A038712 -id:A038712 - cp's OEIS frontend

A385005 The sum of the cubefull divisors of n.

1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 28, 1, 1, 1, 1, 57, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1, 1, 28, 1, 9, 1, 1, 1, 1, 1, 1, 1, 121, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 25, 109, 1, 1, 1

Offset: 1

Amiram Eldar, Jun 15 2025

The sum of the terms in A036966 that divide n.

The number of these divisors is A190867(n), and the largest of them is A360540(n).

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

Cf. A036966, A190867, A183097, A360540.

The sum of divisors d of n such that d is: A000593 (odd), A033634 (exponentially odd), A035316 (square), A038712 (power of 2), A048250 (squarefree), A072079 (3-smooth), A073185 (cubefree), A113061 (cube), A162296 (nonsquarefree), A183097 (powerful), A186099 (5-rough), A353900 (exponentially 2^n), this sequence (cubefull), A385006 (biquadratefree).

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

a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; (p^(e+1)-1)/(p-1) - p - if(e == 1, 0, p^2));}

Multiplicative with a(p^e) = 1 if e <= 2, and a(p^e) = ((p^(e+1)-1) / (p-1)) - p - p^2 if e >= 3.

Dirichlet g.f.: zeta(s-1) * zeta(s) * Product_{p prime} (1 - p^(s-1) + 1/p^(3*s-3)).

A038713 a(n) = n XOR (n-1), i.e., nim-sum of sequential pairs, written in binary.

Original entry on oeis.org

1, 11, 1, 111, 1, 11, 1, 1111, 1, 11, 1, 111, 1, 11, 1, 11111, 1, 11, 1, 111, 1, 11, 1, 1111, 1, 11, 1, 111, 1, 11, 1, 111111, 1, 11, 1, 111, 1, 11, 1, 1111, 1, 11, 1, 111, 1, 11, 1, 11111, 1, 11, 1, 111, 1, 11, 1, 1111, 1, 11, 1, 111, 1, 11, 1, 1111111, 1, 11, 1, 111, 1

Offset: 1

Henry Bottomley, May 02 2000

			a(6) = 11 because 110 XOR 101 = 11 base 2.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for sequences related to Nim-sums

Cf. A038712 translated to binary, A001511 translated to unary (repeated 1's).

[(10^(Valuation(n, 2)+1)-1)/9: n in [1..70]]; // Vincenzo Librandi, Mar 11 2013

Table[(10^IntegerExponent[2*n, 2] - 1)/9, {n, 100}] (* Vincenzo Librandi, Mar 11 2013 *)

a(n)=if(n<1,0, (10^(valuation(n,2)+1)-1)/9) /* Michael Somos, Apr 28 2005 */

a(n) = (10^A001511(n) - 1)/(10 - 1).
Multiplicative with a(2^e) = (10^(e+1) - 1)/9, a(p^e) = 1 if p odd.
G.f.: Sum_{k>=0} 10^k * x^(2^k) / (1 - x^(2^k)). - Ilya Gutkovskiy, Dec 15 2020
Dirichlet g.f.: zeta(s) * 2^s/(2^s - 10). - Amiram Eldar, Sep 21 2023

A091519 G.f.: Sum_{k>=0} (2^kt(1+t)/(1-t)^3, t=x^2^k).

Original entry on oeis.org

1, 6, 9, 28, 25, 54, 49, 120, 81, 150, 121, 252, 169, 294, 225, 496, 289, 486, 361, 700, 441, 726, 529, 1080, 625, 1014, 729, 1372, 841, 1350, 961, 2016, 1089, 1734, 1225, 2268, 1369, 2166, 1521, 3000, 1681, 2646, 1849, 3388, 2025, 3174, 2209, 4464, 2401, 3750

Offset: 1

Ralf Stephan, Jan 18 2004

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

Cf. A000265, A038712, A065442.

nmax:=47: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := 2^p*(2^(p+1) - 1)*(2*n-1)^2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Jan 28 2013

a[n_] := n^2*(2 - 1/2^IntegerExponent[n, 2]); Array[a, 50] (* Amiram Eldar, Nov 29 2022 *)

PARI
```
a(n)=2*n*n-n*n/2^valuation(n,2)
```

a(n)=if(n<1,0,if(n%2==0,2*a(n/2)+n^2,n^2))

a(n) = 2*n^2 - n*A000265(n) = n*A000265(n)*A038712(n).
Recurrence: a(0) = 0, a(2*n) = 2*a(n) + (2*n)^2, a(2*n+1) = (2*n+1)^2.
a((2*n-1)*2^p) = 2^p*(2^(p+1) - 1)*(2*n-1)^2, p >= 0. - Johannes W. Meijer, Jan 28 2013
Sum_{k=1..n} a(k) ~ (4/9) * n^3. - Amiram Eldar, Nov 29 2022
From Amiram Eldar, Jan 05 2023: (Start)
Multiplicative with a(2^e) = 2^e*(2^(e+1)-1), and a(p^e) = p^(2*e) for p >= 3.
Dirichlet g.f.: zeta(s-2)*2^s/(2^s-2).
Sum_{n>=1} 1/a(n) = (c-1)*Pi^2/4, where c = A065442 is Erdős-Borwein constant. (End)

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)

A334045 Bitwise NOR of binary representation of n and n-1.

