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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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A303765 Permutation of nonnegative integers: a(n) = A048675(A303761(n)); see comments for an equivalent alternative description.

Original entry on oeis.org

0, 1, 3, 2, 7, 4, 5, 15, 8, 9, 11, 10, 31, 16, 17, 19, 18, 23, 6, 63, 32, 33, 35, 34, 39, 36, 37, 47, 12, 13, 127, 64, 65, 67, 66, 71, 68, 69, 79, 14, 255, 128, 129, 131, 130, 135, 132, 133, 143, 136, 137, 139, 138, 159, 20, 21, 511, 256, 257, 259, 258, 263, 260, 261, 271, 264, 265, 267, 266, 287, 24, 25, 27, 26, 1023, 512, 513, 515, 514, 519, 516, 517, 527
Offset: 0

Views

Author

Antti Karttunen, May 02 2018

Keywords

Comments

a(0) = 0 and for n > 0, if there are one or more k_i that are not already present in the sequence among terms a(0) .. a(n-1), and for which bitor(k_i,a(n-1)) = a(n-1), then a(n) = that k_i which gives minimum value of A019565(k_i) amongst them; otherwise, when no such k_i exists, a(n) = the least number not already present that can be obtained by cumulatively filling the successive vacant bits of a(n-1) from the least significant end of its binary representation (by toggling 0's to 1's, possibly also one or more leading zeros).
Shares with permutations like A003188, A006068, A300838, A302846, A303763, and A303767 the property that when moving from any a(n) to a(n+1) either a subset of 0-bits are toggled on (changed to 1's), or a subset of 1-bits are toggled off (changed to 0's), but no both kind of changes may occur at the same step.
A300829 gives the positions of records (terms of A000225 in ascending order), that for cases n=2..79 are followed by 2^(n-1).

Examples

			For a(2), a(1) = 1, and the only subset mask (a number k for which bitor(k,1) = k) is 1 itself, already present, so we start toggling 0's to 1's with binary expansion "...00001" of 1, and we get "11" (= binary representation of 3), and 3 is not yet present, thus a(2) = 3.
For a(3), previous a(2) = 3, "...011" in binary, and "10" (= 2) is the least and only submask that is not already present, thus a(3) = 2.
For a(4), previous = 2, "...010" in binary, and there are no submasks that are not already used, thus we start toggling 0's to 1's from the right, and "11" (3) is already present, but "111" (7) is not, thus a(4) = 7.
For a(5), previous = 7, with seven submasks "1", "10", "11", "100", "101", "110", "111" (binary representations for 1 - 7), and of these "100", "101", "110" (4, 5 and 6) are not yet present. A019565(4) = 5, A019565(5) = 10 and A019565(6) = 15, so we choose the first one, thus a(5) = 4.
For a(6), previous = 4, "..0100" in binary, and no submasks that wouldn't have been already used, thus by toggling from the right, we first obtain "...0101" = 5, which is still free, so a(6) = 5.
For a(7), previous = 5, "..0101" in binary, and no submasks that would be free (both 1 and 4 are already present), thus by toggling from the right, we first obtain "...0111" = 7, which also has been used, so we continue filling the zeros, to next obtain "...1111" = 15, which is still free, so a(7) = 15.
For a(8), previous = 15, "..1111" in binary, and the submasks not used are "110" = 6, "1000" = 8, "1001" = 9, "1010" = 10, "1011" = 11, "1100" = 12, "1101" = 13 and "1110" = 14. Applying A019565 to these numbers, A019565(8) = 7 gives the smallest value, thus a(8) = 8.

Links

Crossrefs

Cf. A303766 (inverse).
Cf. A006519, A019565, A048675, A303761.
Cf. A303763, A303767 (variants).
Cf. A300829 (positions of records).

