Md. Towhidul Islam - cp's OEIS frontend

A340672 Number of repeated applications of the '4x+-1' operation needed to reach 1 from n, or -1 if 1 is never reached.

0, 3, 1, 8, 6, 4, 4, 11, 2, 36, 9, 9, 34, 16, 7, 7, 32, 5, 14, 5, 5, 39, 30, 12, 12, 12, 3, 46, 37, 37, 28, 19, 10, 10, 10, 10, 44, 35, 35, 26, 26, 17, 17, 17, 8, 8, 8, 8, 42, 33, 33, 33, 24, 6, 24, 15, 15, 15, 15, 6, 6, 49, 6, 40, 40, 40, 31, 31, 31, 31, 22

Offset: 1

Md. Towhidul Islam, Jan 15 2021

Start with any positive number n. Define f(n) as 'if n mod 3 = 1, f(n) = 4n-1; if n mod 3 = 2, f(n) = 4n+1; if n is divisible by 3, f(n) = n/3'. Apply this procedure repeatedly to reach 1. This sequence gives the number of steps required to reach 1, or -1 if 1 is never reached. The procedure is a modification of the famous Collatz procedure.
We can make a conjecture similar to that of Collatz to claim that the process will always reach 1. However, this is not proven yet.
It seems that more than 65 percent of the terms in this sequence satisfy a(i) = a(i+1). For example, 6655 of the first 10000 terms satisfy this condition.

			a(5)=6 because the trajectory of 5 is (5,21,7,27,9,3,1). That needs 6 repeated applications of the procedure to reach 1.

Md. Towhidul Islam, Table of n, a(n) for n = 1..10001

Cf. A006577.

a:= proc(n) option remember; `if`(n=1, 0, 1 + (r->
      a(`if`(r=0, q, 4*n+2*r-3)))(irem(n, 3, 'q')))
    end:
seq(a(n), n=1..75);  # Alois P. Heinz, Jan 20 2021

a[n_] := a[n] = If[n == 1, 0, {q, r} = QuotientRemainder[n, 3]; 1 +
     a[If[r == 0, q, 4n + 2r - 3]]];
Table[a[n], {n, 1, 75}] (* Jean-François Alcover, May 29 2022, after Alois P. Heinz *)

f(n) = my(m=n%3); if (m==1, 4*n-1, if (m==2, 4*n+1, n/3));
a(n) = my(s=n, c=0); while(s>1, s=f(s); c++); c; \\ Michel Marcus, Jan 18 2021

a(n) = 1+a(b(n)) for n>=2 where b(n) = 0, 3, 9, 1, 15, 21, 2, 27,... for n>=0 = 2* b(n-3)-b(n-6). - R. J. Mathar, Mar 14 2025

A340490 a(n) is the number of digits after the n-th digit of the Champernowne constant A033307 until the next appearance of that digit.

Original entry on oeis.org

9, 13, 14, 15, 16, 17, 18, 19, 20, 2, 20, 1, 1, 2, 15, 2, 20, 2, 20, 2, 20, 2, 20, 2, 20, 2, 20, 5, 20, 2, 20, 2, 20, 1, 1, 2, 13, 2, 20, 2, 20, 2, 20, 2, 20, 2, 20, 7, 20, 2, 20, 2, 20, 2, 20, 1, 1, 2, 11, 2, 20, 2, 20, 2, 20, 2, 20, 9, 20, 2, 20, 2, 20, 2

Offset: 1

Md. Towhidul Islam, Jan 10 2021

In typing the positive integers without leaving spaces between them, it is interesting to see how many places later we press the same number key on the keyboard. This sequence answers that question.

			In concatenating the positive integers, we get 1 first. The next occurrence of 1 is in 10. So 1 occurs 9 places later, which gives a(1)=9. The second digit 2 occurs again in writing 12. So 2 occurs 13 places later and a(2) is 13.

Md. Towhidul Islam, Table of n, a(n) for n = 1..31968

Cf. A033307 (the parent sequence).

