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 1-8 of 8 results.

A241913 Complement of A241912, natural numbers not fixed by A241916.

Original entry on oeis.org

6, 9, 10, 12, 14, 20, 21, 22, 24, 25, 26, 27, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, 44, 46, 48, 49, 51, 52, 54, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 99, 100, 102, 104, 106, 110, 111
Offset: 1

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Author

Antti Karttunen, May 03 2014

Keywords

Comments

Terms that occur in 2-cycles of permutation A241916. (E.g., A241916(6)=9, A241916(9)=6.)
Apart from its initial terms, 1 and 2, all the terms of A088902 occur here because A241909 has no other fixed points than 1 and 2.

Crossrefs

Complement of A241912.
A subsequence apart from its two initial terms: A088902.
Cf. A241916.

Programs

  • Mathematica
    f[n_] := If[n == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ n]; g[w_List] := Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, w]; Complement[Range@ Max@ #, Table[#[[n + 1]]/2, {n, Length@ # - 1}]] &@ Select[Range@ 120, g@ f@ # == g@ Reverse@ f@ # &] (* Michael De Vlieger, Aug 27 2016 *)

A241916 a(2^k) = 2^k, and for other numbers, if n = 2^e1 * 3^e2 * 5^e3 * ... p_k^e_k, then a(n) = 2^(e_k - 1) * 3^(e_{k-1}) * ... * p_{k-1}^e2 * p_k^(e1+1). Here p_k is the greatest prime factor of n (A006530), and e_k is its exponent (A071178), and the exponents e1, ..., e_{k-1} >= 0.

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 7, 8, 6, 25, 11, 27, 13, 49, 15, 16, 17, 18, 19, 125, 35, 121, 23, 81, 10, 169, 12, 343, 29, 75, 31, 32, 77, 289, 21, 54, 37, 361, 143, 625, 41, 245, 43, 1331, 45, 529, 47, 243, 14, 50, 221, 2197, 53, 36, 55, 2401, 323, 841, 59, 375, 61, 961, 175, 64
Offset: 1

Views

Author

Antti Karttunen, May 03 2014

Keywords

Comments

For other numbers than the powers of 2 (that are fixed), this permutation reverses the sequence of exponents in the prime factorization of n from the exponent of 2 to that of the largest prime factor, except that the exponents of 2 and the greatest prime factor present are adjusted by one. Note that some of the exponents might be zeros.
Self-inverse permutation of natural numbers, composition of A122111 & A241909 in either order: a(n) = A122111(A241909(n)) = A241909(A122111(n)).
This permutation preserves both bigomega and the (index of) largest prime factor: for all n it holds that A001222(a(n)) = A001222(n) and A006530(a(n)) = A006530(n) [equally: A061395(a(n)) = A061395(n)].
From the above it follows, that this fixes both primes (A000040) and powers of two (A000079), among other numbers.
Even positions from n=4 onward contain only terms of A070003, and the odd positions only the terms of A102750, apart from 1 which is at a(1), and 2 which is at a(2).

Crossrefs

A241912 gives the fixed points; A241913 their complement.
{A000027, A122111, A241909, A241916} form a 4-group.
The sum of prime indices of a(n) is A243503(n).
Even bisection of A358195 = Heinz numbers of rows of A358172.
A112798 lists prime indices, length A001222, sum A056239.

Programs

  • Mathematica
    nn = 65; f[n_] := If[n == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ n]; g[w_List] := Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, w]; Table[If[IntegerQ@ #, n/4, g@ Reverse@(# - Join[{1}, ConstantArray[0, Length@ # - 2], {1}] &@ f@ n)] &@ Log2@ n, {n, 4, 4 nn, 4}] (* Michael De Vlieger, Aug 27 2016 *)
  • PARI
    A209229(n) = (n && !bitand(n,n-1));
    A241916(n) = if(1==A209229(n), n, my(f = factor(2*n), nbf = #f~, igp = primepi(f[nbf,1]), g = f); for(i=1,nbf,g[i,1] = prime(1+igp-primepi(f[i,1]))); factorback(g)/2); \\ Antti Karttunen, Jul 02 2018
    
