A092391 -id:A092391 - cp's OEIS frontend

A228090 Numbers k for which a sum k + bitcount(k) cannot be obtained as a sum k2 + bitcount(k2) for any other k2<>k . Here bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

0, 1, 2, 5, 6, 7, 8, 9, 10, 13, 18, 21, 22, 23, 24, 25, 26, 30, 33, 37, 38, 39, 40, 41, 42, 45, 50, 53, 54, 55, 56, 57, 58, 61, 63, 64, 66, 69, 70, 71, 72, 73, 74, 77, 82, 85, 86, 87, 88, 89, 90, 94, 97, 101, 102, 103, 104, 105, 106, 109, 114, 117, 118, 119, 120

Offset: 1

Antti Karttunen, Aug 17 2013

In other words, numbers k such that A228085(A092391(k)) = 1.

			0 is in this sequence because the sum 0+A000120(0)=0 cannot be obtained with any other value of k than k=0.
1 is in this sequence because the sum 1+A000120(1)=2 cannot be obtained with any other value of k than k=1.
2 is in this sequence because the sum 2+A000120(2)=3 cannot be obtained with any other value of k than k=2.
3 is not in this sequence because the sum 3+A000120(3)=5 can also be obtained with value k=4, as also 4+A000120(4)=5.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for Colombian or self numbers and related sequences

Sequence A228089 sorted into ascending order. Complement: A228236.

Cf. also A092391, A228085, A228088.

A228236 Numbers k for which a sum k+bitcount(k) can be also obtained as a sum k2 +bitcount(k2) for some other k2<>k . Here bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

Original entry on oeis.org

3, 4, 11, 12, 14, 15, 16, 17, 19, 20, 27, 28, 29, 31, 32, 34, 35, 36, 43, 44, 46, 47, 48, 49, 51, 52, 59, 60, 62, 65, 67, 68, 75, 76, 78, 79, 80, 81, 83, 84, 91, 92, 93, 95, 96, 98, 99, 100, 107, 108, 110, 111, 112, 113, 115, 116, 123, 124, 125, 126, 127, 128

Offset: 1

Antti Karttunen, Aug 17 2013

nonn

In other words, numbers k such that A228085(A092391(k)) > 1.

			0 is not in this sequence because the sum 0+A000120(0)=0 cannot be obtained with any other value of k than k=0.
1 is not in this sequence because the sum 1+A000120(1)=2 cannot be obtained with any other value of k than k=1.
2 is not in this sequence because the sum 2+A000120(2)=3 cannot be obtained with any other value of k than k=2.
3 IS in this sequence because the sum 3+A000120(3)=5 can also be obtained with value k=4, as also 4+A000120(4)=5, and thus also 4 is in this sequence.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Complement: A228090. Subsets: A228091, A228237. Cf. also A092391, A228085.

A230297 a(n) = A010062(n) written in binary: a(n+1) = a(n) + hammingweight(a(n)) in binary.

Original entry on oeis.org

1, 10, 11, 101, 111, 1010, 1100, 1110, 10001, 10011, 10110, 11001, 11100, 11111, 100100, 100110, 101001, 101100, 101111, 110100, 110111, 111100, 1000000, 1000001, 1000011, 1000110, 1001001, 1001100, 1001111, 1010100, 1010111, 1011100, 1100000, 1100010, 1100101, 1101001, 1101101, 1110010, 1110110, 1111011, 10000001, 10000011

Offset: 0

N. J. A. Sloane, Oct 17 2013

Is there any way to tell by looking at a binary number whether or not it is a term of this sequence?

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Index entries for Colombian or self numbers and related sequences.

Cf. A010062.
Essentially the same as A157845.
Cf. A004207 (base-10 analog); A007088 (n in binary), A010062 (this written in base 10), A000120 (Hammingweight), A092391 (A000120(n) + n), A028897 (convert binary to decimal).

s[0] = 1; s[n_] := s[n] = s[n-1] + DigitCount[s[n-1], 2, 1]; Table[FromDigits[IntegerDigits[s[n], 2]], {n, 0, 50}] (* Amiram Eldar, Jul 28 2023 *)

(A230297(n)=A007088(A010062(n))); A230297_vec(N)={vector(N,i, if(i>1, A007088(N+=hammingweight(N)), N=1))} \\ M. F. Hasler, Nov 18 2019

a(n) = A157845(n+1) = A007088(A010062(n)) = A007088(A092391(A028897(a(n-1)))). - M. F. Hasler, Nov 18 2019

A157845 a(0) = 1, a(n) = sum of binary digits of all prior terms, expressed in binary.