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 0, 0, 6, 4, 4, 0, 2, 0, 0, 0, 14, 12, 12, 8, 10, 8, 8, 0, 6, 4, 4, 0, 2, 0, 0, 0, 30, 28, 28, 24, 26, 24, 24, 16, 22, 20, 20, 16, 18, 16, 16, 0, 14, 12, 12, 8, 10, 8, 8, 0, 6, 4, 4, 0, 2, 0, 0, 0, 62, 60, 60, 56, 58, 56, 56, 48, 54, 52, 52

Offset: 1

Christoph Schreier, Apr 13 2020

All terms are even.
a(1) = 0, a(2) = 0, and a(2^n + 1) = 2^n - 2 for n > 0. Are there any other cases where n - a(n) < 4? - Charles R Greathouse IV, Apr 13 2020
The answer to the above question is no. Write n as n = (2m+1)*k, i.e. k = A006519(n) is the highest power of 2 dividing n. If m = 0, a(n) = 0 and n - a(n) = n. If m > 0, then a(n) = 2v*k, where v is the 1's complement of m. Thus n-a(n) = (2(m-v)+1)*k. Since m in binary has a leading 1, m - v >= 1 and thus n - a(n) >= 3 with n - a(n) = 3 when n > 2, k = 1 and m - v = 1, i.e. m is a power of 2 and n is of the form 2^r + 1. - Chai Wah Wu, Apr 13 2020

			a(11) = 11 NOR 10 = bin 1011 NOR 1010 = bin 100 = 4.

Wikipedia, Bitwise operation

Cf. A038712 (n XOR n-1), A086799 (n OR n-1), A129760 (n AND n-1).

a:= n-> Bits[Nor](n, n-1):
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 13 2020

a(n) = my(x=bitor(n-1, n)); bitneg(x, #binary(x)); \\ Michel Marcus, Apr 13 2020

def norbitwise(n):
    a = str(bin(n))[2:]
    b = str(bin(n-1))[2:]
    if len(b) < len(a):
        b = '0' + b
    c = ''
    for i in range(len(a)):
        if a[i] == b[i] and a[i] == '0':
            c += '1'
        else:
            c += '0'
    return int(c,2)

def A334045(n):
    m = n|(n-1)
    return 2**(len(bin(m))-2)-1-m # Chai Wah Wu, Apr 13 2020

A339747 a(n) = (5^(valuation(n, 5) + 1) - 1) / 4.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 31

Offset: 1

Ilya Gutkovskiy, Dec 15 2020

Sum of powers of 5 dividing n.

Denominator of the quotient sigma(5*n) / sigma(n).

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

Cf. A000203, A001620, A038712, A060904, A080278, A088842, A112765, A323921, A339748.

Table[(5^(IntegerExponent[n, 5] + 1) - 1)/4, {n, 1, 100}]
nmax = 100; CoefficientList[Series[Sum[5^k x^(5^k)/(1 - x^(5^k)), {k, 0, Floor[Log[5, nmax]] + 1}], {x, 0, nmax}], x] // Rest

a(n) = (5^(valuation(n, 5) + 1) - 1)/4; \\ Amiram Eldar, Nov 27 2022

G.f.: Sum_{k>=0} 5^k * x^(5^k) / (1 - x^(5^k)).
L.g.f.: -log(Product_{k>=0} (1 - x^(5^k))).
Dirichlet g.f.: zeta(s) / (1 - 5^(1 - s)).
a(n) = sigma(n)/(sigma(5*n) - 5*sigma(n)), where sigma(n) = A000203(n). - Peter Bala, Jun 10 2022
From Amiram Eldar, Nov 27 2022: (Start)
Multiplicative with a(5^e) = (5^(e+1)-1)/4, and a(p^e) = 1 for p != 5.
Sum_{k=1..n} a(k) ~ n*log_5(n) + (1/2 + (gamma - 1)/log(5))*n, where gamma is Euler's constant (A001620). (End)

A339748 a(n) = (6^(valuation(n, 6) + 1) - 1) / 5.

Original entry on oeis.org

1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 43, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 43, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1

Offset: 1

Ilya Gutkovskiy, Dec 15 2020

nonn

Sum of powers of 6 dividing n.

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

Cf. A038712, A080278, A088842, A122841, A234959, A323921, A339747.

Table[(6^(IntegerExponent[n, 6] + 1) - 1)/5, {n, 1, 100}]
nmax = 100; CoefficientList[Series[Sum[6^k x^(6^k)/(1 - x^(6^k)), {k, 0, Floor[Log[6, nmax]] + 1}], {x, 0, nmax}], x] // Rest

G.f.: Sum_{k>=0} 6^k * x^(6^k) / (1 - x^(6^k)).
L.g.f.: -log(Product_{k>=0} (1 - x^(6^k))).
Dirichlet g.f.: zeta(s) / (1 - 6^(1 - s)).