  • Scheme
    (define (A241916 n) (A122111 (A241909 n)))

Formula

a(1)=1, and for n>1, a(n) = A006530(n) * A137502(n)/2.
a(n) = A122111(A241909(n)) = A241909(A122111(n)).
If 2n has prime factorization Product_{i=1..k} prime(x_i), then a(n) = Product_{i=1..k-1} prime(x_k-x_i+1). The opposite version is A000027, even bisection of A246277. - Gus Wiseman, Dec 28 2022

Extensions

Description clarified by Antti Karttunen, Jul 02 2018

A241918 Table of partitions where the ordering is based on the modified partial sums of the exponents of primes in the prime factorization of n.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, May 03 2014, based on Marc LeBrun's Jan 11 2006 message on SeqFan mailing list

Keywords

Comments

a(1) = 0 by convention (stands for an empty partition).
For n >= 2, A203623(n-1)+2 gives the index to the beginning of row n and for n>=1, A203623(n)+1 is the index to the end of row n.

Examples

			Table begins:
Row     Partition
[ 1]    0;         (stands for empty partition)
[ 2]    1;         (as 2 = 2^1)
[ 3]    1,1;       (as 3 = 2^0 * 3^1)
[ 4]    2;         (as 4 = 2^2)
[ 5]    1,1,1;     (as 5 = 2^0 * 3^0 * 5^1)
[ 6]    2,2;       (as 6 = 2^1 * 3^1)
[ 7]    1,1,1,1;   (as 7 = 2^0 * 3^0 * 5^0 * 7^1)
[ 8]    3;         (as 8 = 2^3)
[ 9]    1,2;       (as 9 = 2^0 * 3^2)
[10]    2,2,2;     (as 10 = 2^1 * 3^0 * 5^1)
[11]    1,1,1,1,1;
[12]    3,3;
[13]    1,1,1,1,1,1;
[14]    2,2,2,2;
[15]    1,2,2;     (as 15 = 2^0 * 3^1 * 5^1)
[16]    4;
[17]    1,1,1,1,1,1,1;
[18]    2,3;       (as 18 = 2^1 * 3^2)
etc.
If n is 2^k (k>=1), then the partition is a singleton {k}, otherwise, add one to the exponent of 2 (= A007814(n)), and subtract one from the exponent of the greatest prime dividing n (= A071178(n)), leaving the intermediate exponents as they are, and then take partial sums of all, thus resulting for e.g. 15 = 2^0 * 3^1 * 5^1 the modified sequence of exponents {0+1, 1, 1-1} -> {1,1,0}, whose partial sums {1,1+1,1+1+0} -> {1,2,2} give the corresponding partition at row 15.
		

Crossrefs

For n>=2, the length of row n is given by A061395(n).
Cf. also A067255, A203623, A241914.
Other tables of partitions: A112798 (also based on prime factorization), A227739, A242628 (encoded in the binary representation of n), and A036036-A036037, A080576-A080577, A193073 for various lexicographical orderings.
Permutation A241909 maps between order of partitions employed here, and the order employed in A112798.
Permutation A122111 is induced when partitions in this list are conjugated.
A241912 gives the row numbers for which the corresponding rows in A112798 and here are the conjugate partitions of each other.

Programs

  • Mathematica
    Table[If[n == 1, {0}, Function[s, Function[t, Accumulate[If[Length@ t < 2, {0}, Join[{1}, ConstantArray[0, Length@ t - 2], {-1}]] + ReplacePart[t, Map[#1 -> #2 & @@ # &, s]]]]@ ConstantArray[0, Transpose[s][[1, -1]]]][FactorInteger[n] /. {p_, e_} /; p > 0 :> {PrimePi@ p, e}]], {n, 31}] // Flatten (* Michael De Vlieger, May 12 2017 *)

Formula

If A241914(n)=0 and A241914(n+1)=0, a(n) = A067255(n); otherwise, if A241914(n)=0 and A241914(n+1)>0, a(n) = A067255(n)+1; otherwise, if A241914(n)>0 and A241914(n+1)=0, a(n) = a(n-1) + A067255(n) - 1, otherwise, when A241914(n)>0 and A241914(n+1)>0, a(n) = a(n-1) + A067255(n).