C(nn) = {my(list = List()); for (n=1, nn, my(d=digits(n)); for (k=1, #d, listput(list, d[k]););); list;} \\ A033307
posi(list, i) = {for (j=i+1, #list, if (list[i] == list[j], return (j-i)););}
lista(nn) = {my(list = C(nn)); my(listp = List()); for (i=1, #list, my(pos = posi(list, i)); if (! pos, break); listput(listp, pos);); Vec(listp);} \\ Michel Marcus, Jan 11 2021

def aupton(terms):
    alst, chapnk, k = [], [1], 1
    for n in range(1, terms+1):
        chapn = chapnk.pop(0)
        while chapn not in chapnk:
            k += 1
            chapnk.extend(list(map(int, str(k))))
        alst.append(chapnk.index(chapn) + 1)
    return alst
print(aupton(74)) # Michael S. Branicky, Sep 13 2021

A033307(n - 1 + a(n)) = A033307(n - 1).

A254442 Triangle read by rows: T(n,k) is the total number of parts of denomination k used in all n-part feasible partitions described in A254296.

Original entry on oeis.org

1, 4, 1, 1, 19, 7, 10, 3, 3, 2, 2, 1, 1, 201, 62, 124, 27, 37, 35, 42, 31, 35, 16, 16, 14, 14, 12, 12, 9, 9, 7, 7, 5, 5, 3, 3, 2, 2, 1, 1, 5020, 1271, 3551, 431, 719, 840, 1128, 851, 1051, 255, 303, 327, 369, 370, 408, 358, 387, 340, 366, 309, 330, 262, 280, 248, 264, 226, 238, 183, 183, 173, 173, 162, 162, 150, 150

Offset: 1

Md. Towhidul Islam, May 12 2015

Row n contains 3^(n-1) terms.

Sum of row n equals n*A254430(n).

			Triangle begins:
1;
4, 1, 1;
19, 7, 10, 3, 3, 2, 2, 1, 1;
201, 62, 124, 27, 37, 35, 42, 31, 35, 16, 16, 14, 14, 12, 12, 9, 9, 7, 7, 5, 5, 3, 3, 2, 2, 1, 1;
5020, 1271, 3551, 431, 719, 840, 1128, 851, 1051, 255, 303, 327, 369, 370, 408, 358, 387, 340, 366, 309, 330, 262, 280, 248, 264, 226, 238, 183, 183, 173, 173, 162, 162, 150, 150, 139, 139, 127, 127, 115, 115, 104, 104, 93, 93, 81, 81, 72, 72, 63, 63, 54, 54, 47, 47, 40, 40, 33, 33, 28, 28, 23, 23, 18, 18, 15, 15, 12, 12, 9, 9, 7, 7, 5, 5 ,3, 3, 2, 2, 1, 1;

Md. Towhidul Islam, Table of n, a(n) for n = 1..364
Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254431, A254432, A254433, A254435, A254436, A254437, A254438, A254439, A254440.

A254440 Palindromes in A254296.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 11, 33, 101, 111, 131, 262, 313, 23132, 86068

Offset: 1

Md. Towhidul Islam, Mar 19 2015

Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

A254438 Natural numbers k such that k is a multiple of its number of "feasible" partitions.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 11, 12, 13, 28, 30, 33, 36, 38, 39, 40, 72, 92, 110, 114, 116, 118, 119, 120, 121, 330, 350, 355, 357, 360, 362, 363, 364, 1086, 1088, 1090, 1091, 1092, 1093, 3248, 3270, 3273, 3276, 3278, 3279, 3280, 9792, 9828, 9830, 9834, 9836, 9838, 9839, 9840, 9841, 29376, 29512, 29515, 29517, 29520, 29522, 29523, 29524

Offset: 1

Md. Towhidul Islam, Mar 01 2015

nonn

This sequence lists the natural numbers k that are divisible by A254296(k).

			For n=1,2,3, A254296(n)=1, so they are in the sequence.
For n=4,6,8,10, A254296(n)=2, so they are in the sequence.
For n=5,9, A254296(n)=2, so they are not in the sequence.

Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254430, A254431, A254432, A254433, A254435, A254436, A254437, A254438, A254439, A254440.

