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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A243060 Square array read by antidiagonals: rows are successively recursivized versions of Bulgarian solitaire operation (starting from the usual "first order" version, A243051), as applied to partitions listed in A241918.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 3, 4, 2, 1, 8, 3, 4, 2, 1, 25, 8, 3, 4, 2, 1, 16, 6, 8, 3, 4, 2, 1, 9, 16, 6, 8, 3, 4, 2, 1, 9, 5, 16, 6, 8, 3, 4, 2, 1, 343, 9, 5, 16, 6, 8, 3, 4, 2, 1, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 64, 35, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 14641, 64, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1, 125, 24, 64, 10, 32, 12, 9, 5, 16, 6, 8, 3, 4, 2, 1
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The array is read by antidiagonals: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ... .
Each row is a A241909-conjugate of the corresponding row in A243070.
Rows in both arrays converge towards A122111.
First point where row k differs from row k of A243070 seems to be A000040(k+2): primes from five onward: 5, 7, 11, 13, 17, 19, 23, 29, 31, ...
While the first point where row k differs from A122111 seems to begin as 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, ... A007283 (3*2^n) from the term n=1 onward.
The rows in this table preserve A243503.

Examples

			The top left corner of the array:
  1, 2, 4, 3, 8, 25, 16, 9, 9, 343,  32,  10,  64, 14641,   125,   27, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  35,  64,    24,    18,   25, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  10,  64,    24,    18,    7, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  10,  64,    24,    18,    7, ...
  1, 2, 4, 3, 8,  6, 16, 5, 9,  12,  32,  10,  64,    24,    18,    7, ...
		

Crossrefs

Row 1: A243051, Row 2: A243052, Row 3: A243053.
Rows converge towards A122111.

Programs

  • Scheme
    (define (A243060 n) (A243060bi (A002260 n) (A004736 n)))
    (define (A243060bi row col) (explist->n (ascpart_to_prime-exps (bulgarian-operation-n-th-order (prime-exps_to_ascpart (primefacs->explist col)) row))))
    (define (bulgarian-operation-n-th-order ascpart n) (if (or (zero? n) (null? ascpart)) ascpart (let ((newpart (length ascpart))) (let loop ((newpartition (list)) (ascpart ascpart)) (cond ((null? ascpart) (sort (cons newpart (bulgarian-operation-n-th-order newpartition (- n 1))) <)) (else (loop (if (= 1 (car ascpart)) newpartition (cons (- (car ascpart) 1) newpartition)) (cdr ascpart))))))))
    ;; For other required functions and libraries, please see A243051.

A243072 Second order Bulgarian solitaire operation on partition list A112798: a(1) = 1, a(n) = A000040(A001222(n)) * A242424(A064989(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 12, 5, 9, 12, 20, 10, 28, 18, 18, 7, 44, 15, 52, 20, 27, 30, 68, 14, 36, 42, 25, 30, 76, 30, 92, 11, 45, 66, 54, 21, 116, 78, 63, 28, 124, 45, 148, 50, 50, 102, 164, 22, 81, 60, 99, 70, 172, 35, 90, 42, 117, 114, 188, 42, 212, 138, 75, 13, 126
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The usual Bulgarian Solitaire operation (the "first order" version, cf. A242424) applied to an unordered integer partition means: subtract one from each part, and add a new part as large as there were parts in the old partition.
The "Second Order Bulgarian Solitaire" operation means that after subtracting one from each part of the old partition (and discarding the parts that diminished to zero), we apply the (first order) Bulgarian operation to the remaining partition before adding a new part as large as there were parts in the original partition.
In this context, where the parts of partitions are encoded with the indices of primes in the prime factorization of n (as in A112798), A064989(n) gives the remaining partition after one has been subtracted from each part; A242424 applies the first order Bulgarian operation to it; and multiplying with A000040(A001222(n)) adds a part as large as there originally were parts.

Crossrefs

Row 2 of A243070. Differs from A122111 for the first time at n=7.

Formula

a(1) = 1, a(n) = A000040(A001222(n)) * A242424(A064989(n)) = A105560(n) * A242424(A064989(n)).
a(n) = A241909(A243052(A241909(n))).

A243073 Third-order Bulgarian solitaire operation on partition list A112798: a(1) = 1, a(n) = A000040(A001222(n)) * A243072(A064989(n)).