A242418 Numbers n in whose prime factorization, n = 2^e1 * 3^e2 * 5^e3 * ... * p_k^e_k, the exponents (some of them possibly zero) of prime factors from 2 to p_k form a palindrome, so that e1 = e_k, e2 = e_{k-1}, etc.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 14, 16, 22, 26, 30, 32, 34, 36, 38, 46, 58, 62, 64, 74, 82, 86, 90, 94, 100, 106, 110, 118, 122, 128, 134, 142, 146, 158, 166, 178, 194, 196, 202, 206, 210, 214, 216, 218, 226, 238, 254, 256, 262, 270, 274, 278, 298, 300, 302, 314, 326, 334
Offset: 1

Views

Author

Antti Karttunen, May 20 2014

Keywords

Comments

a(1)=1 is included because 1 has an empty factorization (either no exponents, or all of them are zero), which thus is also a palindrome.

Crossrefs

Fixed points of A137502.
Cf. A241912.
A002110 and A079704 are subsequences.

Programs

  • Mathematica
    f[n_] := If[n == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ n]; g[w_List] := Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, w]; Select[Range@ 336, g@ f@ # == g@ Reverse@ f@ # &] (* Michael De Vlieger, Aug 27 2016 *)

Formula

a(1)=1, and for n > 1, a(n) = 2 * A241912(n-1).

A242413 Numbers in whose prime factorization the first differences of indices of distinct primes form a palindrome; fixed points of A242415.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 21, 23, 24, 25, 27, 29, 30, 31, 32, 36, 37, 41, 43, 47, 48, 49, 53, 54, 59, 60, 61, 63, 64, 65, 67, 70, 71, 72, 73, 79, 81, 83, 89, 90, 96, 97, 101, 103, 107, 108, 109, 113, 120, 121, 125, 127, 128, 131, 133, 137, 139, 140, 144
Offset: 1

Views

Author

Antti Karttunen, May 31 2014

Keywords

Comments

Number n is present, if its prime factorization n = p_a^e_a * p_b^e_b * p_c^e_c * ... * p_i^e_i * p_j^e_j * p_k^e_k where a < b < c < ... < i < j < k, satisfies the condition that the first differences of prime indices (a-0, b-a, c-b, ..., j-i, k-j) form a palindrome.
More formally, numbers n whose prime factorization is either of the form p^e (p prime, e >= 0), i.e., one of the terms of A000961, or of the form p_i1^e_i1 * p_i2^e_i2 * p_i3^e_i3 * ... * p_i_{k-1}^e_{i_{k-1}} * p_{i_k}^e_{i_k}, where p_i1 < p_i2 < ... < p_i_{k-1} < p_k are distinct primes (sorted into ascending order) in the prime factorization of n, and e_i1 .. e_{i_k} are their nonzero exponents (here k = A001221(n) and i_k = A061395(n), the index of the largest prime present), and the indices of primes satisfy the relation that for each index i_j < i_k present, the index i_{k-j} is also present.

Examples

			1 is present because it has an empty factorization, so both the sequence of the prime indices and their first differences are empty, and empty sequences are palindromes as well.
12 = 2*2*3 = p_1^2 * p_2 is present, as the first differences (deltas) of prime indices (1-0, 2-1) = (1,1) form a palindrome.
60 = 2*2*3*5 = p_1^2 * p_2 * p_3 is present, as the deltas of prime indices (1-0, 2-1, 3-2) = (1,1,1) form a palindrome.
61 = p_18 is present, as the deltas of prime indices, (18), form a palindrome.
144 = 2^4 * 3^2 = p_1^4 * p_2^2 is present, as the deltas of prime indices (1-0, 2-1) = (1,1) form a palindrome.
Also, any of the cases mentioned in the Example section of A242417 as being present there, are also present in this sequence.
		

