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.

Showing 1-6 of 6 results.

A292383 Base-2 expansion of a(n) encodes the steps where numbers of the form 4k+3 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 5, 0, 0, 4, 11, 4, 22, 10, 5, 0, 44, 0, 89, 8, 8, 22, 179, 8, 0, 44, 1, 20, 358, 10, 717, 0, 20, 88, 11, 0, 1434, 178, 45, 16, 2868, 16, 5737, 44, 8, 358, 11475, 16, 0, 0, 89, 88, 22950, 2, 17, 40, 176, 716, 45901, 20, 91802, 1434, 17, 0, 40, 40, 183605, 176, 356, 22, 367211, 0, 734422, 2868, 1, 356, 22, 90, 1468845, 32, 0, 5736, 2937691, 32
Offset: 1

Views

Author

Antti Karttunen, Sep 15 2017

Keywords

Examples

			For n = 3, the starting value is of the form 4k+3, after which follows A252463(3) = 2, and A252463(2) = 1, the end point of iteration, and neither 2 nor 1 is of the form 4k+3, thus a(3) = 1*(2^0) + 0*(2^1) + 0*(2^2) = 1.
For n = 5, the starting value is not of the form 4k+3, after which follows A252463(5) = 3 (which is), continuing as before as 3 -> 2 -> 1, thus a(5) = 0*(2^0) + 1*(2^1) + 0*(2^2) + 0*(2^3) = 2.
For n = 10, the starting value is not of the form 4k+3, after which follows A252463(10) = 5 (also not 4k+3), and then A252463(5) = 3 (which is), continuing as before as 3 -> 2 -> 1, thus a(10) = 0*(2^0) + + 0*(2^1) + 1*(2^2) + 0*(2^3) + 0*(2^4) = 4.
		

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] == 3, 1, 0], 2], {n, 84}] (* 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));
    A292383(n) = if(1==n,0,(if(3==(n%4),1,0)+(2*A292383(A252463(n)))));
    
  • Scheme
    (define (A292383 n) (A292373 (A292384 n)))

Formula

a(1) = 0; for n > 1, a(n) = 2*a(A252463(n)) + [n ≡ 3 (mod 4)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 4k+3, and 0 otherwise.
a(n) = A292373(A292384(n)).
a(n) = A292274(A243071(n)).
Other identities. For n >= 1:
a(2n) = 2*a(n).
a(n) + A292385(n) = A243071(n).
a(A163511(n)) = A292274(n).
A000120(a(n)) = A292377(n).

A292381 Base-2 expansion of a(n) encodes the steps where numbers of the form 4k+1 are encountered when map x -> A252463(x) is iterated down to 1, starting from x=n.

Original entry on oeis.org

1, 2, 4, 4, 9, 8, 18, 8, 9, 18, 36, 16, 73, 36, 16, 16, 147, 18, 294, 36, 37, 72, 588, 32, 19, 146, 16, 72, 1177, 32, 2354, 32, 73, 294, 32, 36, 4709, 588, 144, 72, 9419, 74, 18838, 144, 33, 1176, 37676, 64, 39, 38, 292, 292, 75353, 32, 74, 144, 589, 2354, 150706, 64, 301413, 4708, 72, 64, 147, 146, 602826, 588, 1177, 64, 1205652, 72, 2411305, 9418, 36, 1176
Offset: 1

Views

Author

Antti Karttunen, Sep 15 2017

Keywords

Examples

			For n = 1, the starting value (which is also the ending point) is of the form 4k+1, thus a(1) = 1*(2^0) = 1.
For n = 2, the starting value is not of the form 4k+1, but its parent, A252463(2) = 1 is, thus a(2) = 0*(2^0) + 1*(2^1) = 2.
For n = 3, the starting value is not of the form 4k+1, after which follows 2 (also not 4k+1), and then 2 -> 1, and it is only the end-point of iteration which is of the form 4k+1, thus a(3) = 0*(2^0) + 0*(2^1) + 1*(2^2) = 4.
For n = 5, the starting value is of the form 4k+1, after which follows A252463(5) = 3 (which is not), and then continuing as before as 3 -> 2 -> 1, thus a(5) = 1*(2^0) + 0*(2^1) + 0*(2^2) + 1*(2^3) = 9.
		

