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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A372471 Irregular triangle read by rows where row n lists the binary indices of the n-th prime number.

Original entry on oeis.org

2, 1, 2, 1, 3, 1, 2, 3, 1, 2, 4, 1, 3, 4, 1, 5, 1, 2, 5, 1, 2, 3, 5, 1, 3, 4, 5, 1, 2, 3, 4, 5, 1, 3, 6, 1, 4, 6, 1, 2, 4, 6, 1, 2, 3, 4, 6, 1, 3, 5, 6, 1, 2, 4, 5, 6, 1, 3, 4, 5, 6, 1, 2, 7, 1, 2, 3, 7, 1, 4, 7, 1, 2, 3, 4, 7, 1, 2, 5, 7, 1, 4, 5, 7, 1, 6, 7
Offset: 1

Views

Author

Gus Wiseman, May 07 2024

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			We have prime(12) = (2^1 + 2^3 + 2^6)/2, so row 12 is (1,3,6).
Each prime followed by its binary indices:
   2: 2
   3: 1 2
   5: 1 3
   7: 1 2 3
  11: 1 2 4
  13: 1 3 4
  17: 1 5
  19: 1 2 5
  23: 1 2 3 5
  29: 1 3 4 5
  31: 1 2 3 4 5
  37: 1 3 6
  41: 1 4 6
  43: 1 2 4 6
  47: 1 2 3 4 6
		

Crossrefs

Row lengths are A014499.
Second column is A023506(n) + 1.
Final column is A035100.
Prime-indexed rows of A048793.
Row-sums are A372429, restriction of A029931 (sum of binary indices).
A019565 gives Heinz number of binary indices, adjoint A048675.
A029837 gives greatest binary index, least A001511.
A048793 lists binary indices, length A000120, reverse A272020.
A070939 gives length of binary expansion.

Programs

  • Mathematica
    Table[Join@@Position[Reverse[IntegerDigits[Prime[n],2]],1],{n,15}]

A372475 Length of binary expansion (or number of bits) of the n-th squarefree number.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8
Offset: 1

Views

Author

Gus Wiseman, May 09 2024

Keywords

Examples

			The 10th squarefree number is 14, with binary expansion (1,1,1,0), so a(10) = 4.
		

Crossrefs

For prime instead of squarefree we have A035100, 1's A014499, 0's A035103.
Restriction of A070939 to A005117.
Run-lengths are A077643.
For weight instead of length we have A372433 (restrict A000120 to A005117).
For zeros instead of length we have A372472, firsts A372473.
Positions of first appearances are A372540.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.

Programs

  • Mathematica
    IntegerLength[Select[Range[1000],SquareFreeQ],2]
  • Python
    from math import isqrt
    from sympy import mobius
    def A372475(n):
        def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
        m, k = n, f(n)
        while m != k:
            m, k = k, f(k)
        return int(m).bit_length() # Chai Wah Wu, Aug 02 2024

Formula

a(n) = A070939(A005117(n)).
a(n) = A372472(n) + A372433(n).

A124757 Zero-based weighted sum of compositions in standard order.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 3, 3, 4, 5, 6, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14
Offset: 0

Views

Author

Keywords

Comments

The standard order of compositions is given by A066099.
Sum of all positions of 1's except the last in the reversed binary expansion of n. For example, the reversed binary expansion of 14 is (0,1,1,1), so a(14) = 2 + 3 = 5. Keeping the last position gives A029931. - Gus Wiseman, Jan 17 2023

Examples

			Composition number 11 is 2,1,1; 0*2+1*1+2*1 = 3, so a(11) = 3.
The table starts:
  0
  0
  0 1
  0 1 2 3
		

Crossrefs

Cf. A066099, A070939, A029931, A011782 (row lengths), A001788 (row sums).
Row sums of A048793 if we delete the last part of every row.
For prime indices instead of standard comps we have A359674, rev A359677.
Positions of first appearances are A359756.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reverse A030308.
A230877 adds up positions of 1's in binary expansion, length A000120.
A359359 adds up positions of 0's in binary expansion, length A023416.

Programs

  • Mathematica
    Table[Total[Most[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,30}]

Formula

For a composition b(1),...,b(k), a(n) = Sum_{i=1..k} (i-1)*b(i).
For n>0, a(n) = A029931(n) - A070939(n).

A371572 Irregular triangle read by rows: row n lists the 1-based positions (starting from the left) of ones in the binary expansion of n, with row 0 = 0.