Original entry on oeis.org

1, 1, 10, 11, 101, 111, 1010, 1100, 1110, 10001, 10011, 10110, 11001, 11100, 11111, 100100, 100110, 101001, 101100, 101111, 110100, 110111, 111100, 1000000, 1000001, 1000011, 1000110, 1001001, 1001100, 1001111, 1010100, 1010111, 1011100, 1100000, 1100010

Offset: 0

Oliver K. Seet, Mar 07 2009

Equals A230297 = A010062 converted from decimal to binary, prefixed by another initial 1. - M. F. Hasler, Nov 18 2019

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A004207 (base-10 analog); A007088 (n in binary), A010062 (this written in base 10), A000120 (Hammingweight), A092391 (A000120(n) + n), A028897 (convert binary to decimal).

b:= proc(n) option remember; `if`(n<2, 1, b(n-1)+
      add(i, i=convert(a(n-1), base, 10)))
    end:
a:= n-> convert(b(n), binary):
seq(a(n), n=0..44);  # Alois P. Heinz, Nov 18 2019

s[0] = s[1] = 1; s[n_] := s[n] = s[n-1] + DigitCount[s[n-1], 2, 1]; Table[FromDigits[IntegerDigits[s[n], 2]], {n, 0, 50}] (* Amiram Eldar, Jul 28 2023 *)

lista(nn) = {my(s = 1); my(t = 1); print1(t, ", "); for (i=1, nn, sb = binary(s); t = subst(Pol(sb), x, 10); print1(t, ", "); s += hammingweight(sb););}

apply( A157845(n)=fromdigits(binary(A010062(n-!!n))), [0..40]) \\ M. F. Hasler, Nov 18 2019

a(n) = A230297(n-1) = A007088(A010062(n-1)) = A007088(A092391(A028897(a(n-1)))) for n > 0. - M. F. Hasler, Nov 18 2019

a(11) corrected and extended by R. J. Mathar, Mar 12 2009

More terms from Michel Marcus, Apr 19 2014

A228089 Integers k for which a sum k + bitcount(k) cannot be obtained as a sum k2 + bitcount(k2) for any other k2<>k.

Original entry on oeis.org

0, 1, 2, 5, 6, 8, 7, 9, 10, 13, 18, 21, 22, 24, 23, 25, 26, 30, 33, 37, 38, 40, 39, 41, 42, 45, 50, 53, 54, 56, 55, 57, 58, 64, 61, 66, 63, 69, 70, 72, 71, 73, 74, 77, 82, 85, 86, 88, 87, 89, 90, 94, 97, 101, 102, 104, 103, 105, 106, 109, 114, 117, 118, 120, 119

Offset: 1

Antti Karttunen, Aug 17 2013

nonn

The values of k's are sorted here according to the magnitude of the sum k + bitcount(k), where bitcount(k) (= A000120) gives the number of 1's in binary representation of nonnegative integer k; a(n) = A228086(A228088(n)).

			6 is in this sequence because the sum 6+A000120(6)=8 cannot be obtained with any other value of k than k=6.
8 is in this sequence because the sum 8+A000120(8)=9 cannot be obtained with any other value of k than k=8.
7 is in this sequence because the sum 7+A000120(7)=10 cannot be obtained with any other value of k than k=7.
In this sequence 8 becomes before 7 because 8+A000120(8) < 7+A000120(7).

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for Colombian or self numbers and related sequences

A228090 gives the same terms sorted into ascending order.

(define (A228089 n) (A228086 (A228088 n)))

a(n) = A228086(A228088(n)).

A092391(a(n)) = A228088(n).

A228237 Numbers n for which there exists such a natural number k > n that k + bitcount(k) = n + bitcount(n), where bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

Original entry on oeis.org

3, 11, 14, 15, 19, 27, 29, 31, 35, 43, 46, 47, 51, 59, 62, 67, 75, 78, 79, 83, 91, 93, 95, 99, 107, 110, 111, 115, 123, 124, 125, 126, 127, 131, 139, 142, 143, 147, 155, 157, 159, 163, 171, 174, 175, 179, 187, 190, 195, 203, 206, 207, 211, 219, 221, 223, 227

Offset: 1

Antti Karttunen, Sep 11 2013

nonn

In other words, all such terms A228236(n) which satisfy A228236(n) < A228087(A092391(A228236(n))).

Note: 124 is the first term that occurs both here and in A228091.