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 5, 9, 12, 24, 10, 40, 24, 18, 7, 56, 15, 88, 20, 36, 36, 104, 14, 27, 60, 25, 40, 136, 30, 152, 11, 54, 84, 54, 21, 184, 132, 90, 28, 232, 60, 248, 60, 50, 156, 296, 22, 108, 45, 126, 100, 328, 35, 81, 56, 198, 204, 344, 42, 376, 228, 100, 13, 135
Offset: 1

Views

Author

Antti Karttunen, May 29 2014

Keywords

Comments

The usual (first-order) Bulgarian Solitaire operation (cf. A242424) applied to an unordered integer partition means: subtract one from each part, and add a new part as large as there were parts in the old partition.
The "Second-Order Bulgarian Operation" means that after subtracting one from each part of the old partition (and discarding the parts that diminished to zero), we apply the (first order) Bulgarian operation to the remaining partition before adding a new part as large as there were parts in the original partition.
Similarly, in "Third-Order Bulgarian Solitaire Operation", we apply the Second-Order Bulgarian operation to the remaining partition (after we have subtracted one from each part) before adding a new part as large as there were parts in the original partition.
In this context, where the parts of partitions are encoded with the indices of primes in the prime factorization of n (as in A112798), A064989(n) gives the remaining partition after one has been subtracted from each part; A243072 applies the second-order Bulgarian operation to it; and multiplying with A000040(A001222(n)) adds a part as large as there originally were parts.

Crossrefs

Row 3 of A243070. Differs from A122111 for the first time at n=11.

Formula

a(1) = 1, a(n) = A000040(A001222(n)) * A243072(A064989(n)) = A105560(n) * A243072(A064989(n)).
a(n) = A241909(A243053(A241909(n))).

A153212 A permutation of the natural numbers: in the prime factorization of n, swap each prime's index difference (from the previous distinct prime that divides n) and the prime's power.

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 16, 5, 9, 18, 32, 15, 64, 54, 12, 7, 128, 10, 256, 75, 36, 162, 512, 35, 27, 486, 25, 375, 1024, 30, 2048, 11, 108, 1458, 24, 21, 4096, 4374, 324, 245, 8192, 150, 16384, 1875, 45, 13122, 32768, 77, 81, 50, 972, 9375, 65536, 14, 72, 1715, 2916, 39366, 131072, 105, 262144, 118098, 225, 13
Offset: 1

Views

Author

Luchezar Belev (l_belev(AT)yahoo.com), Dec 20 2008

Keywords

Comments

In order for the "index difference" to make sense, we consider the factorization to be sorted with respect to the primes but not the powers to which they are raised; that is, first comes the smallest prime and each subsequent prime is larger than the previous disregarding their powers.
For every n it is true that a(a(n)) = n.
From Antti Karttunen, May 29 2014: (Start)
In other words, this is a self-inverse permutation (involution) of natural numbers.
This permutation maps primes (A000040) to the powers of two larger than one (A000079(n>=1)) and vice versa.
The term a(1) = 1 was added on the grounds that as 1 has an empty prime factorization, there is nothing to swap, thus it stays same. It is also needed as a base case for the given recurrence.
Considered as a function on partitions encoded by the indices of primes in the prime factorization of n (as in table A112798), this implements an operation which exchanges the horizontal and vertical line segment of each "step" in Young (or Ferrers) diagram of a partition. Please see the last example in the example section.
(End)

Examples

			For n = 10 we have 10 = 2^1 * 5^1 = p(1)^1 * p(3)^1 then a(10) = p(1)^1 * p(2)^2 = 2^1 * 3^2 = 18.
For n = 18 we have 18 = 2^1 * 3^2 = p(1)^1 * p(2)^2 then a(18) = p(1)^1 * p(3)^1 = 2^1 * 5^1 = 10.
For n = 19 we have 19 = 19^1 = p(8)^1 then a(19) = p(1)^8 = 2^8 = 256.
For n = 2200, we see that it encodes the partition (1,1,1,3,3,5) in A112798 as 2200 = p_1 * p_1 * p_1 * p_3 * p_3 * p_5 = 2^3 * 5^2 * 11. This in turn corresponds to the following Young diagram in French notation:
   _
  | |
  | |
  | |_ _
  |     |
  |     |_ _
  |_ _ _ _ _|
Exchanging the order of the horizontal and vertical line segment of each "step", results the following Young diagram:
   _ _ _
  |     |_ _
  |         |
  |         |_
  |           |
  |_ _ _ _ _ _|
which represents the partition (3,5,5,6,6), encoded in A112798 by p_3 * p_5^2 * p_6^2 = 5 * 11^2 * 13^2 = 102245, thus a(2200) = 102245.
		