Crossrefs

Fixed points of A242415.
Differs from A243068 for the first time at n=36, where a(36)=60, while A243068(36)=61.

A242417 Numbers in whose prime factorization both the first differences of indices of distinct primes and their exponents form a palindrome; fixed points of A242419.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 16, 17, 19, 21, 23, 25, 27, 29, 30, 31, 32, 36, 37, 41, 43, 47, 49, 53, 59, 61, 64, 65, 67, 70, 71, 73, 79, 81, 83, 89, 90, 97, 101, 103, 107, 109, 113, 121, 125, 127, 128, 131, 133, 137, 139, 149, 151, 154, 157, 163, 165, 167, 169
Offset: 1

Views

Author

Antti Karttunen, May 20 2014

Keywords

Comments

Numbers that are fixed by the permutation A242419, i.e., for which A242419(n) = n. Also, numbers that are fixed by both A069799 and A242415.
Number n is present if its prime factorization n = p_a^e_a * p_b^e_b * p_c^e_c * ... * p_i^e_i * p_j^e_j * p_k^e_k where a < b < c < ... < i < j < k, satisfies the condition, that both the first differences of prime indices (a-0, b-a, c-b, ..., j-i, k-j) and the respective exponents (e_a, e_b, e_c, ... , e_i, e_j, e_k) form a palindrome.
More formally, numbers n whose prime factorization is either of the form p^e (p prime, e >= 0), i.e., one of the terms of A000961, or of the form p_i1^e_i1 * p_i2^e_i2 * p_i3^e_i3 * ... * p_i_{k-1}^e_{i_{k-1}} * p_{i_k}^e_{i_k}, where p_i1 < p_i2 < ... < p_i_{k-1} < p_k are distinct primes (sorted into ascending order) in the prime factorization of n, and e_i1 .. e_{i_k} are their nonzero exponents (here k = A001221(n) and i_k = A061395(n), the index of the largest prime present), and the indices of primes satisfy the relation that for each index i_1 <= i_j < i_k present, the index i_{k-j} is also present, and the exponents e_{i_j} and e_{i_{(k-j)+1}} are equal.

Examples

			1 is present because it has an empty factorization, so both sequences are empty, thus palindromes.
3 = p_2^1 is present, as both the sequence of the first differences (deltas) of prime indices (2-0) = (2) and the exponents (1) are palindromes.
6 = p_1^1 * p_2^1 is present, as both the deltas of prime indices (1-0, 2-1) = (1,1) and the exponents (1,1) form a palindrome.
8 = p_1^3 is present, as both the deltas of prime indices (1) and the exponents (3) form a palindrome.
300 = 4*3*25 = p_1^2 * p_2^1 * p_3^2 is present, as both the deltas of prime indices (1-0, 2-1, 3-2) = (1,1,1) 1, 2 and the exponents (2,1,2), form a palindrome.
144 = 2^4 * 3^2 = p_1^4 * p_2^2 is NOT present, as although the deltas of prime indices (1-0, 2-1) = (1,1) are palindrome, the sequence of exponents (4,2) do NOT form a palindrome.
441 = 9*49 = p_2^2 * p_4^2 is present, as both the deltas of prime indices (2-0, 4-2) = (2,2) and the exponents (2,2) form a palindrome.
30030 = 2*3*5*7*11*13 = p_1 * p_2 * p_3 * p_4 * p_5 * p_6 is present, as the exponents are all ones, and the deltas of indices, (6-5,5-4,4-3,3-2,2-1,1-0) = (1,1,1,1,1,1) likewise are all ones, thus both sequences form a palindrome. This is true for all primorial numbers, A002110.
47775 = 3*5*5*7*7*13 = p_2^1 * p_3^2 * p_4^2 * p_6^1 is present, as the deltas of indices (6-4,4-3,3-2,2-0) = (2,1,1,2) and the exponents (1,2,2,1) both form a palindrome.
90000 = 2*2*2*2*3*3*5*5*5*5 = p_1^4 * p_2^2 * p_3^4 is present, as the deltas of indices (3-2,2-1,1-0) = (1,1,1) and the exponents (4,2,4) both form a palindrome.
		