			For cases 0 + A000120(0) = 0, 1 + A000120(1) = 2, 2 + A000120(2) = 3 there are no larger solutions yielding the same result.
However, for 3 + A000120(3) = 5 there is a larger solution yielding the same result, namely 4 + A000120(4) = 5, thus 3 is the first term of this sequence.
Next time this occurs for 11, as 11 + A000120(11) = 14 = 12 + A000120(12), and 12 > 11.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Subset of A228236. Cf. also A228091.

A230300 a(n) = n + wt(n-1), where wt() = A000120() is the binary weight.

Original entry on oeis.org

1, 3, 4, 6, 6, 8, 9, 11, 10, 12, 13, 15, 15, 17, 18, 20, 18, 20, 21, 23, 23, 25, 26, 28, 27, 29, 30, 32, 32, 34, 35, 37, 34, 36, 37, 39, 39, 41, 42, 44, 43, 45, 46, 48, 48, 50, 51, 53, 51, 53, 54, 56, 56, 58, 59, 61, 60, 62, 63, 65, 65, 67, 68, 70, 66, 68, 69, 71, 71, 73, 74, 76, 75, 77, 78, 80, 80, 82, 83, 85, 83, 85

Offset: 1

N. J. A. Sloane, Oct 23 2013

A228085(m-1) gives the number of times m occurs in this sequence.

Index entries for Colombian or self numbers and related sequences

Cf. A000120, A092391, A228085, A230301 (complement).

a:= n-> n+add(i,i=Bits[Split](n-1)):
seq(a(n), n=1..82);  # Alois P. Heinz, Jul 05 2024

Table[n+Total[IntegerDigits[n-1,2]],{n,100}] (* Harvey P. Dale, May 20 2015 *)

a(n) = n + hammingweight(n-1); \\ Michel Marcus, Jul 05 2024

A320814 Approximation of the 2-adic integer exp(4) up to 2^n.

Original entry on oeis.org

0, 1, 1, 5, 13, 13, 13, 77, 77, 333, 333, 333, 333, 333, 333, 16717, 16717, 16717, 147789, 409933, 934221, 934221, 934221, 934221, 9322829, 9322829, 9322829, 9322829, 143540557, 411976013, 948846925, 948846925, 948846925, 948846925, 9538781517, 9538781517

Offset: 0

Jianing Song, Oct 21 2018

nonn

In p-adic field, the exponential function exp(x) is defined as Sum_{k>=0} x^k/k!.
Let |x|A007814(x)%20be%20the%202-adic%20valuation%20of%20x.%20For%20any%202-adic%20number%20x,%20exp(x)%20=%20Sum">2 be the 2-adic metric of x, and v(x, 2) = A007814(x) be the 2-adic valuation of x. For any 2-adic number x, exp(x) = Sum{i>=0} x^i/i! converges implies that lim_{k->+oo} |x^k/k!|2 = 0, that is, lim{k->+oo} v(x^k/k!, 2) = +oo, or lim_{k->+oo} (k*(v(x, 2) - 1) + A000120(i)) = +oo. So v(x, 2) > 1, x is a 2-adic integer divisible by 4. On the other hand, for any integer n and i >= A320840(n), v(4^i/i!, 2) = A092391(i) >= n, so approximation of exp(4) up to 2^n is wholly determined by Sum_{i=0..A320840(n)-1} 4^i/i! (see formula section below), which is well-defined because it has only finitely many terms.
When extended to a function over the metric completion of the p-adic field, exp(x) has radius of convergence p^(-1/(p-1)).
a(n) is the multiplicative inverse of A321689(n) modulo 2^n. - Jianing Song, Nov 17 2018

			A320840(1) = 1, 4^0/0! = 1, so a(1) = 1.
A320840(4) = 3, Sum_{i=0..2} 4^i/i! = 13, so a(4) = 13.
A320840(6) = 5, Sum_{i=0..4} 4^i/i! = 103/3 == 13 (mod 64), so a(6) = 13.
A320840(8) = 6, Sum_{i=0..5} 4^i/i! = 643/15 == 77 (mod 256), so a(8) = 77.
A320840(9) = 7, Sum_{i=0..6} 4^i/i! = 437/9 == 333 (mod 512), so a(9) = 333.

