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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A292254 a(n) = A292253(A163511(n)).

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 9, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 18, 19, 16, 16, 16, 17, 16, 16, 16, 17, 32, 32, 32, 33, 32, 32, 32, 32, 32, 32, 32, 32, 36, 36, 38, 39, 32, 32, 32, 32, 32, 32, 34, 34, 32, 32, 32, 32, 32, 32, 34, 35, 64, 64, 64, 64, 64, 64, 66, 67, 64, 64, 64, 65, 64, 64, 64, 65, 64, 64, 64, 65, 64, 64, 64, 64, 72, 72, 72, 72, 76, 76
Offset: 0

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

Because A292253(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate the numbers that are either of the form 12k+1 or of the form 12k+11 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).
The AND - XOR formula just restates the fact that J(3|n) = J(-1|n)*J(-3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Cf also A292247, A292248, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292944, A292946 (for similarly constructed sequences).

Programs

Formula

a(n) = A292253(A163511(n)).
a(n) = A292264(n) AND (A292274(n) XOR A292942(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987). [See comments.]
For all n >= 0, a(n) + A292944(n) + A292256(n) = n.

A292255 a(1) = 0, and for n > 1, a(n) = 2*a(A252463(n)) + [n == 1 (mod 2)]*[J(3|n) == -1], where J is the Jacobi-symbol.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 3, 0, 0, 2, 6, 0, 12, 6, 0, 0, 25, 0, 51, 4, 4, 12, 102, 0, 0, 24, 0, 12, 205, 0, 411, 0, 12, 50, 0, 0, 822, 102, 24, 8, 1645, 8, 3291, 24, 0, 204, 6582, 0, 0, 0, 48, 48, 13165, 0, 9, 24, 100, 410, 26330, 0, 52660, 822, 8, 0, 25, 24, 105321, 100, 204, 0, 210642, 0, 421284, 1644, 0, 204, 1, 48, 842569, 16, 0, 3290, 1685138, 16, 48, 6582
Offset: 1

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

Base-2 expansion of a(n) encodes the steps where numbers that are either of the form 12k+5 or of the form 12k+7 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.
The AND - XOR formulas just restate the fact that J(3|n) = J(-1|n)*J(-3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Programs

  • Scheme
    (define (A292255 n) (if (<= n 1) 0 (+ (if (and (odd? n) (= -1 (jacobi-symbol 3 n))) 1 0) (* 2 (A292255 (A252463 n))))))

Formula

a(1) = 0, and for n > 1, a(n) = 2*a(A252463(n)) + [n == 1 (mod 2)]*[J(3|n) == -1], where J is the Jacobi-symbol, and [ ]'s are Iverson brackets, whose product gives 1 only if n is an odd number for which J(3|n) = -1, and 0 otherwise.
a(n) = A292263(n) AND (A292383(n) XOR A292945(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292263(n) AND (A292385(n) XOR A292941(n)). [See comments.]
For n >= 0, a(A163511(n)) = A292256(n).
For n >= 1, a(n) + A292253(n) + A292943(n) = A243071(n).

A292256 a(n) = A292255(A163511(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 6, 6, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 8, 8, 8, 9, 12, 12, 12, 12, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 6, 6, 0, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 2, 16, 16, 16, 16, 16, 16
Offset: 0

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

Because A292255(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate the numbers that are either of the form 12k+5 or of the form 12k+7 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).
The AND - XOR formulas just restate the fact that J(3|n) = J(-1|n)*J(-3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Cf. also A292247, A292248, A292254, A292264, A292271, A292274, A292592, A292593, A292942, A292944, A292946 (for similarly constructed sequences).

Programs

Formula

a(n) = A292255(A163511(n)).
a(n) = A292264(n) AND (A292274(n) XOR A292946(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292264(n) AND (A292271(n) XOR A292942(n)). [See comments].
For all n >= 0, a(n) + A292944(n) + A292254(n) = n.