Crossrefs

Fixed points of A242419. Intersection of A242413 and A242414.
Subsequences: A000961, A002110.

A242421 Fixed points of A153212: After a(1) = 1, numbers of the form p_i1^i1 * p_i2^(i2-i1) * p_i3^(i3-i2) * ... * p_ik^(ik-i_{k-1}), where p_i's are distinct primes present in the prime factorization of n, with i1 < i2 < i3 < ... < ik, and k = A001221(n) and ik = A061395(n).

Original entry on oeis.org

1, 2, 6, 9, 30, 45, 50, 125, 210, 294, 315, 350, 441, 686, 875, 2310, 2401, 3234, 3465, 3630, 3850, 4851, 5445, 6050, 7546, 7986, 9625, 11979, 15125, 26411, 29282, 30030, 35490, 42042, 45045, 47190, 49686, 50050, 53235, 59150, 63063, 65910, 70785, 74529, 78650, 98098, 98865, 103818, 109850, 115934, 125125, 147875, 155727, 161051, 171366, 196625, 257049, 274625, 343343, 380666, 405769, 510510
Offset: 1

Views

Author

Antti Karttunen, May 16 2014

Keywords

Comments

This sequence is closed with respect to A122111, i.e., for any n, A122111(a(n)) is either the same as a(n) or some other term a(k) of the sequence.
These numbers encode partitions in whose Young diagrams all pairs of successive horizontal and vertical segments (those pairs sharing "a common convex corner") are of equal length. Cf. the example-illustration at A153212.
Note: The seventh primorial, 510510 (= A002110(7)) occurs here as a term a(62).

Examples

			2 = p_1^1 is present, as the first prime index delta and exponent are equal.
3 = p_2^1 is not present, as 1 <> 2.
6 = p_1^1 * p_2^(2-1) is present.
9 = p_2^2 is present, as 2 = 2.
30 = p_1^1 * p_2^(2-1) * p_3^(3-2) is present, as all primorials are.
50 = p_1^1 * p_3^(3-1) is present also.
		

Crossrefs

Subsequences: A002110 (primorial numbers), A062457.

A278520 a(n) = A243503(n) - A056239(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, -1, 2, 0, 2, 0, 3, 0, 0, 0, 0, 0, 4, 1, 4, 0, 3, -2, 5, -2, 6, 0, 2, 0, 0, 2, 6, -1, 1, 0, 7, 3, 6, 0, 4, 0, 8, 0, 8, 0, 4, -3, 0, 4, 10, 0, -1, 0, 9, 5, 9, 0, 4, 0, 10, 2, 0, 1, 6, 0, 12, 6, 2, 0, 2, 0, 11, -2, 14, -2, 8, 0, 8, -3, 12, 0, 7, 2, 13, 7, 12, 0, 2, -1, 16, 8, 14, 3, 5, 0, 0, 4, 2, 0, 10, 0, 15, 0, 15, 0, 0, 0, 4, 9, 12, 0, 12
Offset: 1

Views

Author

Antti Karttunen, Dec 11 2016

Keywords

Comments

The 223rd zero in this sequence occurs at n=1078 (a(1078) = 0, 1078 = 2*7*7*11) and it is the first zero that is not included in A241912.

Crossrefs

Cf. A241912 (gives the subset of the indices of zeros).

Formula

a(n) = A243503(n) - A056239(n).
a(A241912(n)) = 0 [but obtains zeros also for other values, see comments.]
Showing 1-8 of 8 results.