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.

A347284 a(n) = Product_{j=1..A089576(n)} p_j^e_j with e_j = floor(e_(j-1)*log(p_(j-1))/log(p_j)) where the first factor is 2^n.

Original entry on oeis.org

1, 2, 12, 24, 720, 151200, 302400, 1814400, 4191264000, 8382528000, 251727315840000, 503454631680000, 3020727790080000, 1542111744113740800000, 3084223488227481600000, 92526704646824448000000, 555160227880946688000000, 1110320455761893376000000, 10769764221549079560253440000000
Offset: 0

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Author

Keywords

Comments

a(n) is the product of the largest prime power divisors p_j^e_j such that p_j^e_j < p_(j-1)^e_(j-1), beginning with p_1^e_1 = 2^n and proceeding with the next prime p until e_j = 0.
{a(n)} is a subset of A025487 which is a subset of A055932. All terms are products of primorials. No primes p_j for 1 <= j <= L have e = 0 with the exception of a(0) = 2^0. Let L = A001221(a(n)).
The largest primorial divisor P(L) = A2110(L).
For n > 0, all terms are even.
The greatest prime divisor p_L has multiplicity e_L = 1.
All multiplicities e are distinct; for 1 <= j <= L, the multiplicity e_j >= L - j + 1.
a(k) | a(n) for 0 <= k <= n.
The numbers q = a(n+1)/a(n) are primorials.
Finite intersection of A002182 and a(n) = {1, 2, 12, 360, 75600}.
Chernoff number A006939(L) | a(n). Quotient K = a(n) | A006939(L) is in A025487.
The prime shape of terms resembles a simplified map of the US state of Idaho.

Examples

			a(0) = 2^0 = 1;
a(1) = 2^1 = 2, since 3^1 > 2^1;
a(2) = 2^2 * 3^1, since 3^1 < 2^2 but 3^2 > 2^2, and since 5^1 > 3^1;
a(3) = 2^3 * 3^1, since 3^1 < 2^3 but 3^2 > 2^3, and 5^1 > 3^1;
a(4) = 2^4 * 3^2 * 5^1, since 3^2 < 2^4 yet 3^3 > 2^4, 5^1 < 3^2 yet 5^2 > 3^2, and 7^1 > 5^1; etc.
Prime shapes of a(n) for 2 <= n <= 5:
                                                     5  o
                                    4  o             4  x
                     3  o           3  x             3  x x
      2  x           2  x           2  x x           2  x x x
a(2)  1  X X   a(3)  1  X X   a(4)  1  X X X   a(5)  1  X X X X
         2 3            2 3            2 3 5            2 3 5 7
This demonstrates that a(n) is in A025487, that A002110(A001221(a(n))) is the greatest primorial divisor of a(n) as a consequence (prime divisors represented by capital X's), and Chernoff A006939(A001221(a(n))) | n, prime divisors represented by x's of any case. a(n) = A006939(A001221(a(n))) * k, k in A025487, represented by o's.
Because each multiplicity e is necessarily distinct, we may compactify a(n) using Sum_{k=1..omega(a(n))} 2^(e-1).
Prime shapes of a(12):
      12  o
      11  o
      10  o
       9  o
       8  o
       7  o o
       6  x o
       5  x x
       4  x x x
       3  x x x x
       2  x x x x x
a(12)  1  X X X X X X
          2 3 5 7 ...
a(12) = A006939(6) * 2^6 * 3^2
      = 5244319080000 * 64 * 9
      = 3020727790080000.
                                                 O
                                       O         x
                            O          x         x
                  O         x          x o       x x
         O        x         x o        x x o     x x x
  O      x o      x x       x x o      x x x o   x x x x
a(1)*6 = a(2)*2 = a(3)*30 = a(4)*210 = a(5)*2 =  a(6), etc., hence a(n) can be generated by a list of indices of primorials {1, 2, 1, 3, 4, 1, 1, 5, ...} and thereby be efficiently compactified.
		