A292382 Base-2 expansion of a(n) encodes the steps where numbers of the form 4k+2 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

Original entry on oeis.org

0, 1, 2, 2, 4, 5, 8, 4, 4, 9, 16, 10, 32, 17, 10, 8, 64, 9, 128, 18, 18, 33, 256, 20, 8, 65, 8, 34, 512, 21, 1024, 16, 34, 129, 20, 18, 2048, 257, 66, 36, 4096, 37, 8192, 66, 20, 513, 16384, 40, 16, 17, 130, 130, 32768, 17, 36, 68, 258, 1025, 65536, 42, 131072, 2049, 36, 32, 68, 69, 262144, 258, 514, 41, 524288, 36, 1048576, 4097, 18, 514, 40
Offset: 1

Views

Author

Antti Karttunen, Sep 15 2017

Keywords

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[Reverse@ NestWhileList[Function[k, Which[k == 1, 1, EvenQ@ k, k/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ k]], n, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 2, 1, 0], 2], {n, 77}] (* Michael De Vlieger, Sep 21 2017 *)
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A252463(n) = if(!(n%2),n/2,A064989(n));
    A292382(n) = if(1==n,0,(if(2==(n%4),1,0)+(2*A292382(A252463(n)))));
    
  • PARI
    a(n) = my(m=factor(n),k=-2); sum(i=1,matsize(m)[1], 1 << (primepi(m[i,1]) + (k+=m[i,2]))); \\ Kevin Ryde, Dec 11 2020
    
  • Python
    from sympy.core.cache import cacheit
    from sympy.ntheory.factor_ import digits
    from sympy import factorint, prevprime
    from operator import mul
    from functools import reduce
    def a292372(n):
        k=digits(n, 4)[1:]
        return 0 if n==0 else int("".join(['1' if i==2 else '0' for i in k]), 2)
    def a064989(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==2 else prevprime(i)**f[i] for i in f])
    def a252463(n): return 1 if n==1 else n//2 if n%2==0 else a064989(n)
    @cacheit
    def a292384(n): return 1 if n==1 else 4*a292384(a252463(n)) + n%4
    def a(n): return a292372(a292384(n))
    print([a(n) for n in range(1, 111)]) # Indranil Ghosh, Sep 21 2017
  • Scheme
    (define (A292382 n) (A292372 (A292384 n)))
    

Formula

a(n) = A292272(A156552(n)).
a(1) = 0; for n > 1, a(n) = 2*a(A252463(n)) + [n == 2 (mod 4)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 4k+2, and 0 otherwise.
a(n) = A292372(A292384(n)).
Other identities. For n >= 1:
a(n) AND A292380(n) = 0, where AND is a bitwise-AND (A004198).
a(n) + A292380(n) = A156552(n).
A000120(a(n)) + A000120(A292380(n)) = A001222(n).

A292941 a(1) = 0, a(2) = 1, and for n > 2, a(n) = 2*a(A252463(n)) + [n == 1 (mod 6)].

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 9, 4, 4, 8, 18, 8, 37, 18, 8, 8, 74, 8, 149, 16, 16, 36, 298, 16, 9, 74, 8, 36, 596, 16, 1193, 16, 36, 148, 16, 16, 2387, 298, 72, 32, 4774, 32, 9549, 72, 16, 596, 19098, 32, 19, 18, 148, 148, 38196, 16, 33, 72, 296, 1192, 76392, 32, 152785, 2386, 32, 32, 72, 72, 305571, 296, 596, 32, 611142, 32, 1222285, 4774, 16, 596, 32
Offset: 1

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

Base-2 expansion of a(n) encodes the steps where numbers of the form 6k+1 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n. An exception is the most significant bit of a(n) which corresponds with the final 1, but is shifted one bit-position towards right (less significant end).
The AND - XOR formulas just restate the fact that J(-3|n) = J(-1|n)*J(3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Programs

  • Scheme
    (define (A292941 n) (if (<= n 2) (- n 1) (+ (if (= 1 (modulo n 6)) 1 0) (* 2 (A292941 (A252463 n))))))

Formula

a(1) = 0, a(2) = 1, and for n > 2, a(n) = 2*a(A252463(n)) + [n == 1 (mod 6)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 6k+1, and 0 otherwise.
Also, for n > 2, a(n) = 2*a(A252463(n)) + [n == 1 (mod 2)]*[J(-3|n) = 1], where J is the Jacobi-symbol.
a(n) = A292263(n) AND (A292253(n) XOR A292383(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292263(n) AND (A292255(n) XOR A292385(n)). [See comments.]
For n >= 0, a(A163511(n)) = A292942(n).
For n >= 1, a(n) + A292943(n) + A292945(n) = A243071(n).

A292942 a(n) = A292941(A163511(n)).

