A333970 - OEIS

%I #99 Aug 05 2023 21:19:50

%S 2,2,3,2,4,2,5,2,3,6,2,3,7,2,3,4,8,2,3,9,2,5,10,2,11,2,3,4,6,12,2,4,

%T 13,2,4,7,14,2,3,4,5,15,2,3,4,8,16,2,3,17,2,3,6,9,18,2,3,19,2,3,4,5,

%U 10,20,2,3,5,7,21,2,3,5,11,22,2,3,5,23,2,3,4,5,6,8,12,24

%N Irregular triangle read by rows where the n-th row lists the bases 2<=b<=n+1 where n in base b contains the digit b-1.

%C If a number n has base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)) contains digit b-1, where b = q*(k+1)/k, k>=1 , and Sum_{i>=0} ((A(i)(mod b-q))*((b-q)^i)) > 0 then there exists  n' < n such that that n' in base b-q = b' contains digit b'-1 at the same place as n in base b and 0 <= (A(i)-A'(i))/b' <= (k+1)-((A'(i)+1)/b') (A'(i) is digit of n' in base b')for all i>=0.*

%C This condition is necessary and sufficient.

%C Proof that Condition is Necessary:

%C Since b-1 = b-q+q-1 and b' = q/k (as b = q*(k+1)/k). Therefore (b-1) (mod b') = (b'+q-1) (mod b') = (q-1) (mod b') = b'-1 :-(1).

%C n in base 'b' representation = (... (b-1) A(j-1) ...A(3) A(2) A(1) A(0)).Then n = Sum_{i>=0} (A(i)*(b^i)) = Sum_{i>=0} (A(i)*((b-q+q)^i)).

%C n = Sum_{i>=0} (A(i)*(b'^i)) +

%C     Sum_{i>=1} (A(i)*(b^i - b'^i))

%C    = Sum_{i>=0} (A'(i)*(b'^i)) + Sum_{i>=0} ((A(i)-A'(i))* (b'^i)) + Sum_{i>=1} (A(i)*(b^i - b'^i)),

%C          where A'(i) = A(i) (mod b').

%C Now n-Sum_{i>=0} ((A(i)-A'(i))*(b'^i))

%C        - Sum_{i>=1} (A(i)*(b^i - b'^i))

%C        = Sum_{i>=0} (A'(i)*(b'^i)).

%C Since A'(j) = A(j) (mod b') =  (b-1) (mod b') = b'-1(due to equation (1) above and A(j) = b-1.

%C Hence there exists n'  = Sum_{i>=0} (A'(i)*(b'^i)) > 0 containing digit b'-1 in base b'.

%C Table of n/b with cell containing T(n, b) = (n', b') for q = b/2. n' = Sum_{i>=0} (A'(i)*(b'^i))

%C n/b|  4  |  6  |  8  |  10 | 12

%C 3  |(1,2)|     |     |     |

%C 4  |     |     |     |     |

%C 5  |     |(2,3)|     |     |

%C 6  |     |     |     |     |

%C 7  |(3,2)|     |(3,4)|     |

%C 8  |     |     |     |     |

%C 9  |     |     |     |(4,5)|

%C 10 |     |     |     |     |

%C 11 |(1,2)|(5,3)|     |     |(5,6)

%C Example: For table n/b in comments containing (n',b') in its cells.

%C For n = 7:

%C In base b = 4, n = 13 :- q = b' = 4/2 = 2, and n' = (3 mod (2))*(2)^0 + (1 mod(2))*(2)^1 = 1+2 = 3.

%C In base b = 8, n = 7 :- q = b' = 8/2 = 4, and n' = (7 mod (4))*(4)^0 = 3.

%C There are no other bases b >= 4 except 4, 8 for n = 7.

%C (n, b) maps to (0, 1) if b is prime. Following this and comment in A337536 we can say that all of the terms of A337536 will map to (0, 1) only, except A337536(2).

%C For above (n, b) -> (n', b') one possible (n, b) pair for (n', b') is { Sum_{i>=0} ((A'(i)+b') *((2*b')^i)), 2*b'}.

%H Rémy Sigrist, <a href="/A333970/b333970.txt">Table of n, a(n) for n = 1..10027</a> (rows for n = 1..1014, flattened)

%e Triangle begins

%e   Row    Bases

%e   n=1:   2

%e   n=2:   2  3

%e   n=3:   2  4

%e   n=4:   2  5

%e   n=5:   2  3  6

%e   n=6:   2  3  7

%e   n=7:   2  3  4  8

%e   n=8:   2  3  9

%e   n=9:   2  5  10

%e   n=10:  2  11

%o (PARI) row(n) = {my(list = List()); for (b=2, n+1, if (vecmax(digits(n, b)) == b-1, listput(list, b));); Vec(list);} \\ _Michel Marcus_, Sep 11 2020

%Y Cf. A337535 (second column), A338295 (penultimate column), A337496 (row widths), A337536 (width 2), A337143 (width 3).

%Y Rows containing bases 3..11 respectively: A074940, A337250, A337572, A333656, A337141, A337239, A338090, A011539, A095778.

%K nonn,base,tabf,uned

%O 1,1

%A _Devansh Singh_, Sep 03 2020

%E More terms from _Michel Marcus_, Sep 11 2020