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.

User: Joseph Rozhenko

Joseph Rozhenko's wiki page.

Joseph Rozhenko has authored 3 sequences.

A348154 Number of inequivalent strip arrangements.

Original entry on oeis.org

1, 3, 11, 100, 1063, 15686, 271975, 5509456, 126604661, 3256687324, 92655915831, 2888838414540, 97940953019995, 3587315304010374, 141162897496953263, 5939167862427259456, 266046178356979847881, 12641661811772879875640, 635092155152649300232063, 33633813271235206436451100
Offset: 1

Views

Author

Joseph Rozhenko, Oct 04 2021

Keywords

Comments

Given n strips, each of length n squares (dimensions 1 X n), a(n) is the number of distinct shapes that can be created by setting the strips side by side while satisfying the condition that the shape must include at least one row of length L=n squares (row considered to be a direction measured perpendicular to the strips). Shapes differing only by a rotation are considered to be equivalent.

Links

Crossrefs

Cf. A045531 (when rotations are considered distinct).

Programs

  • PARI
    a(n) = (n^n - (n-1)^n + n^(n\2) + !(n%2)*(n-1)^(n\2))/2; \\ Jinyuan Wang, Oct 08 2021

Formula

From Jinyuan Wang, Oct 08 2021: (Start)
a(2*k+1) = ((2*k+1)^(2*k+1) - (2*k)^(2*k+1) + (2*k+1)^k) / 2.
a(2*k) = ((2*k)^(2*k) - (2*k-1)^(2*k) + (2*k)^k + (2*k-1)^k) / 2.
(End)

Extensions

More terms from Jinyuan Wang, Oct 08 2021

A347688 Let c (resp. C) be the smallest (resp. largest) number that can be obtained by cyclically permuting the digits of n; a(n) = C - c.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0, 9, 18, 27, 36, 45, 54, 63, 72, 18, 9, 0, 9, 18, 27, 36, 45, 54, 63, 27, 18, 9, 0, 9, 18, 27, 36, 45, 54, 36, 27, 18, 9, 0, 9, 18, 27, 36, 45, 45, 36, 27, 18, 9, 0, 9, 18, 27, 36, 54, 45, 36, 27, 18, 9, 0, 9, 18, 27, 63, 54, 45, 36, 27, 18, 9, 0, 9, 18, 72, 63, 54, 45, 36, 27, 18, 9, 0, 9, 81, 72, 63, 54, 45, 36, 27, 18, 9, 0, 99, 99, 189, 279, 369, 459, 549, 639, 729, 819, 99, 0, 99
Offset: 0

Views

Author

N. J. A. Sloane, Sep 22 2021, following a suggestion from Joseph Rozhenko

Keywords

Comments

Agrees with A151949 for n <= 101.
All terms are multiples of 9 (cf. A347689).
Repeatedly applying the operation of subtracting the smallest cyclic permutation from the largest produces interesting sequences. It appears that n-digit numbers converge to a small number m << n of loops. Any 3-digit number converges to one of two loops of length 3: (189, 729, 675) and (378, 459, 486); any 4-digit number to either of (189, 729, 675) or a loop of length 25; any 5-digit number to one of three loops of length 3: (38007, 79335, 59778), (48780, 82926, 65853), or (29889, 69003, 89667), and so on. For 6-digit numbers there are 10 loops (of lengths 1, 2, 3, 10, 12 or 60). - Joseph Rozhenko, Oct 05 2021
For any cyclic number N, there must exist a cyclic permutation P of N for which a(P) = P. - Joseph Rozhenko, Oct 07 2021
An interesting plot of this sequence comes up when the axes are not n vs. a(n), but rather C vs. c. In other words, each number a(n) is represented on the X axis by C (largest number that can be obtained by cyclically permuting the digits of n) and on the Y axis as c (smallest number that can be obtained by cyclically permuting the digits of n). - Joseph Rozhenko, Jan 26 2022

Examples

			If n = 102, c = 21, C = 210, a(102) = 210 - 21 = 189.

Links

Crossrefs

Cf. A010785, A056965, A151949, A163380, A163381, A347689.

Programs

  • Maple
    A347688 := proc(n)
        local dgs,C,c,ndgs,r ;
        dgs := convert(n,base,10) ;
        ndgs := nops(dgs) ;
        C := digcatL(dgs) ;
        c := C ;
        for r from 0 to ndgs-1 do
                C := max(C,digcatL(dgs)) ;
                c := min(c,digcatL(dgs)) ;
                dgs := ListTools[Rotate](dgs,1) ;
        end do:
        C-c ;
end proc: # R. J. Mathar, Sep 27 2021
  • Mathematica
    {0}~Join~Table[First@Differences@MinMax[FromDigits/@(RotateLeft[IntegerDigits@n,#]&/@Range@IntegerLength@n)],{n,112}] (* Giorgos Kalogeropoulos, Sep 22 2021 *)
  • PARI
    a(n, base=10) = { my (c=n, C=n, d=digits(n, base)); for (k=1, #d, my (r=fromdigits(concat(d[k+1..#d], d[1..k]), base)); c=min(c, r); C=max(C, r)); C-c } \\ Rémy Sigrist, Sep 22 2021
  • Python
    def a(n):
    s = str(n)
    cyclic_perms = [int("".join(s[i:] + s[:i])) for i in range(len(s))]
    c, C = min(cyclic_perms), max(cyclic_perms)
    return C - c
print([a(n) for n in range(113)]) # Michael S. Branicky, Sep 26 2021

