cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A067590 Number of partitions of n into odious numbers (A000069).

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 7, 11, 12, 16, 19, 25, 29, 37, 44, 56, 64, 79, 92, 112, 130, 157, 182, 217, 250, 296, 341, 401, 461, 539, 615, 717, 817, 945, 1076, 1240, 1409, 1614, 1830, 2091, 2365, 2692, 3040, 3451, 3885, 4398, 4942, 5581, 6260, 7050, 7896, 8870, 9916
Offset: 0

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Author

Naohiro Nomoto, Jan 31 2002

Keywords

Crossrefs

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<1, 0,
         `if`(add(i, i=Bits[Split](n))::odd, n, h(n-1)))
        end:
    b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
          b(n, h(i-1))+b(n-i, h(min(n-i, i))))
        end:
    a:= n-> b(n, h(n)):
    seq(a(n), n=0..100);  # Alois P. Heinz, Sep 08 2019
  • Mathematica
    f[n_] := f[n] = DigitCount[n, 2][[1]] // OddQ // Boole;
    etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d p[d], {d, Divisors[j]}] b[n - j], {j, 1, n}]/n]; b];
    a[n_] := etr[f][n];
    a /@ Range[0, 100] (* Jean-François Alcover, Nov 23 2020 *)

Formula

Euler Transform of characteristic function of A000069.
Conjecture: a(n) ~ c * exp(Pi*sqrt(n/3)) / n^(5/8), where c = 0.221864833... - Vaclav Kotesovec, Jul 06 2025

Extensions

a(0) corrected by Alois P. Heinz, Sep 08 2019

A116490 Numbers that can be written as a product of evil numbers (A001969) and of odious numbers (A000069).

Original entry on oeis.org

1, 25, 50, 69, 77, 81, 87, 100, 115, 117, 138, 145, 147, 154, 162, 169, 171, 174, 175, 200, 209, 213, 230, 231, 234, 245, 261, 265, 275, 276, 285, 287, 289, 290, 294, 308, 324, 325, 329, 333, 338, 339, 342, 343, 345, 348, 350, 351, 355, 357, 369, 375, 385
Offset: 1

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Author

Christian G. Bower, Feb 17 2006

Keywords

Comments

Numbers such that A116488(n) != 0 and A116489(n) != 0.
If m is in the sequence then so is 2*m for m > 1. - David A. Corneth, Mar 27 2021

Examples

			77 is in the sequence as it's the product of odious numbers via 7*11 and of evil numbers via 1*77. - _David A. Corneth_, Mar 27 2021
		

Crossrefs

A143579 Permutation of the natural numbers (0,1,2,3,...): Odious numbers (A000069) interleaved with Evil numbers (A001969).

Original entry on oeis.org

1, 0, 2, 3, 4, 5, 7, 6, 8, 9, 11, 10, 13, 12, 14, 15, 16, 17, 19, 18, 21, 20, 22, 23, 25, 24, 26, 27, 28, 29, 31, 30, 32, 33, 35, 34, 37, 36, 38, 39, 41, 40, 42, 43, 44, 45, 47, 46, 49, 48, 50, 51, 52, 53, 55, 54, 56, 57, 59, 58, 61, 60, 62, 63, 64, 65, 67, 66, 69, 68, 70, 71, 73
Offset: 0

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Author

Gary W. Adamson, Aug 24 2008

Keywords

Comments

This is the lodumo_2 of the Thue-Morse sequence A010059, where lodumo_k of sequences is defined in A159970. [Philippe Deléham, Apr 29 2009]

Crossrefs

Formula

Antidiagonals of an array, Odious numbers (A000069) in row 1 and Evil numbers (A001969) in row 2: 1, 2, 4, 7, 8, 11, 13,... 0, 3, 5, 6, 9, 10, 12,...

Extensions

Extended by R. J. Mathar, Sep 05 2008
Edited by R. J. Mathar, Feb 04 2010

A178253 a(n) is the least positive integer m such that (A001969(n))^m is in A000069.