Crossrefs

Fixed points: A242421.
{A000027, A122111, A153212, A242419} form a 4-group.

Programs

  • PARI
    a(n) = {my(f = factor(n)); my(g = f); for (i=1, #f~, if (i==1, g[i,1] = prime(f[i,2]), g[i,1] = prime(f[i,2]+ primepi(g[i-1,1]))); if (i==1, g[i,2] = primepi(f[i,1]), g[i,2] = primepi(f[i,1]) - primepi(f[i-1,1]));); factorback(g);} \\ Michel Marcus, Dec 16 2014

Formula

Denote the i-th prime with p(i): p(1)=2, p(2)=3, p(3)=5, p(4)=7, etc. Let n = p(a1)^b1 * p(a2)^b2 * ... * p(ak)^bk is the factorization of n where p(i)^j is the i-th prime raised to power j. As mentioned above, we assume that the primes are sorted, i.e., a1 < a2 < a3 ... < ak. Then a(n) = p(c1)^d1 * p(c2)^d2 * ... * p(ck)^dk where c1 = b1 and c(i) = b(i) + c(i-1) for i > 1 d1 = a1 and d(i) = a(i) - a(i-1) for i > 1.
From Antti Karttunen, May 16 2014: (Start)
a(1) = 1 and for n>1, let r = a(A051119(n)). Then a(n) = r * (A000040(A061395(r)+A071178(n)) ^ A241919(n)).
a(n) = A122111(A242419(n)) = A242419(A122111(n)).
(End)

Extensions

Term a(1)=1 prepended, and also more terms computed by Antti Karttunen, May 16 2014

A243504 Product of parts of integer partitions as ordered by the table A241918: a(n) = Product_{i=A203623(n-1)+2..A203623(n)+1} A241918(i).

Original entry on oeis.org

1, 1, 1, 2, 1, 4, 1, 3, 2, 8, 1, 9, 1, 16, 4, 4, 1, 6, 1, 27, 8, 32, 1, 16, 2, 64, 3, 81, 1, 18, 1, 5, 16, 128, 4, 12, 1, 256, 32, 64, 1, 54, 1, 243, 9, 512, 1, 25, 2, 12, 64, 729, 1, 8, 8, 256, 128, 1024, 1, 48, 1, 2048, 27, 6, 16, 162, 1, 2187, 256, 36, 1, 20, 1, 4096
Offset: 1

Views

Author

Antti Karttunen, Jun 05 2014

Keywords

Crossrefs

The positions of ones after a(1)=1 is given by A000040 (primes).
Cf. A243503 (the sum of parts), A241918, A227184, A075158, A003963, A241909.

Formula

a(n) = Product_{i=A203623(n-1)+2..A203623(n)+1} A241918(i).
a(n) = A003963(A241909(n)).
a(n) = A227184(A075158(n-1)).
a(A000040(n)) = 1 for all n.
a(A000079(n)) = n for all n.

A245454 Self-inverse permutation of nonnegative integers, A075158-conjugate of blue code: a(n) = 1 + A075157(A193231(A075158(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 6, 5, 18, 8, 9, 25, 11, 16, 64, 14, 27, 12, 96, 7, 288, 21, 20, 243, 891, 45, 10, 405, 15, 162, 33750, 30, 78650, 75, 625, 2025, 35, 81, 390390, 224, 875, 250, 41, 375, 16384, 270, 24, 300125, 24576, 150, 125, 54, 6125, 1350, 73728, 50, 108, 350, 594, 140777, 5845851, 98, 221433750, 1446445, 343, 13
Offset: 1

Views

Author

Antti Karttunen, Jul 22 2014

Keywords

Crossrefs

Programs

Formula

a(n) = 1 + A075157(A193231(A075158(n-1))).

A331600 a(n) = A002487(A331595(n)).