Original entry on oeis.org

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

Views

Author

Paolo Xausa, Mar 28 2024

Keywords

Examples

			Triangle begins:
  [0] 0;
  [1] 1;
  [2] 1;
  [3] 1, 2;
  [4] 1;
  [5] 1, 3;
  [6] 1, 2;
  [7] 1, 2, 3;
  [8] 1;
  ...
Row n = 50 is 1, 2, 5:
  binary expansion of 50: 1 1 0 0 1 0
  positions of ones:      1 2 - - 5 -
		

Crossrefs

Cf. A230877 (row sums), A371571 (position of zeros).
Cf. A048793.

Programs

  • Mathematica
    Join[{{0}}, Array[Flatten[Position[IntegerDigits[#, 2], 1]] &, 50]]

A372540 Least k such that the k-th squarefree number has binary expansion of length n. Index of the smallest squarefree number >= 2^n.

Original entry on oeis.org

1, 2, 4, 7, 12, 21, 40, 79, 158, 315, 625, 1246, 2492, 4983, 9963, 19921, 39845, 79689, 159361, 318726, 637462, 1274919, 2549835, 5099651, 10199302, 20398665, 40797328, 81594627, 163189198, 326378285, 652756723, 1305513584, 2611027095, 5222054082, 10444108052
Offset: 0

Views

Author

Gus Wiseman, May 10 2024

Keywords

Examples

			The squarefree numbers A005117(a(n)) together with their binary expansions and binary indices begin:
       1:                  1 ~ {1}
       2:                 10 ~ {2}
       5:                101 ~ {1,3}
      10:               1010 ~ {2,4}
      17:              10001 ~ {1,5}
      33:             100001 ~ {1,6}
      65:            1000001 ~ {1,7}
     129:           10000001 ~ {1,8}
     257:          100000001 ~ {1,9}
     514:         1000000010 ~ {2,10}
    1027:        10000000011 ~ {1,2,11}
    2049:       100000000001 ~ {1,12}
    4097:      1000000000001 ~ {1,13}
    8193:     10000000000001 ~ {1,14}
   16385:    100000000000001 ~ {1,15}
   32770:   1000000000000010 ~ {2,16}
   65537:  10000000000000001 ~ {1,17}
  131073: 100000000000000001 ~ {1,18}
		

Crossrefs

Counting zeros instead of length gives A372473, firsts of A372472.
For prime instead of squarefree we have:
- zeros A372474, firsts of A035103
- ones A372517, firsts of A014499
- bits A372684, firsts of A035100
Positions of first appearances in A372475, run-lengths A077643.
For weight instead of length we have A372541, firsts of A372433.
Indices of the squarefree numbers listed by A372683.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A005117 lists squarefree numbers.
A030190 gives binary expansion, reversed A030308.
A070939 counts bits, binary length, or length of binary expansion.

Programs

  • Mathematica
    nn=1000;
    ssnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[Max@@#]&];
    dcs=IntegerLength[Select[Range[nn],SquareFreeQ],2];
    Table[Position[dcs,i][[1,1]],{i,ssnm[dcs]}]
  • Python
    from itertools import count
    from math import isqrt
    from sympy import mobius, factorint
    def A372540(n): return next(sum(mobius(a)*(k//a**2) for a in range(1, isqrt(k)+1)) for k in count(1<Chai Wah Wu, May 12 2024

Formula

A005117(a(n)) = A372683(n).
a(n) = A143658(n)+1 for n > 1. - Chai Wah Wu, Aug 26 2024

Extensions

a(24)-a(34) from Chai Wah Wu, May 12 2024

A372473 Least k such that the k-th squarefree number has exactly n zeros in its binary expansion.

Original entry on oeis.org

1, 2, 7, 12, 21, 40, 79, 158, 315, 1247, 1246, 2492, 4983, 9963, 19921, 39845, 79689, 159361, 318726, 637462, 1274919, 2549835, 5099651, 10199302, 20398665, 40797328, 81594627, 163189198, 326378285, 652756723, 1305513584, 2611027095, 5222054082, 10444108052
Offset: 0

Views

Author

Gus Wiseman, May 09 2024

Keywords

Comments

Note that the data is not strictly increasing.

Examples

			The squarefree numbers A005117(a(n)) together with their binary expansions and binary indices begin:
     1:              1 ~ {1}
     2:             10 ~ {2}
    10:           1010 ~ {2,4}
    17:          10001 ~ {1,5}
    33:         100001 ~ {1,6}
    65:        1000001 ~ {1,7}
   129:       10000001 ~ {1,8}
   257:      100000001 ~ {1,9}
   514:     1000000010 ~ {2,10}
  2051:   100000000011 ~ {1,2,12}
  2049:   100000000001 ~ {1,12}
  4097:  1000000000001 ~ {1,13}
  8193: 10000000000001 ~ {1,14}
		

Crossrefs

Positions of first appearances in A372472.
For prime instead of squarefree we have A372474, A035103, A372517, A014499.
Counting bits (length) gives A372540, firsts of A372475, runs A077643.
Counting 1's (weight) instead of 0's gives A372541, firsts of A372433.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A005117 lists squarefree numbers.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A070939 gives length of binary expansion (number of bits).
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.

