A147991 -id:A147991 - cp's OEIS frontend

A368239 Irregular table of nonnegative integers T(n, k), n >= 0, k = 1..A080100(n), read by rows; the 1's in the binary expansion of n exactly match the 1's in the balanced ternary expansions of the terms in the n-th row.

0, 1, 2, 3, 4, 5, 6, 8, 9, 7, 10, 11, 12, 13, 14, 15, 17, 18, 23, 24, 26, 27, 16, 19, 25, 28, 20, 21, 29, 30, 22, 31, 32, 33, 35, 36, 34, 37, 38, 39, 40, 41, 42, 44, 45, 50, 51, 53, 54, 68, 69, 71, 72, 77, 78, 80, 81, 43, 46, 52, 55, 70, 73, 79, 82, 47, 48, 56, 57, 74, 75, 83, 84

Offset: 0

Rémy Sigrist, Dec 18 2023

As a flat sequence, this is a permutation of the nonnegative integers with inverse A368240.

			Table T(n, k) begins:
    0;
    1;
    2, 3;
    4;
    5, 6, 8, 9;
    7, 10;
    11, 12;
    13;
    14, 15, 17, 18, 23, 24, 26, 27;
    16, 19, 25, 28;
    20, 21, 29, 30;
    22, 31;
    32, 33, 35, 36;
    34, 37;
    38, 39;
    40;
    ...

Rémy Sigrist, Table of n, a(n) for n = 0..9841 (rows for n = 0..2^9-1 flattened)
Index entries for sequences that are permutations of the natural numbers

See A368225 for a similar sequence.

Cf. A005836, A080100, A147991, A343228, A368240 (inverse).

row(n) = { my (r = [0], b = binary(n)); for (k = 1, #b, r = [3*v+b[k]|v<-r]; if (b[k]==0, r = concat(r, [v-1|v<-r]););); Set(r); }

T(n, 1) = A147991(n) for any n > 0.
T(n, A080100(n)) = A005836(n + 1).
A343228(T(n, k)) = n.

A308364 a(0) = 0, a(3n) = a(n), a(3n+1) = a(n)3 + 1, a(3n-1) = a(n)3 - 1.

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 2, 7, 2, 1, 4, 11, 4, 13, 14, 5, 16, 5, 2, 7, 20, 7, 22, 5, 2, 7, 2, 1, 4, 11, 4, 13, 32, 11, 34, 11, 4, 13, 38, 13, 40, 41, 14, 43, 14, 5, 16, 47, 16, 49, 14, 5, 16, 5, 2, 7, 20, 7, 22, 59, 20, 61, 20, 7, 22, 65, 22, 67, 14, 5, 16, 5, 2, 7, 20, 7, 22, 5, 2, 7, 2, 1

Offset: 0

Peter Munn, May 22 2019

Defines a function on all the integers, but only nonnegative terms are in the data. A147991 gives the nonnegative fixed points of the function and the nonnegative part of its image.
Consider a Sierpinski arrowhead curve to be formed of vectors placed head to tail and numbered consecutively from 0 at its axis of symmetry. Vector a(n) equals vector n.
Removing all 0's from the balanced ternary expansion of n yields a(n). - Charlie Neder, Jun 03 2019

			As 6 is congruent to 0 modulo 3, a(6) = a(3*2) = a(2).
As 2 is congruent to -1 modulo 3, a(2) = a(3*1 - 1) = a(1)*3 - 1.
As 1 is congruent to 1 modulo 3, a(1) = a(0*1 + 1) = a(0)*3 + 1 = 0*3 + 1 = 1.
So a(2) = a(1)*3 - 1 = 1*3 - 1 = 2. So a(6) = a(2) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..19683
Peter Munn, Sierpiński arrowhead with numbered vectors.
Wikipedia, Sierpiński arrowhead curve

Cf. A147991, A307672.

a:=[1]; for n in [2..80] do  if n mod 3 eq 2 then a[n]:= 3*a[(n+1) div 3]-1; end if; if n mod 3 eq 1 then a[n]:=3*a[(n-1) div 3]+1; end if; if n mod 3 eq 0 then a[n]:=a[n div 3]; end if; end for; [0] cat a; // Marius A. Burtea, Nov 14 2019

a(n) = if (n == 0, 0, r = n%3; if (r==0, a(n/3), if (r==1, 3*a((n-1)/3)+1, 3*a((n+1)/3)-1))); \\ Michel Marcus, May 29 2019

a(-n) = - a(n).
A307672(a(n)) = A307672(n).
a(A147991(n)) = A147991(n).
a(9n-1) = a(9n-3). a(9n+1) = a(9n+3).