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 4, 3, 1, 4, 1, 3, 4, 2, 1, 3, 2, 2, 3, 3, 1, 4, 1, 5, 4, 2, 4, 3, 1, 2, 4, 3, 1, 4, 1, 3, 7, 2, 1, 5, 2, 12, 4, 3, 1, 3, 8, 3, 4, 2, 1, 3, 1, 2, 7, 5, 8, 4, 1, 3, 4, 12, 1, 5, 1, 2, 4, 3, 4, 4, 1, 5, 3, 2, 1, 3, 8, 2, 4, 3, 1, 3, 8, 3, 4, 2, 8, 5, 1, 16, 7, 3, 1, 4, 1, 3, 18
Offset: 1

Views

Author

Antti Karttunen, Jan 22 2020

Keywords

Crossrefs

Programs

  • Mathematica
    Array[If[# == 1, 1, NestWhile[If[OddQ[#3], {#1, #1 + #2, #4}, {#1 + #2, #2, #4}] & @@ Append[#, Floor[#[[-1]]/2]] &, {1, 0, #}, #[[-1]] > 0 &][[2]] &@ Apply[GCD, {Block[{k = #, m = 0}, Times @@ Power @@@ Table[k -= m; k = DeleteCases[k, 0]; {Prime@ Length@ k, m = Min@ k}, Length@ Union@ k]] &@ Catenate[ConstantArray[PrimePi[#1], #2] & @@@ #], Function[t, Times @@ Prime@ Accumulate[If[Length@ t < 2, {0}, Join[{1}, ConstantArray[0, Length@ t - 2], {-1}]] + ReplacePart[t, Map[#1 -> #2 & @@ # &, #]]]]@ ConstantArray[0, Transpose[#][[1, -1]]] &[# /. {p_, e_} /; p > 0 :> {PrimePi@ p, e}]}] &@ FactorInteger[#]] &, 105] (* Michael De Vlieger, Jan 25 2020, after JungHwan Min at A122111 *)
  • PARI
    A002487(n) = { my(a=1, b=0); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); (b); }; \\ From A002487
    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)};
    A122111(n) = if(1==n,n,prime(bigomega(n))*A122111(A064989(n)));
    A241909(n) = if(1==n||isprime(n),2^primepi(n),my(f=factor(n),h=1,i,m=1,p=1,k=1); while(k<=#f~, p = nextprime(1+p); i = primepi(f[k,1]); m *= p^(i-h); h = i; if(f[k,2]>1, f[k,2]--, k++)); (p*m));
    A331595(n) = gcd(A122111(n), A241909(n));
    A331600(n) = A002487(A331595(n));

Formula

a(n) = A002487(A331595(n)) = A002487(gcd(A122111(n), A241909(n))).
a(n) = A002487(A331731(n)).

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

Views

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 *)

A245451 Self-inverse permutation of nonnegative integers, A075158-conjugate of gray code: a(n) = 1 + A075157(A003188(A075158(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 8, 9, 16, 6, 5, 27, 32, 18, 64, 81, 25, 12, 128, 7, 256, 54, 125, 243, 512, 36, 10, 729, 15, 162, 1024, 49, 2048, 24, 625, 2187, 50, 14, 4096, 6561, 3125, 108, 8192, 343, 16384, 486, 75, 19683, 32768, 72, 20, 21, 15625, 1458, 65536, 35, 250, 324, 78125, 59049, 131072, 98, 262144, 177147, 375, 48
Offset: 1

Views

Author

Antti Karttunen, Jul 22 2014

Keywords

Crossrefs

Inverse: A245452.
Similar permutations: A245454, A122111, A241909, A241916.

Programs

Formula

a(n) = 1 + A075157(A003188(A075158(n-1))).

A245452 Self-inverse permutation of nonnegative integers, A075158-conjugate of the inverse of gray code: a(n) = 1 + A075157(A006068(A075158(n-1))).

Original entry on oeis.org

1, 2, 4, 3, 9, 8, 18, 5, 6, 25, 75, 16, 150, 36, 27, 7, 735, 12, 1470, 49, 50, 245, 12705, 32, 15, 300, 10, 72, 25410, 125, 195195, 11, 225, 4235, 54, 24, 390390, 2940, 490, 121, 4339335, 100, 8678670, 847, 81, 65065, 92147055, 64, 30, 35, 2205, 600, 184294110, 20, 147, 144, 8470, 50820, 2565568005, 343, 5131136010, 1446445, 98, 13
Offset: 1

Views

Author

Antti Karttunen, Jul 22 2014

Keywords

Crossrefs

Inverse: A245451.
Similar permutations: A245454, A122111, A241909, A241916.

Programs

Formula

a(n) = 1 + A075157(A006068(A075158(n-1))).
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