Programs

  • Mathematica
    nn=10000;
    spnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[0,Max@@#]&];
    dcs=DigitCount[Select[Range[nn],SquareFreeQ],2,0];
    Table[Position[dcs,i][[1,1]],{i,0,spnm[dcs]}]
  • Python
    from math import isqrt
    from itertools import count
    from sympy import factorint, mobius
    from sympy.utilities.iterables import multiset_permutations
    def A372473(n):
        if n==0: return 1
        for l in count(n):
            m = 1<Chai Wah Wu, May 10 2024

Extensions

a(23)-a(33) from Chai Wah Wu, May 10 2024

A372474 Least k such that the k-th prime number has exactly n zeros in its binary expansion.

Original entry on oeis.org

2, 1, 8, 7, 19, 32, 99, 55, 174, 310, 565, 1029, 1902, 3513, 6544, 6543, 23001, 43395, 82029, 155612, 295957, 564164, 1077901, 3957811, 3965052, 7605342, 14630844, 28194383, 54400029, 105097568, 393615809, 393615807, 762939128, 1480206930, 2874398838, 5586502349
Offset: 0

Views

Author

Gus Wiseman, May 11 2024

Keywords

Examples

			The prime numbers A000040(a(n)) together with their binary expansions and binary indices begin:
         3:                          11 ~ {1,2}
         2:                          10 ~ {2}
        19:                       10011 ~ {1,2,5}
        17:                       10001 ~ {1,5}
        67:                     1000011 ~ {1,2,7}
       131:                    10000011 ~ {1,2,8}
       523:                  1000001011 ~ {1,2,4,10}
       257:                   100000001 ~ {1,9}
      1033:                 10000001001 ~ {1,4,11}
      2053:                100000000101 ~ {1,3,12}
      4099:               1000000000011 ~ {1,2,13}
      8209:              10000000010001 ~ {1,5,14}
     16417:             100000000100001 ~ {1,6,15}
     32771:            1000000000000011 ~ {1,2,16}
     65539:           10000000000000011 ~ {1,2,17}
     65537:           10000000000000001 ~ {1,17}
    262147:         1000000000000000011 ~ {1,2,19}
    524353:        10000000000001000001 ~ {1,7,20}
   1048609:       100000000000000100001 ~ {1,6,21}
   2097169:      1000000000000000010001 ~ {1,5,22}
   4194433:     10000000000000010000001 ~ {1,8,23}
   8388617:    100000000000000000001001 ~ {1,4,24}
  16777729:   1000000000000001000000001 ~ {1,10,25}
  67108913: 100000000000000000000110001 ~ {1,5,6,27}
  67239937: 100000000100000000000000001 ~ {1,18,27}
		

Crossrefs

Positions of first appearances in A035103.
For squarefree instead of prime we have A372473, firsts of A372472.
Counting ones (weight) gives A372517, firsts of A014499.
Counting squarefree bits gives A372540, firsts of A372475, runs A077643.
Counting squarefree ones gives A372541, firsts of A372433.
Counting bits (length) gives A372684, firsts of A035100.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A070939 gives length of binary expansion (number of bits).

Programs

  • Mathematica
    nn=10000;
    spnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[0,Max@@#]&];
    dcs=DigitCount[Select[Range[nn],PrimeQ],2,0];
    Table[Position[dcs,i][[1,1]],{i,0,spnm[dcs]}]
  • Python
    from itertools import count
    from sympy import isprime, primepi
    from sympy.utilities.iterables import multiset_permutations
    def A372474(n):
        for l in count(n):
            m = 1<Chai Wah Wu, May 13 2024

Formula

a(n) = A000720(A066195(n)). - Robert Israel, May 13 2024

Extensions

a(22)-a(35) from and offset corrected by Chai Wah Wu, May 13 2024

A359402 Numbers whose binary expansion and reversed binary expansion have the same sum of positions of 1's, where positions in a sequence are read starting with 1 from the left.