A351243 Counterexamples to a conjecture of Selfridge and Lacampagne.

Original entry on oeis.org

247, 277, 967, 977, 1211, 1219, 1895, 1937, 1951, 1961, 2183, 2191, 2911, 2921, 3029, 3641, 3649

Offset: 1

Jeffrey Shallit, Feb 05 2022

The conjecture was that every natural number k not divisible by 3 can be written as the quotient of two terms chosen from A147991.
For every specific k, the problem of representing k as the quotient of two terms of A147991 can be decided by using a queue-based breadth-first search algorithm on the transition diagram of a finite automaton that on input j in base 3 computes j*k and checks to see if both j and j*k are in A147991.
It is not known if there are infinitely many counterexamples to the conjecture, but perhaps 3^m+4, for m >= 5 and odd, are.

R. K. Guy, Unsolved Problems in Number Theory, Springer, 2004. In Section F31, the conjecture of Selfridge and Lacampagne is mentioned, and it is stated that Don Coppersmith found the counterexample 247.

James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, Quotients of Palindromic and Antipalindromic Numbers, INTEGERS 22 (2022), #A96.
J. H. Loxton and A. J. van der Poorten, An Awful Problem About Integers in Base Four, Acta Arithmetica, volume 49, 1987, pages 193-203. In section 7, Selfridge and Lacampagne ask whether every k != 0 (mod 3) is the quotient of two terms of this sequence.

Cf. A147991.

A359925 Numbers with easy multiplication table - the first 9 multiples of these numbers can be derived by either incrementing or decrementing the corresponding digits from the previous multiple.

Original entry on oeis.org

1, 9, 11, 89, 91, 109, 111, 889, 891, 909, 911, 1089, 1091, 1109, 1111, 8889, 8891, 8909, 8911, 9089, 9091, 9109, 9111, 10889, 10891, 10909, 10911, 11089, 11091, 11109, 11111, 88889, 88891, 88909, 88911, 89089, 89091, 89109, 89111, 90889, 90891, 90909, 90911

Offset: 1

Kiran Ananthpur Bacche, Jan 25 2023

This is also the list of numbers having exactly one dot or one antidot in each box in the Decimal Exploding Dots notation.

			a(4) = 89. The first nine multiples of 89 are {089, 178, 267, 356, 445, 534, 623, 712, 801}. The digits in the hundreds place increment by 1, while the digits in the tens and units place decrement by 1. In the Decimal Exploding Dots notation, 89 is represented as DOT-ANTIDOT-ANTIDOT = 100 - 10 - 1 = 89

Global Math Week, Exploding Dots.
Gevorg Hmayakyan, Generalizing a Trig Identity [mentions this sequence]

Cf. A006257, A147991, A147992, A153777, A147993, A256290.

Column k=10 of A360099.

a:= proc(n) option remember;
      `if`(n=0, 0, 10*a(iquo(n, 2, 'm'))+2*m-1)
    end:
seq(a(n), n=1..44);   # Alois P. Heinz, Jan 25 2023

a(n) = 10*a(floor(n/2))+2*(n mod 2)-1 for n>0, a(0)=0. - Alois P. Heinz, Jan 25 2023

a(n) = 2*A256290(n-1) + 1 for n>1. - Hugo Pfoertner, Jan 28 2023

cp's OEIS Frontend

A368239 Irregular table of nonnegative integers T(n, k), n >= 0, k = 1..A080100(n), read by rows; the 1's in the binary expansion of n exactly match the 1's in the balanced ternary expansions of the terms in the n-th row.

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A308364 a(0) = 0, a(3n) = a(n), a(3n+1) = a(n)*3 + 1, a(3n-1) = a(n)*3 - 1.

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Magma

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A351243 Counterexamples to a conjecture of Selfridge and Lacampagne.

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A359925 Numbers with easy multiplication table - the first 9 multiples of these numbers can be derived by either incrementing or decrementing the corresponding digits from the previous multiple.

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Maple

Formula

A308364 a(0) = 0, a(3n) = a(n), a(3n+1) = a(n)3 + 1, a(3n-1) = a(n)3 - 1.