Original entry on oeis.org

4, 2, 4, 2, 2, 4, 4, 2, 2, 2, 2, 4, 5, 2, 4, 2, 2, 2, 2, 2, 2, 3, 2, 4, 2, 4, 5, 5, 2, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 4, 3, 2, 2, 5, 4, 5, 3, 2, 5, 2, 3, 4, 3, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 3, 4, 2, 2, 2, 2, 4, 2, 3, 3, 2, 3, 2, 2, 4, 3, 2, 3, 2, 2, 2, 3, 2, 5
Offset: 2

Views

Author

Vladimir Shevelev, May 24 2010, May 25 2010

Keywords

Comments

Conjecture: a(n) exists for all n >= 2.
The values of A001969(n)^m are given in A178373. [From Jeremy Gardiner, May 26 2010]

Crossrefs

Programs

  • PARI
    a(n)=my(e=2*n--+hammingweight(n)%2,m=1); e>>=valuation(e,2);  while(hammingweight(e^m++)%2==0,);m \\ Charles R Greathouse IV, Mar 22 2013

Formula

If k=A001969(n)=2^m*A001969(s), where A001969(s) is odd, then a(n)=a(s).

Extensions

Extended by Jeremy Gardiner, May 26 2010
Edited by N. J. A. Sloane, Jun 08 2010

A227932 Evil numbers (A001969) which can be written as a product of two odious numbers (A000069).

Original entry on oeis.org

77, 147, 154, 169, 175, 209, 231, 245, 275, 287, 294, 308, 325, 329, 338, 343, 350, 399, 407, 413, 418, 427, 441, 455, 462, 469, 483, 490, 525, 533, 550, 553, 567, 574, 588, 605, 609, 616, 649, 650, 658, 676, 679, 686, 700, 703, 715, 735, 759, 763, 777, 798
Offset: 1

Views

Author

Vladimir Shevelev, Oct 15 2013

Keywords

Examples

			Evil number 275 = 25*11. Since 25 and 11 are odious, then 275 is in the sequence.
		

Crossrefs

Programs

  • Mathematica
    evilQ[n_] := EvenQ[DigitCount[n, 2][[1]]]; odiousQ[n_] := OddQ[DigitCount[n, 2][[1]]]; fQ[n_] := Module[{f, i}, If[PrimeQ[n], False, f = Select[Divisors[n], # > 1 && # <= Sqrt[n] &]; i = 1; While[i <= Length[f] && ! (odiousQ[f[[i]]] && odiousQ[n/f[[i]]]), i++]; i <= Length[f]]]; Select[Range[1000], evilQ[#] && fQ[#] &] (* T. D. Noe, Oct 16 2013 *)

Extensions

Extended by T. D. Noe, Oct 16 2013

A230070 a(n) is the number of odious integers (A000069) not exceeding n and respectively prime to n.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 2, 5, 2, 5, 3, 6, 3, 8, 4, 9, 4, 9, 5, 8, 5, 12, 5, 12, 6, 13, 5, 15, 5, 15, 8, 14, 8, 12, 8, 18, 9, 17, 8, 20, 8, 22, 10, 19, 11, 23, 11, 18, 11, 24, 12, 27, 12, 21, 10, 25, 14, 29, 11, 30, 15, 24, 16, 26, 13, 33, 17, 32, 12, 36, 16, 36
Offset: 1

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Author

Vladimir Shevelev, Oct 10 2013

Keywords

Comments

Let b(n) is the number of evil integers (A001969) not exceeding n and respectively prime to n. Then a(n) + b(n) = phi(n) (phi = A000010). For which numbers a(n) < b(n)? This sequence begins 28,... . For n = 1,2,3,15, we have a(n) = phi(n). What other solutions has this equation? When a(n) = phi(n)/2, we call n a balanced number. The sequence of balanced numbers begins 4,6,7,8,10,11,13,14,16,19,22,...