A379007 a(n) = (n^2) XOR ((n^2)-1).

Original entry on oeis.org

1, 7, 1, 31, 1, 7, 1, 127, 1, 7, 1, 31, 1, 7, 1, 511, 1, 7, 1, 31, 1, 7, 1, 127, 1, 7, 1, 31, 1, 7, 1, 2047, 1, 7, 1, 31, 1, 7, 1, 127, 1, 7, 1, 31, 1, 7, 1, 511, 1, 7, 1, 31, 1, 7, 1, 127, 1, 7, 1, 31, 1, 7, 1, 8191, 1, 7, 1, 31, 1, 7, 1, 127, 1, 7, 1, 31, 1, 7, 1, 511, 1, 7, 1, 31, 1, 7, 1, 127, 1, 7, 1, 31, 1, 7, 1, 2047

Offset: 1

Antti Karttunen, Dec 16 2024

Antti Karttunen, Table of n, a(n) for n = 1..20000
Index entries for sequences related to binary expansion of n.

Cf. A000225, A000290, A003987, A007814, A037227, A038712.

Cf. also A283750.

Map[BitXor[#, # - 1] &, Range[100]^2] (* Paolo Xausa, Dec 18 2024 *)

PARI
```
A379007(n) = bitxor(n^2, ((n^2)-1));
```

def A379007(n): return (m:=n**2)^m-1 # Chai Wah Wu, Dec 17 2024

Multiplicative with a(p^e) = 2^(1+2*e)-1 if p = 2; 1 if p > 2.
a(n) = A038712(A000290(n)).
a(n) = A000225(A037227(n)) = (2^(1+2*A007814(n))) - 1.
Dirichlet g.f.: zeta(s) * (2^s + 2)/(2^s - 4). - Amiram Eldar, Jan 12 2025

A059249 Tersum n + (n-1); write n and n-1 in base 3 and add mod 3 with no carries.

Original entry on oeis.org

1, 0, 5, 7, 6, 2, 4, 3, 17, 19, 18, 23, 25, 24, 20, 22, 21, 8, 10, 9, 14, 16, 15, 11, 13, 12, 53, 55, 54, 59, 61, 60, 56, 58, 57, 71, 73, 72, 77, 79, 78, 74, 76, 75, 62, 64, 63, 68, 70, 69, 65, 67, 66, 26, 28, 27, 32, 34, 33, 29, 31, 30, 44, 46, 45, 50, 52, 51, 47, 49, 48, 35

Offset: 1

Henry Bottomley, Jan 22 2001

			a(21)=14 since 21 and 20 are written in base 3 as 210 and 202 and so their tersum is 112 in base 3, i.e. 9+3+2=14.

Alois P. Heinz, Table of n, a(n) for n = 1..19682

Cf. A004482, A004488, A004489, A038712.

A283750 a(n) = n^2 XOR (n + 1)^2.

Original entry on oeis.org

1, 5, 13, 25, 9, 61, 21, 113, 17, 53, 29, 233, 57, 109, 37, 481, 33, 101, 45, 249, 41, 93, 1013, 81, 49, 213, 125, 457, 89, 205, 69, 1985, 65, 197, 77, 473, 73, 253, 85, 945, 209, 117, 477, 169, 121, 4013, 229, 417, 97, 165, 1005, 185, 105, 413, 181, 1937, 241, 405, 189, 905, 153, 397, 133, 8065, 129, 389, 141, 921

Offset: 0

Ilya Gutkovskiy, Mar 15 2017

nonn

XOR the binary representations of n^2 and (n + 1)^2.

Antti Karttunen, Table of n, a(n) for n = 0..20000
Ilya Gutkovskiy, Extended graphical example
Eric Weisstein's World of Mathematics, XOR
Index entries for sequences related to binary expansion of n

Cf. A000290, A003987, A016813 (records), A038712, A112591, A169810.

Cf. also A379007.

Table[BitXor[n^2, (n + 1)^2], {n, 0, 67}]

A283750(n) = bitxor(n^2, (n+1)^2); \\ Indranil Ghosh, Mar 15 2017, Antti Karttunen, Dec 16 2024

def A283750(n): return (n**2)^(n + 1)**2 # Indranil Ghosh, Mar 15 2017

a(n) = A000290(n) XOR A000290(n+1).

cp's OEIS Frontend

A385005 The sum of the cubefull divisors of n.

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A038713 a(n) = n XOR (n-1), i.e., nim-sum of sequential pairs, written in binary.

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A091519 G.f.: Sum_{k>=0} (2^k*t*(1+t)/(1-t)^3, t=x^2^k).

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

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A334045 Bitwise NOR of binary representation of n and n-1.

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A339747 a(n) = (5^(valuation(n, 5) + 1) - 1) / 4.

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A339748 a(n) = (6^(valuation(n, 6) + 1) - 1) / 5.

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A091519 G.f.: Sum_{k>=0} (2^kt(1+t)/(1-t)^3, t=x^2^k).