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] == 1, 1, 0], 2], {n, 76}] (* 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));
    A292381(n) = if(1==n,n,(if(1==(n%4),1,0)+(2*A292381(A252463(n)))));
    
  • 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 a292371(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)
    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 a292371(a292384(n))
    print([a(n) for n in range(1, 111)]) # Indranil Ghosh, Sep 21 2017
  • Scheme
    (define (A292381 n) (A292371 (A292384 n)))
    

Formula

a(1) = 1; for n > 1, a(n) = 2*a(A252463(n)) + [n ≡ 1 (mod 4)], where the last part of the formula is Iverson bracket, giving 1 only if n is of the form 4k+1, and 0 otherwise.
a(n) = A292371(A292384(n)).
Other identities. For n >= 1:
a(2n) = 2*a(n).
A000120(a(n)) = A292375(n).
For n >= 2, a(n) = A004754(A292385(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).

A292243 a(1) = 1; for n > 1, a(n) = 3*a(A253889(n)) + (n mod 3).

Original entry on oeis.org

1, 5, 3, 16, 17, 9, 49, 11, 33, 160, 50, 156, 52, 53, 147, 88, 29, 27, 82, 149, 474, 457, 35, 453, 106, 101, 441, 151, 482, 303, 265, 152, 483, 250, 470, 1449, 1441, 158, 480, 1429, 161, 1407, 469, 443, 1371, 298, 266, 318, 1348, 89, 969, 961, 83, 954, 910, 248, 897, 268, 449, 1455, 322, 1424, 99, 808, 1373, 738, 1366, 107
Offset: 1

Views

Author

Antti Karttunen, Sep 15 2017

Keywords

Comments

a(n) encodes in its base-3 representation the succession of modulo 3 residues obtained when map x -> A253889(x), starting from x=n, is iterated down to the eventual 1.

Crossrefs

Programs

  • Mathematica
    f[n_] := Times @@ Power[If[# == 1, 1, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger[2 n - 1]; g[n_] := (Times @@ Power[If[# == 1, 1, NextPrime@ #] & /@ First@ #, Last@ #] + 1)/2 &@ Transpose@ FactorInteger@ n; a[1] = 1; a[n_] := a[n] = 3 a[Floor@ g[Floor[f[n]/2]]] + Mod[n, 3]; Array[a, 68] (* Michael De Vlieger, Sep 16 2017 *)
  • PARI
    A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From Michel Marcus
    A048673(n) = (A003961(n)+1)/2;
    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)};
    A064216(n) = A064989((2*n)-1);
    A253889(n) = if(1==n,n,A048673(A064216(n)\2));
    A292243(n) = if(1==n,n,((n%3) + 3*A292243(A253889(n))));
    
  • Scheme
    ;; With memoization-macro definec.
    (definec (A292243 n) (if (= 1 n) n (+ (modulo n 3) (* 3 (A292243 (A253889 n))))))

Formula

a(1) = 1; for n > 1, a(n) = 3*a(A253889(n)) + A010872(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).

A292603 Doudna-tree reduced modulo 4: a(n) = A005940(1+n) mod 4.

Original entry on oeis.org

1, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 1, 2, 1, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 1, 2, 3, 0, 1, 2, 1, 0, 3, 2, 3, 0, 1, 2, 3, 0, 3, 2, 1, 0, 3, 2, 1, 0, 1
Offset: 0

Views

Author

Antti Karttunen, Dec 01 2017

Keywords

Examples

			The first six levels of the binary tree (compare also to the illustrations given at A005940 and A292602):
                               1
                               |
                               2
                ............../ \..............
               3                               0
        ....../ \......                 ....../ \......
       1               2               1               0
      / \             / \             / \             / \
     /   \           /   \           /   \           /   \
    3     2         3     0         1     2         3     0
   / \   / \       / \   / \       / \   / \       / \   / \
  3   2 1   0     3   2 1   0     1   2 3   0     1   2 1   0
		

Crossrefs

Cf. A004767 (gives the positions of 0's), A016813 (of 2's).

Programs

Formula

a(n) = A010873(A005940(1+n)).
a(n) + 4*A292602(n) = A005940(1+n).
a(2n+1) = 2*a(n) mod 4.
a(A004767(n)) = 0.
a(A016813(n)) = 2.
a(2*A156552(A246261(n))) = 1.
a(2*A156552(A246263(n))) = 3.
a(n * 2^(1+A246271(A005940(1+n)))) = 1.
Showing 1-6 of 6 results.