Formula

a(n) = 0 iff n belongs to A010785. - Rémy Sigrist, Sep 22 2021

Extensions

More than the usual number of terms are shown in order to distinguish this from similar sequences.

A342585 Inventory sequence: record the number of zeros thus far in the sequence, then the number of ones thus far, then the number of twos thus far and so on, until a zero is recorded; the inventory then starts again, recording the number of zeros.

Original entry on oeis.org

0, 1, 1, 0, 2, 2, 2, 0, 3, 2, 4, 1, 1, 0, 4, 4, 4, 1, 4, 0, 5, 5, 4, 1, 6, 2, 1, 0, 6, 7, 5, 1, 6, 3, 3, 1, 0, 7, 9, 5, 3, 6, 4, 4, 2, 0, 8, 9, 6, 4, 9, 4, 5, 2, 1, 3, 0, 9, 10, 7, 5, 10, 6, 6, 3, 1, 4, 2, 0, 10, 11, 8, 6, 11, 6, 9, 3, 2, 5, 3, 2, 0, 11, 11, 10
Offset: 1

Views

Author

Joseph Rozhenko, Mar 16 2021

Keywords

Comments

To get started we ask: how many zero terms are there? Since there are no terms in the sequence yet, we record a '0', and having recorded a '0', we begin again: How many zero terms are there? There is now one 0, so we record a '1' and continue. How many 1's are there? There's currently one '1' in the sequence, so we record a '1' and continue. How many 2's are there? There are no 2's yet, so we record a '0', and having recorded a 0, we begin again with the question "how many zero terms are there?" And so on.
a(46) = 0 because no 8's appear before it; but note a higher number, namely 9, has appeared. - Michael S. Branicky, Mar 16 2021
A similar situation occurs at n=124, where 14 has not yet appeared in the sequence, although 15 has appeared.
Reminiscent of Van Eck's sequence A181391. - N. J. A. Sloane, May 02 2021
From Jan Ritsema van Eck, May 02 2021: (Start)
The first 1000 terms seem to grow more or less in saw-tooth fashion with the largest terms (= the number of 0's), as well as the distance between the 0's, both approximately equal to the inverse triangular numbers A003056 (see attached graph #1).
But the picture changes when we go out to 10000 terms. Around the 1700th term, the 1's become more frequent than the 0's and the largest values are consistently somewhat larger than the inverse triangular numbers. Around the 2500th term the 2's become the most frequent number. Also after some 4000 terms, the largest values become much larger than the inverse triangular numbers. See graph #2. (End)
Comment on the colored plot of the first 1000467 terms, from Hans Havermann, May 02 2021: (Start)
If one is drawing a points-joined graph, it will obscure some of the inherent large-number dynamics. To get around that, this plot joins the points with a green line, superimposing the actual points in blue. This plot was created by Mathematica.
Your browser will likely compress the very large image to window size, so click on it to expand.
The points fall into linear features of the various counts of the various integers. The count for each integer changes as we move towards infinity and hence crosses over (changes place with) other counts unpredictably.
I decided to chart (see the blue text) the twenty largest counts at the rightmost spike which runs from the zero at 997010 to the zero at 1000467. These largest values are for the counts of integers 2 to 21 and appear at a(997013) for the 2-count; a(997014) for the 3-count, ..., and a(997032) for the 21-count.
The counts are 15275, 26832, 40162, 48539, 56364, 54372, 53393, 43588, 37288, 27396, 22425, 16735, 13099, 11460, 9466, 8386, 7191, 6478, 5777, and 5208, respectively. In my text they are sorted largest-to-smallest and written "count @ integer-being-counted": 56364 @ 6, 54372 @ 7, 53393 @ 8, 48539 @ 5, ... 5208 @ 21. (End)
A useful view may be gained by plotting the sequence against itself with an offset. Using the "Plot 2" link in the web page footer, enter "A342585" as sequences 1 and 2. Select "Plot Seq2(n+shift) vs Seq1(n)" and "Draw line segments". Start with "1" as the shift. The sequence appears somewhat like a fan, the first 4 or 5 sectors showing clearly, later sectors overlying each other. Larger shift values effectively compress early sectors into the vertical axis, making later sectors more visible. - Peter Munn, May 08 2021
For a version where a row ends not at the first zero, but rather at the last zero, see A347317. - N. J. A. Sloane, Sep 10 2021
For n around 2.5*10^9, the upper envelope of the sequence seems to be growing roughly like n/50, or maybe like O(n/log(n)). - N. J. A. Sloane, Feb 10 2023