Crossrefs

Programs

  • Mathematica
    Array[Times @@ NestWhile[Append[#1, #2^Floor@ Log[#2, #1[[-1]]]] & @@ {#, Prime[Length@ # + 1]} &, {2^#}, Last[#] > 1 &] &, 18, 0] (* or *)
    Block[{nn = 2^5, a = {}, b, e, i, m, p}, Array[Set[e[#], 0] &, Floor[2^# If[# <= 4, 1/2, -1 + 2^(7/(3 #))]] &[Ceiling@ Log2@ nn]]; Do[e[1]++; b = {2^e[1]}; Do[If[Last[b] == 1, Break[], i = e[j]; p = Prime[j]; While[p^i < b[[j - 1]], i++]; AppendTo[b, p^(i - 1)]; If[i > e[j], e[j]++]], {j, 2, k}]; AppendTo[a, Times @@ b], {k, nn}]; Prepend[a, 1]]
    (* Generate up to 4096 terms from the bitmap image *)
    With[{r = ImageData@ Import["https://oeis.org/A347284/a347284.png"]}, {1}~Join~Table[Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, Reverse@ Position[r[[i]], 0.][[All, 1]]], {i, 20}]]
    (* Generate up to 10000 terms using b-file at A347354 (numbers are large as n increases, limit nn is set to 120): *)
    Block[{nn = 120, s, m}, s = Import["https://oeis.org/A347354/b347354.txt", "Data"][[1 ;; nn, -1]]; m = Prime@ Range@ Max[s]; {1}~Join~FoldList[Times, Map[Times @@ m[[1 ;; #]] &, s]]] (* Michael De Vlieger, Sep 25 2021 *)

Formula

a(n) = Product_{j=1..k} p_j^T(n,j) where T = A347285 and k = A089576(n).
Row n of A347285 yields row a(n) of A067255.
a(n) = product of row n of A347288.

Extensions

Definition edited by Peter Munn, May 19 2023

A347285 Irregular triangle T(n,k) starting with n followed by e_k = floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0.

Original entry on oeis.org

0, 1, 2, 1, 3, 1, 4, 2, 1, 5, 3, 2, 1, 6, 3, 2, 1, 7, 4, 2, 1, 8, 5, 3, 2, 1, 9, 5, 3, 2, 1, 10, 6, 4, 3, 2, 1, 11, 6, 4, 3, 2, 1, 12, 7, 4, 3, 2, 1, 13, 8, 5, 4, 3, 2, 1, 14, 8, 5, 4, 3, 2, 1, 15, 9, 6, 4, 3, 2, 1, 16, 10, 6, 4, 3, 2, 1, 17, 10, 6, 4, 3, 2, 1
Offset: 0

Views

Author

Michael De Vlieger, Aug 26 2021

Keywords

Comments

Irregular triangle T(n,k) starting with n followed by e_k corresponding to the largest 1 < p_k^e_k < p_(k-1)^e_(k-1).
T(0,1) = 0 by convention; 0 is not allowed for n > 0.
T(n,k) > T(n,k+1). The least first difference among row n is 1.
Conjecture: let S be the sum of the absolute values of the first differences of terms in row n. For all n > 0, n - S = 1. - Michael De Vlieger, Aug 27 2021

Examples

			Row 0 contains {0} by convention.
Row 1 contains {1} since we can find no nonzero exponent e such that 3^e < 2^1.
Row 2 contains {2,1} since 3^1 < 2^2 yet 3^2 > 2^2. (We assume hereinafter that the powers listed are the largest possible smaller than the immediately previous term.)
Row 3 contains {3,1} since 2^3 > 3^1.
Row 4 contains {4,2,1} since 2^4 > 3^2 > 5^1, etc.
Triangle begins:
   0
   1
   2   1
   3   1
   4   2  1
   5   3  2  1
   6   3  2  1
   7   4  2  1
   8   5  3  2  1
   9   5  3  2  1
  10   6  4  3  2  1
  11   6  4  3  2  1
  12   7  4  3  2  1
  13   8  5  4  3  2  1
  14   8  5  4  3  2  1
  15   9  6  4  3  2  1
  16  10  6  4  3  2  1
  ...
		

Crossrefs

Cf. A000217, A000961, A089576 (row lengths).

Programs

  • Mathematica
    Array[NestWhile[Block[{p = Prime[#2]}, Append[#1, {p^#, #} &@ Floor@ Log[p, #1[[-1, 1]]]]] & @@ {#, Length@ # + 1} &, {{2^#, #}}, #[[-1, -1]] > 1 &][[All, -1]] &, 18, 0] // Flatten

Formula

T(n,1) = n; T(n,k) = floor(log_p_k(p_(k-1)^T(n,k-1))).
A000217(A089576(n)) <= sum of terms in row n.