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 9, 8, 8, 8, 9, 16, 16, 16, 16, 16, 16, 18, 19, 16, 16, 16, 16, 16, 16, 18, 18, 32, 32, 32, 33, 32, 32, 32, 33, 32, 32, 32, 32, 36, 36, 38, 39, 32, 32, 32, 33, 32, 32, 32, 32, 32, 32, 32, 33, 36, 36, 36, 37, 64, 64, 64, 64, 64, 64, 66, 67, 64, 64, 64, 64, 64, 64, 66, 66, 64, 64, 64, 65, 64, 64, 64, 65, 72, 72, 72, 72, 76, 76
Offset: 0

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

Because A292941(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are of the form 6k+1 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).
The AND - XOR formula is just a restatement of the fact that J(-3|n) = J(-1|n)*J(3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Cf. also A292247, A292248, A292254, A292256, A292264, A292271, A292274, A292592, A292593, A292944, A292946 (for similarly constructed sequences).

Programs

Formula

a(n) = A292941(A163511(n)).
a(n) = A292264(n) AND (A292254(n) XOR A292274(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987). [See comments.]
For all n >= 0, a(n) + A292944(n) + A292946(n) = n.

A292946 a(n) = A292945(A163511(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 4, 5, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 8, 8, 8, 8, 8, 8, 10, 10, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 4, 5, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0
Offset: 0

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

Because A292945(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are of the form 6k+5 in binary tree A163511 (or its mirror image tree A005940) on that trajectory which leads from the root of the tree to the node containing A163511(n).
The AND - XOR formulas just restate the fact that J(-3|n) = J(-1|n)*J(3|n), as the Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Cf. also A292247, A292248, A292254, A292256, A292264, A292271, A292274, A292592, A292593, A292942, A292944 (for similarly constructed sequences).

Programs

Formula

a(n) = A292945(A163511(n)).
a(n) = A292264(n) AND (A292256(n) XOR A292274(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292264(n) AND (A292254(n) XOR A292271(n)). [See comments.]
For all n >= 0, A292942(n) + A292944(n) + a(n) = n.

A292371 A binary encoding of 1-digits in the base-4 representation of n.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Sep 15 2017

Keywords

Examples

			   n      a(n)     base-4(n)  binary(a(n))
                  A007090(n)  A007088(a(n))
  --      ----    ----------  ------------
   1        1          1           1
   2        0          2           0
   3        0          3           0
   4        2         10          10
   5        3         11          11
   6        2         12          10
   7        2         13          10
   8        0         20           0
   9        1         21           1
  10        0         22           0
  11        0         23           0
  12        0         30           0
  13        1         31           1
  14        0         32           0
  15        0         33           0
  16        4        100         100
  17        5        101         101
  18        4        102         100
		

Crossrefs

Cf. A289813 (analogous sequence for base 3).

Programs

  • Mathematica
    Table[FromDigits[IntegerDigits[n, 4] /. k_ /; IntegerQ@ k :> If[k == 1, 1, 0], 2], {n, 0, 112}] (* Michael De Vlieger, Sep 21 2017 *)
  • Python
    from sympy.ntheory.factor_ import digits
    def a(n):
        k=digits(n, 4)[1:]
        return 0 if n==0 else int("".join('1' if i==1 else '0' for i in k), 2)
    print([a(n) for n in range(116)]) # Indranil Ghosh, Sep 21 2017
    
  • Python
    def A292371(n): return int(bin(n&~(n>>1))[:1:-2][::-1],2) # Chai Wah Wu, Jun 30 2022

Formula

a(n) = A059905(A292272(n)) = A059905(n AND A003188(n)), where AND is bitwise-AND (A004198).
For all n >= 0, A000120(a(n)) = A160381(n).

A292380 Base-2 expansion of a(n) encodes the steps where multiples of 4 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 3, 2, 0, 0, 1, 0, 0, 0, 7, 0, 4, 0, 1, 0, 0, 0, 3, 4, 0, 6, 1, 0, 0, 0, 15, 0, 0, 0, 9, 0, 0, 0, 3, 0, 0, 0, 1, 2, 0, 0, 7, 8, 8, 0, 1, 0, 12, 0, 3, 0, 0, 0, 1, 0, 0, 2, 31, 0, 0, 0, 1, 0, 0, 0, 19, 0, 0, 8, 1, 0, 0, 0, 7, 14, 0, 0, 1, 0, 0, 0, 3, 0, 4, 0, 1, 0, 0, 0, 15, 0, 16, 2, 17, 0, 0, 0, 3, 0
Offset: 1