Examples

			As an irregular triangle this begins:
   0;
   1,  1,  0;
   2,  2,  2,  0;
   3,  2,  4,  1,  1,  0;
   4,  4,  4,  1,  4,  0;
   5,  5,  4,  1,  6,  2,  1,  0;
   6,  7,  5,  1,  6,  3,  3,  1,  0;
   7,  9,  5,  3,  6,  4,  4,  2,  0;
   8,  9,  6,  4,  9,  4,  5,  2,  1,  3,  0;
   9, 10,  7,  5, 10,  6,  6,  3,  1,  4,  2,  0;
  10, 11,  8,  6, 11,  6,  9,  3,  2,  5,  3,  2,  0;
  ...
For row lengths see A347299. - _N. J. A. Sloane_, Aug 27 2021
From _David James Sycamore_, Oct 18 2021: (Start)
a(1) is 0 because the count is reset, and as yet there is no zero term immediately following another term. a(2) = 1 since the count is reset, a(1) = 0 and a(0) precedes it. The count now increments to terms equal to 1.
a(3) = 1 since a(2) = 1 and a(1) precedes it. a(4) = 0 because there is no term equal to 2 which is immediately preceded by another term.
a(5) = 2 since the count is reset, a(1) = a(4) = 0 and a(0), a(3) respectively, precede them. (End)

Links

Crossrefs

Records: A347305 and A348782.
Cf. A181391, A343878, A343880, A347299, A347315, A347316, A347317.
Other inventory-type sequences: A030717, A174382, A333867, A358066, A357443, A356784.
A012257 (cf. also A011784) reverses the inventory process.
See A347062, A347738, A355916, A355917, A355918, A357317 for variants.

Programs

  • AWK
    # See Links section. - Luc Rousseau, May 02 2021
  • MATLAB
    function [val,arr]=invSeq(N) % val = Nth term, arr = whole array up to N
k=0;
arr=zeros(1,N); % pre-allocate array
for i=1:N
    an=sum((k==arr(2:i)));
    arr(i)=an;
    if an == 0
        k = 0;
    else
        k=k+1;
    end
end
val=arr(end);
end % Ben Cha, Nov 11 2022
  • Maple
    a:= proc(n) option remember; local t;
      t:= `if`(a(n-1)=0, 0, b(n-1)+1);
      b(n):=t; add(`if`(a(j)=t, 1, 0), j=1..n-1)
    end: b(1), a(1):= 0$2:
seq(a(n), n=1..120);  # Alois P. Heinz, Mar 16 2021
  • Mathematica
    a[n_] := a[n] = Module[{t}, t = If[a[n-1] == 0, 0, b[n-1]+1];
     b[n] = t; Sum[If[a[j] == t, 1, 0], {j, 1, n-1}]];
b[1] = 0; a[1] = 0;
Array[a, 120] (* Jean-François Alcover, May 03 2021, after Alois P. Heinz *)
  • PARI
    A342585_vec(N,c=[],i)=vector(N,j, while(#c<=i||#c<=c[i+1], c=concat(c,0)); c[i+=1]+if(c[1+c[i]]++&&!c[i]||j==1,i=0)) \\ M. F. Hasler, Nov 13 2021
  • PARI
    \\ See Links section.
  • Python
    def calc(required_value_number):
    values_lst = []
    current_count = 0
    new_value = 0
    for i in range(required_value_number):
        new_value = values_lst.count(current_count)
        values_lst.append(new_value)
        if new_value == 0:
            current_count = 0
        else:
            current_count += 1
    return new_value # Written by Gilad Moyal
  • Python
    from collections import Counter
def aupton(terms):
  num, alst, inventory = 0, [0], Counter([0])
  for n in range(2, terms+1):
    c = inventory[num]
    num = 0 if c == 0 else num + 1; alst.append(c); inventory.update([c])
  return alst
print(aupton(84)) # Michael S. Branicky, Jun 12 2021
  • R
    # Prints the first 10,068 terms
library("dplyr")
options(max.print=11000)
inventory <- data.frame(1, 0)
colnames(inventory) <- c("n", "an")
value_to_count = 0
n = 1
for(x in 1:128) # Increase the 128 for more terms. The number of terms
                # given is on the order of x^1.9 in the region around 128.
  {
  status <- TRUE
  while(status)
    {
    count <- length(which(inventory$an == value_to_count))
    n = n + 1
    inventory <- rbind(inventory, c(n, count))
    status <- isTRUE(count != 0)
    value_to_count = value_to_count + 1
    }
  value_to_count = 0
  }
inventory # Damon Lay, Nov 10 2023

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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