A347354 a(n) = sum of T(n,k) - T(n-1,k) for row n of A347285.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 2, 5, 1, 6, 1, 2, 7, 1, 3, 2, 1, 8, 1, 4, 2, 1, 9, 10, 1, 2, 11, 1, 3, 1, 2, 12, 1, 4, 13, 1, 2, 1, 14, 15, 1, 2, 3, 1, 6, 2, 1, 4, 1, 16, 2, 1, 17, 18, 1, 2, 1, 3, 5, 1, 2, 19, 1, 4, 2, 1, 20, 1, 3, 21, 1, 2, 22, 1, 9, 1, 2, 4, 1, 8, 2, 1, 3
Offset: 1

Views

Author

Michael De Vlieger, Sep 16 2021

Keywords

Comments

If for row n, k > A089576(n-1), we interpret T(n-1,k) = 0.
Also the length of d = T(n,k) - T(n-1,k) in row n of A347285 such that d > 1, with 0 <= d <= 1.
Compactification of A347284 via indices k of primorials A002110(k). This is the most efficient compactification of A347284, superior to binary compactification via A347287. It employs the fact that A347284 concerns products of primorials, i.e., is a subset of A025487.
We can construct row n of A347285 by summing a constant array of a(k) 1's for 1 <= k <= n-1.

Examples

			Relation of a(n) and irregular triangle A347285, placing "." after the term in the current row where T(n,k) no longer exceeds T(n-1,k). Since the rows of A347285 reach a fixed point of 0, we interpret T(n,k) for vacant T(n-1,k) as exceeding same.
n    Row n of A347285   a(n)
-----------------------------
0:    0
1:    1.                   1
2:    2  1.                2
3:    3. 1                 1
4:    4  2  1.             3
5:    5  3  2  1.          4
6:    6. 3  2  1           1
7:    7  4. 2  1           2
8:    8  5  3  2  1.       5
9:    9. 5  3  2  1        1
10:  10  6  4  3  2  1.    6
...
a(3) = 1 since row 3 of A347285 has {3,1} while row 2 has {2,1}; only the first term of the former exceeds the analogous term in the latter.
a(4) = 3 since row 4 = {4,2,1} and row 3 = {3,1}; all 3 terms of the former are larger than the analogous term in the latter, etc.
		

Crossrefs

Programs

  • Mathematica
    Block[{nn = 84, a = {0}, c, e, m}, e[1] = 0; Do[c = 1; e[1]++; Do[Set[m, j]; Which[e[j - 1] == 1, Break[], IntegerQ@ e[j], If[e[j] < #, e[j]++; c++] &@ Floor@ Log[Prime[j], Prime[j - 1]^e[j - 1]], True, Set[e[j], 1]], {j, 2, k}]; AppendTo[a, c + Boole[c == m - 2]], {k, 2, nn}]; MapAt[# - 1 &, a, 4]]

Formula

A347284(n) = Product_{k=1..n} A002110(a(k)).

A347288 Irregular triangle T(n,k) starting with 2^n followed by p_k^e_k = p_k^floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0.

Original entry on oeis.org

1, 2, 4, 3, 8, 3, 16, 9, 5, 32, 27, 25, 7, 64, 27, 25, 7, 128, 81, 25, 7, 256, 243, 125, 49, 11, 512, 243, 125, 49, 11, 1024, 729, 625, 343, 121, 13, 2048, 729, 625, 343, 121, 13, 4096, 2187, 625, 343, 121, 13, 8192, 6561, 3125, 2401, 1331, 169, 17
Offset: 0

Views

Author

Michael De Vlieger, Aug 28 2021

Keywords

Comments

T(0,1) = 1 by convention.
T(n,1) = 2^n. T(n,k) = p_k^e_k such that p_k^T(n,k) is the largest 1 < p_k^e_k < p_(k-1)^e_(k-1).

Examples

			Row 0 contains {1} by convention.
Row 1 contains {2} since no nonzero exponent e exists such that 3^e < 2^1.
Row 2 contains {4,3} since 3^1 < 2^2 yet 3^2 > 2^2. (We assume hereinafter that the powers listed are the largest possible smaller than the immediately previous term.)
Row 3 contains {8,3} since 2^3 > 3^1.
Row 4 contains {16,9,5} since 2^4 > 3^2 > 5^1, etc.
Triangle begins:
           2      3      5      7     11    13    17  ...
  --------------------------------------------------
  0:       1
  1:       2
  2:       4      3
  3:       8      3
  4:      16      9      5
  5:      32     27     25      7
  6:      64     27     25      7
  7:     128     81     25      7
  8:     256    243    125     49     11
  9:     512    243    125     49     11
  10:   1024    729    625    343    121    13
  11:   2048    729    625    343    121    13
  12:   4096   2187    625    343    121    13
  13:   8192   6561   3125   2401   1331   169   17
  14:  16384   6561   3125   2401   1331   169   17
  ...
		