Views

Author

Antti Karttunen, Sep 15 2017

Keywords

Examples

			For n = 4, the starting value is a multiple of four, after which follows A252463(4) = 2, and A252463(2) = 1, the end point of iteration, and neither 2 nor 1 is a multiple of four, thus a(4) = 1*(2^0) + 0*(2^1) + 0*(2^2) = 1.
For n = 8, the starting value is a multiple of four, after which follows A252463(8) = 4 (also a multiple), continuing as before as 4 -> 2 -> 1, thus a(8) = 1*(2^0) + 1*(2^1) + 0*(2^2) + 0*(2^3) = 3.
For n = 9, the starting value is not a multiple of four, after which follows A252463(9) = 4 (which is), continuing as before as 4 -> 2 -> 1, thus a(9) = 0*(2^0) + 1*(2^1) + 0*(2^2) + 0*(2^3) = 2.
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[Reverse@ NestWhileList[Function[k, Which[k == 1, 1, EvenQ@ k, k/2, True, Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ k]], n, # > 1 &] /. k_ /; IntegerQ@ k :> If[Mod[k, 4] == 0, 1, 0], 2], {n, 105}] (* Michael De Vlieger, Sep 21 2017 *)
  • PARI
    a(n) = my(m=factor(n),k=-1,ret=0); for(i=1,matsize(m)[1], ret += bitneg(0,m[i,2]-1) << (primepi(m[i,1])+k); k+=m[i,2]); ret; \\ Kevin Ryde, Dec 11 2020
  • Python
    from sympy.core.cache import cacheit
    from sympy.ntheory.factor_ import digits
    from sympy import factorint, prevprime
    from operator import mul
    from functools import reduce
    def a292370(n):
        k=digits(n, 4)[1:]
        return 0 if n==0 else int("".join(['1' if i==0 else '0' for i in k]), 2)
    def a064989(n):
        f=factorint(n)
        return 1 if n==1 else reduce(mul, [1 if i==2 else prevprime(i)**f[i] for i in f])
    def a252463(n): return 1 if n==1 else n//2 if n%2==0 else a064989(n)
    @cacheit
    def a292384(n): return 1 if n==1 else 4*a292384(a252463(n)) + n%4
    def a(n): return a292370(a292384(n))
    print([a(n) for n in range(1, 111)]) # Indranil Ghosh, Sep 21 2017
    
  • Scheme
    (define (A292380 n) (A292370 (A292384 n)))
    

Formula

a(n) = A048735(A156552(n)).
a(n) = A292370(A292384(n)).
Other identities. For n >= 1:
a(n) AND A292382(n) = 0, where AND is a bitwise-AND (A004198).
a(n) + A292382(n) = A156552(n).
A000120(a(n)) + A000120(A292382(n)) = A001222(n).
A000035(a(n)) = A121262(n).

A292945 Base-2 expansion of a(n) encodes the steps where numbers of the form 6k+5 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 2, 0, 0, 2, 5, 0, 10, 4, 0, 0, 21, 0, 42, 4, 4, 10, 85, 0, 0, 20, 0, 8, 171, 0, 342, 0, 8, 42, 1, 0, 684, 84, 20, 8, 1369, 8, 2738, 20, 0, 170, 5477, 0, 0, 0, 40, 40, 10955, 0, 8, 16, 84, 342, 21911, 0, 43822, 684, 8, 0, 17, 16, 87644, 84, 168, 2, 175289, 0, 350578, 1368, 0, 168, 3, 40, 701156, 16, 0, 2738, 1402313, 16, 40, 5476, 340, 40
Offset: 1

Views

Author

Antti Karttunen, Sep 28 2017

Keywords

Comments

The AND - XOR formulas are just a restatement of the fact that J(-3|n) = J(-1|n)*J(3|n), i.e., that Jacobi-symbol is multiplicative (also) with respect to its upper argument.

Crossrefs

Programs

  • Scheme
    (define (A292945 n) (if (<= n 1) 0 (+ (if (= 5 (modulo n 6)) 1 0) (* 2 (A292945 (A252463 n))))))

Formula

a(1) = 0, and for n > 1, a(n) = 2*a(A252463(n)) + [n == 5 (mod 6)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 6k+5, and 0 otherwise.
Also, for n > 1, a(n) = 2*a(A252463(n)) + [n == 1 (mod 2)]*[J(-3|n) = -1], where J is the Jacobi-symbol.
a(n) = A292263(n) AND (A292255(n) XOR A292383(n)), where AND is bitwise-and (A004198) and XOR is bitwise-XOR (A003987).
a(n) = A292263(n) AND (A292253(n) XOR A292385(n)). [See comments.]
For n >= 0, a(A163511(n)) = A292946(n).
For n >= 1, A292941(n) + A292943(n) + a(n) = A243071(n).
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