Examples

			For n = 30, we have the following numbers respectively prime to n: 1, 7, 11, 13, 17, 19, 23, 29, from which only 5 numbers 1, 7, 11, 13 and 19 are odious. So, a(30) = 5.
		

Crossrefs

Programs

  • Mathematica
    odiouses=Select[Range[rng=100],OddQ[DigitCount[#,2][[1]]]&]; tmp=1; Table[Count[Map[CoprimeQ[n,#]&, Take[odiouses, tmp=NestWhile[#+1&,tmp+1, odiouses[[#]]
    				
  • PARI
    a(n) = sum(k = 1, n, gcd(k, n) == 1 && hammingweight(k) % 2); \\ Amiram Eldar, Nov 10 2024

Formula

For odd prime p, a(p) = (p + 1 or - 1)/2. Primes p for which a(p) = (p+1)/2 are 3, 5, 17, 23, 29,..., i.e., evil primes (A027699), while odd primes p for which a(p) = (p-1)/2 are 7,11,13,19,..., i.e., odious primes (A027697).

A230095 Odious numbers (A000069) that are the product of exactly two distinct primes.

Original entry on oeis.org

14, 21, 22, 26, 35, 38, 55, 62, 69, 74, 82, 87, 91, 93, 94, 115, 118, 122, 133, 134, 143, 145, 146, 155, 158, 161, 185, 194, 203, 205, 206, 213, 214, 217, 218, 247, 253, 254, 259, 262, 265, 274, 295, 299, 301, 302, 309, 314, 319, 321, 327, 334, 339, 341, 346
Offset: 1

Views

Author

N. J. A. Sloane, Oct 11 2013

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Range[400],OddQ[DigitCount[#,2,1]]&&PrimeNu[#]==PrimeOmega[#]==2&] (* Harvey P. Dale, May 21 2024 *)
  • PARI
    isodious(n)=b = binary(n); sum(i=1, #b, b[i]==1) % 2;
    isok(n) = isodious(n) && (bigomega(n)==2) && (omega(n)==2); \\ Michel Marcus, Oct 12 2013
    
  • PARI
    list(lim)=my(v=List()); forprime(p=2,lim\2, forprime(q=2,min(lim\p,p-1), if(hammingweight(p*q)%2, listput(v,p*q)))); Set(v) \\ Charles R Greathouse IV, Jan 31 2017

Extensions

More terms from Michel Marcus, Oct 12 2013

A230384 Table read by rows: Least set of n odious numbers (A000069) such that any two or more add to an evil number (A001969); ordered by total sum of elements, then by largest element(s).

Original entry on oeis.org

1, 1, 2, 2, 7, 8, 4, 19, 49, 67, 42, 84, 138, 174, 357, 168, 348, 372, 702, 906, 1407
Offset: 1

Views

Author

M. F. Hasler, Oct 17 2013

Keywords

Comments

Row sums are given in A230387. See A230385 for a "dual" version.
Is this sequence finite, or is there for any n at least one admissible set of n odious numbers, i.e., such that any sum of two or more elements add up to an evil number?

Examples

			For n=1 to 4, we have the sets
n=1: {1} with sum = 1,
n=2: {1, 2} with sum = 3
n=3: {2, 7, 8} with sum = 17,
n=4: {4, 19, 49, 67} with sum = 139.
E.g., for n=3, the numbers 2, 7 and 8 have an odd bit sum, but 2+7, 2+8, 7+8 and 2+7+8 all have an odd bit sum.
For n=4, we also have the admissible set {14, 31, 44, 61} which has a smaller maximal element, but a larger total sum.
n=5: {42, 84, 138, 174, 357} with sum = 795.
n=6: {168, 348, 372, 702, 906, 1407} with sum = 3903.
		

Crossrefs

Extensions

a(11)-a(21) from Charles R Greathouse IV, Oct 18 2013

A231271 Numbers k such that k and k+d are both odious (A000069) or both evil (A001969) for every divisor d of k.

