range[2,500,2]],#>0&]
Select[Range[280], OddQ @ IntegerExponent[#, 4] &] (* Amiram Eldar, Sep 20 2020 *)

def A217319(n):
    def f(x): return n+x-sum((k:=x>>(m<<1))-(k>>2) for m in range(0,(x.bit_length()+1>>1),2))
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return m<<2 # Chai Wah Wu, May 25 2025

Conjecture. For n>=1, a(n) = A171949(n+1).

Named edited by David A. Corneth, Sep 22 2020

A025646 Exponent of 4 (value of i) in n-th number of form 4^i*5^j.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 2, 1, 0, 5, 4, 3, 2, 1, 0, 6, 5, 4, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 8, 0, 7, 6, 5, 4, 3, 2, 9, 1, 8, 0, 7, 6, 5, 4, 3, 10, 2, 9, 1, 8, 0, 7, 6, 5, 4, 11, 3, 10, 2, 9, 1, 8, 0, 7, 6, 5, 12, 4, 11, 3, 10, 2, 9, 1, 8, 0, 7, 6, 13, 5, 12, 4, 11, 3, 10, 2, 9, 1, 8, 0, 7, 14, 6, 13

Offset: 1

David W. Wilson

nonn

Cf. A025617, A025650, A235127.

Differs from A025661 at a(1881).

With[{nn=15},IntegerExponent[#,4]&/@Union[Flatten[Table[4^x 5^y,{x,0,nn},{y,0,nn}]]]] (* Harvey P. Dale, May 10 2019 *)
With[{max = 10^10}, IntegerExponent[Sort[Flatten[Table[4^i*5^j, {i, 0, Log[4, max]}, {j, 0, Log[5, max/4^i]}]]], 4]] (* Amiram Eldar, Jul 09 2025 *)

a(n) = A235127(A025617(n)). - Amiram Eldar, Jul 09 2025

A025647 Exponent of 4 (value of i) in n-th number of form 4^i*6^j.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 3, 2, 1, 5, 0, 4, 3, 2, 6, 1, 5, 0, 4, 3, 7, 2, 6, 1, 5, 0, 4, 8, 3, 7, 2, 6, 1, 5, 9, 0, 4, 8, 3, 7, 2, 6, 10, 1, 5, 9, 0, 4, 8, 3, 7, 11, 2, 6, 10, 1, 5, 9, 0, 4, 8, 12, 3, 7, 11, 2, 6, 10, 1, 5, 9, 0, 13, 4, 8, 12, 3, 7, 11, 2, 6, 10, 1, 14, 5, 9, 0, 13, 4, 8, 12, 3, 7, 11, 2

Offset: 1

David W. Wilson

nonn

Cf. A025618, A025658, A235127.

Differs from A025653 at a(2805).

With[{max = 10^10}, SortBy[Flatten[Table[{i, j}, {i, 0, Log[4, max]}, {j, 0, Log[6, max/4^i]}], 1], 4^#[[1]] * 6^#[[2]] &][[;; , 1]]] (* Amiram Eldar, Jul 09 2025 *)

a(n) = A235127(A025618(n)/6^A025658(n)). - Amiram Eldar, Jul 09 2025

A214682 Remove 2's that do not contribute to a factor of 4 from the prime factorization of n.

Original entry on oeis.org

1, 1, 3, 4, 5, 3, 7, 4, 9, 5, 11, 12, 13, 7, 15, 16, 17, 9, 19, 20, 21, 11, 23, 12, 25, 13, 27, 28, 29, 15, 31, 16, 33, 17, 35, 36, 37, 19, 39, 20, 41, 21, 43, 44, 45, 23, 47, 48, 49, 25, 51, 52, 53, 27, 55, 28, 57, 29

Offset: 1

Tyler Ball, Jul 25 2012

In this sequence, the number 4 exhibits some characteristics of a prime number since all extraneous 2's have been removed from the prime factorizations of all other numbers.

			For n=8, v_4(8)=1, v_2(8)=3, so a(8)=(8*4^1)/(2^3)=4.
For n=12, v_4(12)=1, v_2(12)=2, so a(12)=(12*4^1)/(2^2)=12.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A214681, A214685.
Range of values: A003159.
Missing values: A036554.
A056832, A059895, A073675 are used in a formula defining this sequence.
A059897 is used to express relationship between terms of this sequence.
Cf. A007814 (v_2(n)), A235127 (v_4(n)).

a[n_] := n/(2^Mod[IntegerExponent[n, 2], 2]); Array[a, 100] (* Amiram Eldar, Dec 09 2020 *)

a(n)=n>>(valuation(n,2)%2) \\ Charles R Greathouse IV, Jul 26 2012

def A214682(n): return n>>1 if (~n&n-1).bit_length()&1 else n # Chai Wah Wu, Jan 09 2023