(* This program is not suitable to compute a large number of terms. *)
okQ[v_] := Module[{s=0}, For[i=1, i <= Length[v], i++, If[v[[i]] > 2s+1, Return[False], s += v[[i]]]]; Return[True]];
b[n_] := b[n] = With[{k = Ceiling[Log[3, 2 n]]}, Select[Reverse /@ IntegerPartitions[n, {k}], okQ] // Length];
Reap[Do[If[Divisible[k, b[k]], Print[k]; Sow[k]], {k, 1, 120}]][[2, 1]] (* Jean-François Alcover, Nov 03 2018 *)

a(48)-a(64) added by Md. Towhidul Islam, Apr 18 2015

A254436 A component sequence of A254296.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 4, 3, 6, 3, 6, 5, 8, 7, 10, 7, 12, 9, 14, 11, 16, 14, 19, 17, 22, 20, 28, 23, 31, 26, 34, 32, 40, 35, 43, 38, 51, 46, 59, 51, 64, 61, 74, 71, 84, 76, 94, 86, 104, 96, 114, 108, 126, 120, 138, 132, 157, 146, 171

Offset: 1

Md. Towhidul Islam, Feb 28 2015

nonn

This sequence is a component of the formula for counting A254296.

If m=ceiling(log_3(2k)), define n=(3^(m-1)+1)/2+(3^(m-2))-k for k in the range (3^(m-1)+1)/2<=k<=(3^(m-1)-1)/2+(3^(m-2)). Then this sequence gives the first 3^(m-2) terms.

Md. Towhidul Islam, Table of n, a(n) for n = 1..6561
Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254430, A254431, A254432, A254433, A254435, A254437, A254438, A254439, A254440.

If m=ceiling(log_3(2k)), define n=(3^(m-1)+1)/2+(3^(m-2))-k for k in the range (3^(m-1)+1)/2<=k<=(3^(m-1)-1)/2+(3^(m-2)).

Then a(n)=Sum_{d=ceiling((3k+2)/5)..(3^(m-1)-1)/2} Sum_{p=ceiling((d-1)/3..2d-k-1} A254296(p).

A254437 Natural number that is a factor of its number of "feasible" partition(s).

Original entry on oeis.org

1, 72, 184, 254, 539, 1743, 2874, 5589, 21316, 37581, 49829, 61047

Offset: 1

Md. Towhidul Islam, Mar 01 2015

This sequence lists the natural numbers m such that A254296(m) is divisible by m.

			A254296(1) = 1 and A254296(72) = 72, so 1 and 72 are in this sequence.
A254296(184) = 2208 = 12*184, so 184 is here too.

Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254430, A254431, A254433, A254434, A254435, A254436, A254438, A254439, A254440.

a(9) added by Md. Towhidul Islam, Apr 18 2015

a(10)-a(12) from Robert Price, Mar 28 2019

A254435 Squares in A254296.

Original entry on oeis.org

1, 9, 81, 729, 1296, 23532201

Offset: 1

Md. Towhidul Islam, Feb 03 2015

They seem to be in the ((3^(n-1) + 1)/2)-th to ((3^(n-1)-1)/2 + 3^(n-2))-th values of A254296, though this is not proved. The terms are the squares of: 1, 3, 9, 27, 36, 4851.

Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254430, A254431, A254432, A254433, A254436, A254437, A254438, A254439, A254440, A254442.

A254439 Median of terms of A254296 in the range (3^(n-1)+1)/2 to (3^n-1)/2.

Original entry on oeis.org

1, 1, 2, 7, 47, 682, 23132, 1913821, 397731998, 212521309666, 297464368728296

Offset: 1

Md. Towhidul Islam, Mar 01 2015

As described in A254296, all the 'feasible' partitions of natural numbers (3^(n-1)+1)/2 to (3^n-1)/2 has n parts. A254439 lists the "median of the range ((3^(n-1)+1)/2)-th to ((3^n-1)/2)-th terms of A254296".

From conjectured formula, it appears that next terms are 1107102779611719118, 11090084422457163934046, 302002529294596303158583642. - Benedict W. J. Irwin, Nov 16 2016

			As described in sequence A254296, "feasible" partitions of the integers 41 through 121 consist of 5 parts. The number 3^(5-1) = 81 has 47 "feasible" partitions, which is the median of the range from the 41st to the 121st term of A254296.

Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254430, A254431, A254432, A254433, A254435, A254436, A254437, A254438, A254440.

/* C Code to make Mathematica Code for conjectured n-th term n>3 */
#include 
int main(int argc, char* argv[]){
int i, n=atoi(argv[1])-3;
printf("F[a_,x_,k_]:=Sum[x,{a,1,k}]\n");
for(i=1; i<=n; i++)printf("F[i%d,",i);
printf("3i%d-1,",n);
for(i=n-1; i>0; i--)printf("3i%d-1],",i);
printf("2]\n");
return 0;
}
/* Benedict Irwin, Nov 16 2016 */

F[a_, x_, k_] := Sum[x, {a, 1, k}]
F[i1, 3*i1 - 1, 2]
F[i1, F[i2, 3*i2 - 1, 3*i1 - 1], 2]
F[i1, F[i2, F[i3, 3*i3 - 1, 3*i2 - 1], 3*i1 - 1], 2]
F[i1, F[i2, F[i3, F[i4, 3*i4 - 1, 3*i3 - 1], 3*i2 - 1], 3*i1 - 1], 2] (* Examples of how to get first few terms, use the C code to generate the n-th term of the conjectured formula, Benedict W. J. Irwin, Nov 16 2016 *)

a(n) = A254296(3^(n-1)).

Conjecture: for n>3, a(n+3) = Sum_{i_1=1..2} Sum_{i_2=1..3*i_1-1} ... Sum_{i_n..3*i_(n-1)-1} (3*i_n - 1). - Benedict W. J. Irwin, Nov 16 2016

A254433 Maximum number of "feasible" partitions of length n.

Original entry on oeis.org

1, 1, 3, 12, 140, 3950, 263707, 42285095, 16825391023, 17095967464466, 45375565948693336

Offset: 1

Md. Towhidul Islam, Feb 03 2015

a(n) gives the highest value in the (3^(n-1)+1)/2-th through the (3^n-1)/2-th terms of the sequence A254296. It lists the highest possible number of "feasible" partitions into n parts.

			The numbers 2, 3 and 4 are "feasibly" partitionable into 2 parts. Each of them has 1 feasible partitions. So a(2)=1.
The numbers 14 to 40 are "feasibly" partitionable into 4 parts. Among them 16, 18, 19 and 22 each has the highest 12 "feasible" partitions. So a(4)=12.
The numbers 122 to 364 are "feasibly" partitionable into 6 parts. Among them 124 has the highest 3950 "feasible" partitions. So a(6)=3950.

Md Towhidul Islam & Md Shahidul Islam, Number of Partitions of an n-kilogram Stone into Minimum Number of Weights to Weigh All Integral Weights from 1 to n kg(s) on a Two-pan Balance, arXiv:1502.07730 [math.CO], 2015.

Cf. A254296, A254430, A254431, A254432, A254435, A254436, A254437, A254438, A254439, A254440, A254442.

The first term is 1. For n>=2, a(n) = A254296((3^(n-1)+5)/2).

a(9) corrected and a(10)-a(11) added by Md. Towhidul Islam, Apr 18 2015

cp's OEIS Frontend

User: Md. Towhidul Islam

A340672 Number of repeated applications of the '4x+-1' operation needed to reach 1 from n, or -1 if 1 is never reached.

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A340490 a(n) is the number of digits after the n-th digit of the Champernowne constant A033307 until the next appearance of that digit.

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A254442 Triangle read by rows: T(n,k) is the total number of parts of denomination k used in all n-part feasible partitions described in A254296.

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A254440 Palindromes in A254296.

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A254438 Natural numbers k such that k is a multiple of its number of "feasible" partitions.

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A254436 A component sequence of A254296.

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A254437 Natural number that is a factor of its number of "feasible" partition(s).

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A254435 Squares in A254296.

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A254439 Median of terms of A254296 in the range (3^(n-1)+1)/2 to (3^n-1)/2.

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