Original entry on oeis.org

1, 5, 7, 9, 13, 17, 23, 29, 31, 37, 41, 49, 53, 61, 71, 73, 85, 89, 97, 101, 103, 109, 113, 119, 125, 127, 129, 133, 137, 149, 151, 157, 167, 173, 181, 193, 197, 199, 217, 223, 229, 233, 241, 249, 257, 263, 269, 277, 281, 293, 311, 313, 317, 321, 325, 337, 341
Offset: 1

Views

Author

Vladimir Shevelev, Nov 06 2013

Keywords

Comments

A prime p is a term iff its binary expansion ends in odd number of 1's (A095283). All terms are in A079523.

Examples

			The odious number k = 341 has divisors {1, 11, 31, 341}. Since the numbers 341 + 1 = 342, 341 + 11 = 352, 341 + 31 = 372, 341 + 341 = 682 are all odious, then 341 is a term.
		

Crossrefs

Programs

  • Mathematica
    odiousQ[n_] := OddQ[DigitCount[n, 2][[1]]];selQ[n_] := Length[Union[Map[odiousQ, Flatten[{n, Map[n+#&, Divisors[n]]}]]]] == 1; Select[Range[200], selQ] (* Peter J. C. Moses, Nov 08 2013 *)
  • PARI
    is(k) = {my(hw = hammingweight(k) % 2); fordiv(k, d, if(hammingweight(k+d) % 2 != hw, return(0))); 1;} \\ Amiram Eldar, Aug 12 2024

Extensions

More terms from Peter J. C. Moses, Nov 08 2013

A232637 Odious numbers of order 2: a(n) = A000069(A000069(n)).

Original entry on oeis.org

1, 2, 7, 13, 14, 21, 25, 26, 31, 37, 41, 42, 49, 50, 55, 61, 62, 69, 73, 74, 81, 82, 87, 93, 97, 98, 103, 109, 110, 117, 121, 122, 127, 133, 137, 138, 145, 146, 151, 157, 161, 162, 167, 173, 174, 181, 185, 186, 193, 194, 199, 205, 206, 213, 217, 218, 223, 229, 233, 234, 241, 242, 247, 253
Offset: 1

Views

Author

Gerasimov Sergey, Nov 27 2013

Keywords

Comments

Odious numbers with odious subscripts.
From Antti Karttunen, Nov 29 2013: (Start)
Starting from 4 and iterating A000069(4), A000069(A000069(4)), A000069(A000069(A000069(4))), etc. gives A004119 from its second term onward: 4, 7, 13, 25, 49, 97, 193, ..., which is thus a subsequence of this sequence from the term 7 onward.
Proof: All of the terms A004119(n) are odious although A004119(n)-1 is evil, and the formula for A000069(n) reduces to a(n) = 2n - 1 when n-1 is evil, and iterating that formula starting from 4 gives A004119 from 7 onward (cf. Philippe Deléham's formula there dated Feb 20 2004).
(End)
These numbers are never multiples of 4. Probably there are infinitely many multiples of m in this sequence for any m not divisible by 4. Equivalently, A233419(n) > 0 for all n. - Charles R Greathouse IV, Dec 05 2013

Examples

			The first odious number, A000069(1) = 1, and A000069(1) = 1, so a(1) = 1.
The second odious number, A000069(2) = 2, and A000069(2) = 2, so a(2) = 2.
Those were the only fixed points of A000069, and after that, we have:
The third odious number, A000069(3) = 4, and A000069(4) = 7, thus a(3) = 7.
The fourth odious number, A000069(4) = 7, and A000069(7) = 13, thus a(4) = 13.
		

Crossrefs

A004119 from term 7 onward is a subsequence.
Subsequence of A042968.

Programs

Formula

a(n) = A000069(A000069(n)).
4n-6 <= a(n) <= 4n-3, see PARI script. - Charles R Greathouse IV, Dec 05 2013
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