Programs

  • PARI
    up_to = (2^7)-1;
A006519(n) = (2^valuation(n, 2));
A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
v303765 = vector(up_to);
m303766 = Map();
w=1; for(n=1,up_to,s = Set([]); for(m=1,w, if((bitor(w,m)==w) && !mapisdefined(m303766,m), s = setunion(Set([A019565(m)]),s))); if(length(s)>0, w = A048675(vecmin(s)), while(mapisdefined(m303766,w), w += A006519(1+w))); v303765[n] = w; mapput(m303766,w,n));
A303765(n) = if(!n,n,v303765[n]);
A303766(n) = if(!n,n,mapget(m303766,n));

Formula

a(n) = A048675(A303761(n)).
For all n >= 0, A019565(a(n)) = A303761(n).

A303761 Divisor-or-multiple permutation of squarefree numbers: a(0) = 1, and for n >= 1, a(n) is either the least divisor of a(n-1) not already present, or (if all divisors already used), a(n) is obtained by iterating the map x -> x*A053669(x), starting from x = a(n-1), until x is found which is not already present in the sequence.

Original entry on oeis.org

1, 2, 6, 3, 30, 5, 10, 210, 7, 14, 42, 21, 2310, 11, 22, 66, 33, 330, 15, 30030, 13, 26, 78, 39, 390, 65, 130, 2730, 35, 70, 510510, 17, 34, 102, 51, 510, 85, 170, 3570, 105, 9699690, 19, 38, 114, 57, 570, 95, 190, 3990, 133, 266, 798, 399, 43890, 55, 110, 223092870, 23, 46, 138, 69, 690, 115, 230, 4830, 161, 322, 966, 483, 53130, 77
Offset: 0

Views

Author

Antti Karttunen, May 02 2018

Keywords

Comments

Each a(n+1) is either a divisor or a multiple of a(n).
The primorials (A002110) occur in ascending order, in positions given by A300829, and each is then followed by the least unused term up to that point. For n = 2 .. 79 this is the highest prime factor of the said primorial, but note that for A300829(80) = 4965, a(4965) = A002110(80), but a(4966) = 407 = 11*37, instead of prime(80) = 409. Note that 409 occurs at a(5043), where 5043 = 1+A300829(81).
For example, 11 comes after a(A300829(5)) = a(12) = 2310 = 2*3*5*7*11, and all squarefree numbers < 11: {1, 2, 3, 5, 6, 7, 10} occur before a(13).

Links

Crossrefs

Cf. A005117, A019565, A053669, A300829 (gives the positions of records), A303765.
Cf. also A303751, A303760, A303762.

Programs

  • PARI
    default(parisizemax,2^31);
up_to = 2^8;
A053669(n) = forprime(p=2, , if (n % p, return(p))); \\ From A053669
v303761 = vector(up_to);
m_inverses = Map();
prev=1;for(n=1,up_to,fordiv(prev,d,if(!mapisdefined(m_inverses,d),v303761[n] = d;mapput(m_inverses,d,n);break)); if(!v303761[n], while(mapisdefined(m_inverses,prev), prev *= A053669(prev)); v303761[n] = prev; mapput(m_inverses,prev,n)); prev = v303761[n]);
A303761(n) = v303761[n+1];

Formula

a(n) = A019565(A303765(n)).
For n >= 0, a(A300829(n)) = A002110(n) [primorials are the records].
For n = 2 .. 79, a(1+A300829(n)) = A000040(n).

A303766 Inverse permutation to A303765.

Original entry on oeis.org

0, 1, 3, 2, 5, 6, 18, 4, 8, 9, 11, 10, 28, 29, 39, 7, 13, 14, 16, 15, 54, 55, 88, 17, 70, 71, 73, 72, 128, 129, 420, 12, 20, 21, 23, 22, 25, 26, 156, 24, 110, 111, 113, 112, 178, 179, 209, 27, 151, 152, 154, 153, 274, 275, 822, 155, 384, 385, 387, 386, 1943, 1944, 4580, 19, 31, 32, 34, 33, 36, 37, 252, 35, 242, 243, 245, 244, 316, 317, 497, 38, 247
Offset: 0

Views

Author

Antti Karttunen, May 02 2018

Keywords

Links

Crossrefs

Cf. A303765 (inverse).
Cf. A300829, A303761.

Programs

Formula

For n >= 0, a(A000225(n)) = A300829(n).
Showing 1-3 of 3 results.

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