Crossrefs

Programs

  • Mathematica
    {{1}}~Join~Array[Most@ NestWhile[Block[{p = Prime[#2]}, Append[#1, p^Floor@ Log[p, #1[[-1]]]]] & @@ {#, Length@ # + 1} &, {2^#}, #[[-1]] > 1 &] &, 13] (* Michael De Vlieger, Aug 28 2021 *)

Formula

T(n,1) = 2^n; T(n,k) = p_k^floor(log_p_k(p_(k-1)^T(n,k-1))).
A347385(n,k) = p_k^T(n,k).
A089576(n) = row lengths.
A347284(n) = product of row n.

A347355 Index of first n in A347354.

Original entry on oeis.org

1, 2, 4, 5, 8, 10, 13, 18, 23, 24, 27, 32, 35, 39, 40, 50, 53, 54, 62, 67, 70, 73, 85, 89, 94, 99, 100, 104, 105, 115, 129, 132, 134, 140, 143, 153, 157, 159, 170, 173, 175, 180, 184, 188, 192, 194, 199, 229, 233, 235, 238, 248, 249, 254, 267, 275, 283, 289, 294
Offset: 1

Views

Author

Michael De Vlieger, Sep 28 2021

Keywords

Comments

Indices of records in A347354, where the record is A000027(n).
List of k such that A089576(k) = A089576(k-1) + 1.

Examples

			Relation of A347354 and irregular triangle A347285, placing "." after the last term in the current row where T(n,k) exceeds T(n-1,k). Since the rows of A347285 reach a fixed point of 0, we interpret T(n,k) for vacant T(n-1,k) as exceeding same. The indices n that are highlighted with parentheses are the terms in this sequence.
    n     Row n of A347285     A347354(n)
   --------------------------------------
    0:    0
   (1):   1.                   1
   (2):   2  1.                2
    3:    3. 1                 1
   (4):   4  2  1.             3
   (5):   5  3  2  1.          4
    6:    6. 3  2  1           1
    7:    7  4. 2  1           2
   (8):   8  5  3  2  1.       5
    9:    9. 5  3  2  1        1
  (10):  10  6  4  3  2  1.    6
  ...
		

Crossrefs

Programs

  • Mathematica
    Block[{nn = 300, a = {1}, c, e, m}, e[1] = 0; Do[c = 1; e[1]++; Do[Set[m, j]; Which[e[j - 1] == 1, Break[], IntegerQ@ e[j], If[e[j] < #, e[j]++; c++] &@ Floor@ Log[Prime[j], Prime[j - 1]^e[j - 1]], True, Set[e[j], 1]], {j, 2, k}]; If[c == m - 2, AppendTo[a, k - 1]], {k, 2, nn}]; Delete[a, {3}]]

A347356 a(n) = m/A006939(A001221(m)) with m = A347284(n).

Original entry on oeis.org

1, 1, 2, 2, 2, 4, 24, 24, 48, 48, 96, 576, 576, 1152, 34560, 207360, 414720, 414720, 829440, 174182400, 1045094400, 2090188800, 2090188800, 2090188800, 4180377600, 25082265600, 25082265600, 50164531200, 1504935936000, 3009871872000, 18059231232000
Offset: 1

Views

Author

Michael De Vlieger, Oct 02 2021

Keywords

Examples

			Diagram of prime power decomposition of A347284(12) = 2^12 * 3^7 * 5^4 * 7^3 * 11^2 * 13, showing Chernoff number A006939(6) with "x" and "X", A002110(A347354(12)) with "X", and a(12) with "o":
          12  o
          11  o
          10  o
           9  o
           8  o
           7  o o
           6  x o
           5  x x
           4  x x x
           3  x x x x
           2  x x x x x
           1  X X X X X X
              2 3 5 7 ...
A347284(12) = A006939(6) * a(12)
          = 5244319080000 * 576
          = 3020727790080000.
		

Crossrefs

Programs

  • Mathematica
    Block[{nn = 31, a = {}, b, c, e, i, p}, Array[Set[e[#], 0] &, Floor[2^# If[# <= 4, 1/2, -1 + 2^(7/(3 #))]] &[Ceiling@ Log2@ nn]]; Do[e[1]++; b = {2^e[1]}; c = {e[1]}; Do[If[Last[b] == 1, Break[], i = e[j]; p = Prime[j]; While[p^i < b[[j - 1]], i++]; AppendTo[b, p^(i - 1)]; AppendTo[c, (i - 1)]; If[i > e[j], e[j]++]], {j, 2, k}]; AppendTo[a, If[First[#] == 0, 1, Times @@ MapIndexed[Prime[First[#2]]^#1 &, TakeWhile[#, # > 0 &]]] &[# - Range[Length[#], 1, -1]] &@ If[k > 2, Most@ c, c]], {k, nn}]; a]

Formula

a(n) = A347284(n)/A006939(A089576(n)).
Showing 1-6 of 6 results.