C = []
for i in [1..n]:
    C.append(i*4^(Integer(i).valuation(4))/2^(Integer(i).valuation(2)))

a(n) = (n*4^(v_4(n)))/(2^(v_2(n))) where v_k(n) is the k-adic valuation of n. That is, v_k(n) is the largest power of k, a, such that k^a divides n.
For n odd, a(n)=n since n has no factors of 2 (or 4).
From Peter Munn, Nov 29 2020: (Start)
a(A003159(n)) = n.
a(A036554(n)) = n/2.
a(n) = n/A056832(n) = n/A059895(n, 2) = min(n, A073675(n)).
a(A059897(n, k)) = A059897(a(n), a(k)). (End)
Multiplicative with a(2^e) = 2^(2*floor(e/2)), and a(p^e) = p^e for odd primes p. - Amiram Eldar, Dec 09 2020
Sum_{k=1..n} a(k) ~ (5/12) * n^2. - Amiram Eldar, Nov 10 2022
Dirichlet g.f.: zeta(s-1)*(2^s+1)/(2^s+2). - Amiram Eldar, Dec 30 2022

A369180 Alternating sum of the k-adic valuations (ruler functions) of n.

Original entry on oeis.org

0, 1, -1, 3, -1, 1, -1, 5, -3, 1, -1, 4, -1, 1, -3, 8, -1, 0, -1, 4, -3, 1, -1, 7, -3, 1, -5, 4, -1, 1, -1, 10, -3, 1, -3, 5, -1, 1, -3, 7, -1, 1, -1, 4, -6, 1, -1, 11, -3, 0, -3, 4, -1, -1, -3, 7, -3, 1, -1, 6, -1, 1, -6, 14, -3, 1, -1, 4, -3, 1, -1, 9, -1, 1, -6, 4, -3, 1

Offset: 1

Friedjof Tellkamp, Jan 15 2024

sign

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Cf. A007814, A007949, A222171, A235127, A309891.

a:= n-> add((-1)^i*padic[ordp](n, i), i=2..n):
seq(a(n), n=1..78);  # Alois P. Heinz, Jan 15 2024

z = 70; Sum[(-1)^k IntegerExponent[Range[z], k], {k, 2, z}]

a(n) = sum(k=2, n, (-1)^k * valuation(n,k)); \\ Michel Marcus, Jan 18 2024

a(n)=sumdiv(n,k, if(k>1, (-1)^k * valuation(n, k))) \\ Charles R Greathouse IV, Jan 23 2025

a(n) = Sum_{k=2..n} (-1)^k * valuation(n,k).
a(n) = A007814(n) - A007949(n) + A235127(n) - (...).
G.f.: Sum_{k>=2, j>=1} (-1)^k x^(k^j)/(1-x^(k^j)).
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{n=1..m} a(n) = log(2).
Dirichlet g.f.: zeta(s) * Sum_{k>=1} (1 - eta(ks)).
Sum_{n>=1} a(n)/n^2 = Pi^2/24.

A323921 a(n) = (4^(valuation(n, 4) + 1) - 1) / 3.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 21, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 21, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 21, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 85, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 21, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 21, 1, 1, 1, 5

Offset: 1

Ilya Gutkovskiy, Dec 15 2020

Sum of powers of 4 dividing n.

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

Cf. A001620, A038712, A080278, A088842, A234957, A235127, A339747, A339748.

Table[(4^(IntegerExponent[n, 4] + 1) - 1)/3, {n, 1, 100}]
nmax = 100; CoefficientList[Series[Sum[4^k x^(4^k)/(1 - x^(4^k)), {k, 0, Floor[Log[4, nmax]] + 1}], {x, 0, nmax}], x] // Rest

a(n) = (4^(valuation(n, 4) + 1) - 1) / 3; \\ Michel Marcus, Jul 09 2022

def A323921(n): return ((1<<((~n&n-1).bit_length()&-2)+2)-1)//3 # Chai Wah Wu, Jul 09 2022

G.f.: Sum_{k>=0} 4^k * x^(4^k) / (1 - x^(4^k)).
L.g.f.: -log(Product_{k>=0} (1 - x^(4^k))).
Dirichlet g.f.: zeta(s) / (1 - 4^(1 - s)).
From Amiram Eldar, Nov 27 2022: (Start)
Multiplicative with a(2^e) = (4^floor((e+2)/2)-1)/3, and a(p^e) = 1 for p != 2.
Sum_{k=1..n} a(k) ~ n*log_4(n) + (1/2 + (gamma - 1)/log(4))*n, where gamma is Euler's constant (A001620). (End)

A258059 Let n = Sum_{i=0..k} d_i*4^i be the base-4 expansion of n, with 0 <= d_i < 4. Then a(n) = minimal i such that d_i is not 1, or k+1 if there is no such i.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1

Offset: 1

Richard C. Webster, May 17 2015

This is the "General Ruler Sequence Base 4 Focused at 1" of Webster (2015).

			1 = 0*4+1, so a(1)=1.
7 = 1*4+3, so a(7)=0.
21 = 0*4^3+1*4^2+1*4+1, so a(21)=3.
523 base 10 is 20023 in base 4, so a(523)=0.
1365 base 10 is 111111 in base 4, so a(1365)=6.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
Richard C. Webster, One Sequence to Rule Them All: The Ruler Sequence and Its Relation to Odd Perfect Numbers and Multiplicative Order, MS Thesis, California State Polytechnic University, Pomona, CA, 2015.