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 70, 73, 78, 85, 93, 99, 107, 119, 127, 129, 150, 153, 165, 189, 195, 219, 231, 255, 257, 266, 273, 282, 294, 297, 310, 313, 325, 334, 341, 350, 355, 365, 371, 381, 387, 397, 403, 413, 427, 443, 455, 471
Offset: 1

Views

Author

Gus Wiseman, Jan 05 2023

Keywords

Comments

Also numbers whose binary expansion and reversed binary expansion have the same sum of partial sums.
Also numbers whose average position of a 1 in their binary expansion is (c+1)/2, where c is the number of digits.
Conjecture: Also numbers whose binary expansion has as least squares fit a line of zero slope, counted by A222955.

Examples

			The binary expansion of 70 is (1,0,0,0,1,1,0), with positions of 1's {1,5,6}, while the reverse positions are {2,3,7}. Both sum to 12, so 70 is in the sequence.
		

Crossrefs

Binary words of this type appear to be counted by A222955.
For greater instead of equal sums we have A359401.
These are the indices of 0's in A359495.
A030190 gives binary expansion, reverse A030308.
A048793 lists partial sums of reversed standard compositions, sums A029931.
A070939 counts binary digits, 1's A000120.
A326669 lists numbers with integer mean position of a 1 in binary expansion.

Programs

  • Mathematica
    Select[Range[0,100],#==0||Mean[Join@@Position[IntegerDigits[#,2],1]]==(IntegerLength[#,2]+1)/2&]
  • Python
    from functools import reduce
    from itertools import count, islice
    def A359402_gen(startvalue=0): # generator of terms
        return filter(lambda n:(r:=reduce(lambda c, d:(c[0]+d[0]*(e:=int(d[1])),c[1]+e),enumerate(bin(n)[2:],start=1),(0,0)))[0]<<1==(n.bit_length()+1)*r[1],count(max(startvalue,0)))
    A359402_list = list(islice(A359402_gen(),30)) # Chai Wah Wu, Jan 08 2023

Formula

A230877(a(n)) = A029931(a(n)).

A372472 Number of zeros in the binary expansion of the n-th squarefree number.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, May 09 2024

Keywords

Examples

			The 12th squarefree number is 17, with binary expansion (1,0,0,0,1), so a(12) = 3.
		

Crossrefs

Positions of first appearances are A372473.
Restriction of A023416 to A005117.
For prime instead of squarefree we have A035103, ones A014499, bits A035100.
Counting 1's instead of 0's (so restrict A000120 to A005117) gives A372433.
For binary length we have A372475, run-lengths A077643.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.

Programs

Formula

a(n) = A023416(A005117(n)).
a(n) + A372433(n) = A070939(A005117(n)) = A372475(n).

A359495 Sum of positions of 1's in binary expansion minus sum of positions of 1's in reversed binary expansion, where positions in a sequence are read starting with 1 from the left.

Original entry on oeis.org

0, 0, -1, 0, -2, 0, -2, 0, -3, 0, -2, 1, -4, -1, -3, 0, -4, 0, -2, 2, -4, 0, -2, 2, -6, -2, -4, 0, -6, -2, -4, 0, -5, 0, -2, 3, -4, 1, -1, 4, -6, -1, -3, 2, -5, 0, -2, 3, -8, -3, -5, 0, -7, -2, -4, 1, -9, -4, -6, -1, -8, -3, -5, 0, -6, 0, -2, 4, -4, 2, 0, 6
Offset: 0

Views

Author

Gus Wiseman, Jan 05 2023

Keywords

Comments

Also the sum of partial sums of reversed binary expansion minus sum of partial sums of binary expansion.

Examples

			The binary expansion of 158 is (1,0,0,1,1,1,1,0), with positions of 1's {1,4,5,6,7} with sum 23, reversed {2,3,4,5,8} with sum 22, so a(158) = 1.
		

Crossrefs

Indices of positive terms are A359401.
Indices of 0's are A359402.
A030190 gives binary expansion, reverse A030308.
A070939 counts binary digits.
A230877 adds up positions of 1's in binary expansion, reverse A029931.

Programs

  • Maple
    a:= n-> (l-> add(i*(l[-i]-l[i]), i=1..nops(l)))(Bits[Split](n)):
    seq(a(n), n=0..127);  # Alois P. Heinz, Jan 09 2023
  • Mathematica
    sap[q_]:=Sum[q[[i]]*(2i-Length[q]-1),{i,Length[q]}];
    Table[sap[IntegerDigits[n,2]],{n,0,100}]
  • Python
    def A359495(n):
        k = n.bit_length()-1
        return sum((i<<1)-k for i, j in enumerate(bin(n)[2:]) if j=='1') # Chai Wah Wu, Jan 09 2023

Formula

a(n) = A029931(n) - A230877(n).
If n = Sum_{i=1..k} q_i * 2^(i-1), then a(n) = Sum_{i=1..k} q_i * (2i-k-1).
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