Jianing Song, Table of n, a(n) for n = 0..2000
Wikipedia, p-adic number

Cf. A000120, A007814, A092391, A320815, A320840, A321689.

a(n) = lift(sum(i=0, n-1-(n>=2), Mod(4^i/i!, 2^n)))

a(n) = lift(exp(4 + O(2^n))); \\ Andrew Howroyd, Nov 05 2018

If Sum_{i=0..A320840(n)-1} 4^i/i! = p/q, gcd(p, q) = 1, then a(n) = p*q^(-1) mod 2^n.

a(n) = Sum_{i=0..n-1} A320815(i)*2^i.

A348367 a(n) = w(n + w(n)), where w(n) is the binary weight of n, A000120(n).

Original entry on oeis.org

1, 2, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 1, 2, 3, 2, 3, 2, 3, 3, 2, 3, 4, 3, 3, 4, 5, 5, 2, 2, 2, 2, 3, 2, 3, 3, 2, 3, 4, 3, 3, 4, 5, 5, 3, 3, 3, 3, 3, 4, 5, 5, 4, 4, 4, 5, 5, 5, 1, 1, 2, 3, 3, 2, 3, 2, 3, 3, 2, 3, 4, 3, 3, 4, 5, 5, 3, 3, 3, 3, 3, 4, 5, 5, 4, 4, 4, 5, 5, 5, 2, 2, 3, 4, 4, 3, 3, 4, 5

Offset: 1

Ctibor O. Zizka, Oct 15 2021

			n = 5; a(5) = A000120(5 + A000120(5)) = 3.

Cf. A000120, A092391.

h[n_] := DigitCount[n, 2, 1]; a[n_] := h[n + h[n]]; Array[a, 100] (* Amiram Eldar, Oct 15 2021 *)

a(n) = hammingweight(n + hammingweight(n)); \\ Michel Marcus, Oct 17 2021

def h(n): return bin(n).count('1')
def a(n): return h(n + h(n))
print([a(n) for n in range(1, 100)]) # Michael S. Branicky, Oct 15 2021

a(n) = A000120(n + A000120(n)); a(n) = A000120(A092391(n)).

A350229 a(n) is the sum of n and the balanced ternary digits in n.

Original entry on oeis.org

0, 2, 2, 4, 6, 4, 6, 8, 8, 10, 12, 12, 14, 16, 12, 14, 16, 16, 18, 20, 20, 22, 24, 22, 24, 26, 26, 28, 30, 30, 32, 34, 32, 34, 36, 36, 38, 40, 40, 42, 44, 38, 40, 42, 42, 44, 46, 46, 48, 50, 48, 50, 52, 52, 54, 56, 56, 58, 60, 58, 60, 62, 62, 64, 66, 66, 68

Offset: 0

Rémy Sigrist, Jan 09 2022

The image of this sequence is the set of nonnegative even numbers (A005843).

			For n = 42:
- the balanced ternary representation of 42 is "1TTT0",
- so a(42) = 42 + 1 - 1 - 1 - 1 + 0 = 40.

See A062028, A092391, A230641 for similar sequences.

Cf. A005843, A065363, A174658 (fixed points).

Array[# + Total[If[First@ # == 0, Rest@ #, #] &[Prepend[IntegerDigits[#, 3], 0] //. {x___, y_, k_ /; k > 1, z___} :> {x, y + 1, k - 3, z}]] &, 70, 0] (* Michael De Vlieger, Jan 15 2022 *)

a(n) = my (v=n, d); while (n, n=(n-d=[0,1,-1][1+n%3])/3; v+=d); v

a(n) = n + A065363(n).

a(n) = n iff n belongs to A174658.

cp's OEIS Frontend

A228090 Numbers k for which a sum k + bitcount(k) cannot be obtained as a sum k2 + bitcount(k2) for any other k2<>k . Here bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

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A228236 Numbers k for which a sum k+bitcount(k) can be also obtained as a sum k2 +bitcount(k2) for some other k2<>k . Here bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

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A230297 a(n) = A010062(n) written in binary: a(n+1) = a(n) + hammingweight(a(n)) in binary.

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A157845 a(0) = 1, a(n) = sum of binary digits of all prior terms, expressed in binary.

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A228089 Integers k for which a sum k + bitcount(k) cannot be obtained as a sum k2 + bitcount(k2) for any other k2<>k.

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A228237 Numbers n for which there exists such a natural number k > n that k + bitcount(k) = n + bitcount(n), where bitcount(k) (A000120) gives the number of 1's in binary representation of nonnegative integer k.

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A230300 a(n) = n + wt(n-1), where wt() = A000120() is the binary weight.

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A320814 Approximation of the 2-adic integer exp(4) up to 2^n.

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