The nonzero terms give A263845.
This sequence and A263845 are analogs of the pair of ruler sequences A007814 and A001511.
Cf. A007090, A235127.
Cf. A030386.

a258059 = f 0 . a030386_row where
   f i [] = i
   f i (t:ts) = if t == 1 then f (i + 1) ts else i
-- Reinhard Zumkeller, Nov 08 2015

f:= proc(n)
  if n mod 4 = 1 then procname((n-1)/4) + 1 else 0 fi
end proc:
map(f, [$1..1000]); # Robert Israel, Jun 08 2015

a(n) = {v = Vecrev(digits(n, 4)); for (i=1, #v, if (v[i] != 1, return (i-1));); return(#v);}

Recurrence: a(1)=1; thereafter a(4*n+1) = a(n)+1, a(4*n+j) = 0 for j = 0,2,3. G.f. g(x) = Sum_{k>=0} k * x^((4^k-1)/3) * (1 + x^(2*4^k) + x^(3*4^k))/(1 - x^(4*4^k)) satisfies g(x) = x*g(x^4) + x/(1-x^4). - Robert Israel, Jun 08 2015

Edited by N. J. A. Sloane, Oct 31 2015 and Nov 06 2015.

A339455 a(n) is the X-coordinate of the n-th point of the space filling curve H defined in Comments section; A339456 gives Y-coordinates.

Original entry on oeis.org

0, 1, 0, 1, 1, 1, 2, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 3, 3, 3, 4, 3, 4, 3, 2, 3, 2, 3, 3, 3, 4, 3, 4, 3, 2, 1, 0, 1, 0, 1, 1, 1, 2, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 2, 3, 3, 3, 4, 3, 4, 3, 2, 3, 2, 3, 3, 3, 4, 3, 4, 4, 4

Offset: 0

Rémy Sigrist, Dec 06 2020

nonn

We consider a hexagonal lattice with X-axis and Y-axis as follows:
           Y
          /
         /
        0 ---- X
We define the family {H_n, n > 0} as follows:
- T_1 contains the origin (0, 0) and (1, 0), in that order:
          +-->--+
         O
- for any n > 0, H_{n+1} is built from 4 copies of H_n connected with 2^(n+1) unit segments as follows:
               +->-2->-+
                \     /
                 ^   v
                  \ /
           +->-1->-+->-4->-+
          O       / \
                 v   ^
                /     \
               +->-3->-+
- H is the limit of H_n as n tends to infinity,
- H visits once every unit segment (u, v) where u and v are lattice points and at least one of u or v belongs to the region { (x, y) | x > 0 or x + y > 0 },
- the n-th segment of curve H has length 2^A235127(n).

Rémy Sigrist, Table of n, a(n) for n = 0..3008
Rémy Sigrist, Illustration of H_3
Rémy Sigrist, PARI program for A339455
Index entries for sequences related to coordinates of 2D curves

Cf. A235127, A339456.

PARI
```
See Links section.
```

A358345 a(n) is the number of even square divisors of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 2

Offset: 1

Amiram Eldar, Nov 11 2022

nonn

First differs from A235127 at n = 36.

The first position of k >= 0 in this sequence is A187941(k)^2.

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

Cf. A016742, A046951, A187941, A222171, A235127, A298735.

f1[p_, e_] := Floor[e/2] + 1; f2[p_, e_] := If[p == 2, 1, Floor[e/2] + 1]; a[1] = 0; a[n_] := Times @@ f1 @@@ (fct = FactorInteger[n]) - Times @@ f2 @@@ fct; Array[a, 100]

a(n) = {my(f = factor(n)); prod(i=1, #f~, 1+f[i,2]\2) - prod(i=1, #f~, if(f[i,1] == 2, 1, 1+f[i,2]\2))};

a(n) = sumdiv(n, d, if (issquare(d) && !(d%2), 1)); \\ Michel Marcus, Nov 11 2022

a(n) = A046951(n) - A298735(n).
a(n) = 2 * A046951(n) - A046951(4*n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi^2/24 (A222171).

cp's OEIS Frontend

A309891 a(n) is the total number of trailing zeros in the representations of n over all bases b >= 2.

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A217319 Numbers with binary representation ending in 4*k+2 or 4*k+3 zeros.

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A025646 Exponent of 4 (value of i) in n-th number of form 4^i*5^j.

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A025647 Exponent of 4 (value of i) in n-th number of form 4^i*6^j.

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A214682 Remove 2's that do not contribute to a factor of 4 from the prime factorization of n.

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A369180 Alternating sum of the k-adic valuations (ruler functions) of n.

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A323921 a(n) = (4^(valuation(n, 4) + 1) - 1) / 3.

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A217319 Numbers with binary representation ending in 4k+2 or 4k+3 zeros.