login
Irregular triangle T(n,b) (n >= 3, 2 <= b <= A230624(n)/2+1) read by rows. Let m = A230624(n). Then T(n,b) is the smallest nonnegative number k such that k+S_b(k)=m, where S_b(k) is the sum of the digits of k in base b.
1

%I #16 Jul 17 2024 04:17:44

%S 7,7,8,7,5,11,11,10,9,12,10,7,19,17,20,17,16,17,18,15,20,16,11,35,34,

%T 31,33,29,31,33,31,28,29,30,25,32,26,34,27,36,28,19,58,58,55,57,56,55,

%U 52,51,49,51,53,49,57,52,46,47,48,49,50,41,52,42,54,43,56,44,58,45,60,46,31

%N Irregular triangle T(n,b) (n >= 3, 2 <= b <= A230624(n)/2+1) read by rows. Let m = A230624(n). Then T(n,b) is the smallest nonnegative number k such that k+S_b(k)=m, where S_b(k) is the sum of the digits of k in base b.

%C T(n,b) must be nonzero for all b >= 2, and so this triangle is actually the upper left corner of an array with infinitely long rows (it is believed that there are also infinitely many rows).

%C Since T(n,b) = m/2 for all b > m/2, we may truncate row n after m/2 terms. The rows do not change beyond that point.

%e Triangle begins as follows:

%e n m Row n

%e 3 10 [7, 7, 8, 7, 5],

%e 4 14 [11, 11, 10, 9, 12, 10, 7],

%e 5 22 [19, 17, 20, 17, 16, 17, 18, 15, 20, 16, 11],

%e 6 38 [35, 34, 31, 33, 29, 31, 33, 31, 28, 29, 30, 25, 32, 26, 34, 27, 36, 28, 19],

%e 7 62 [58, 58, 55, 57, 56, 55, 52, 51, 49, 51, 53, 49, 57, 52, 46, 47, 48, 49, 50, 41, 52, 42, 54, 43, 56, 44, 58, 45, 60, 46, 31],

%e 8 94 [90, 89, 89, 87, 87, 83, 89, 79, 83, 82, 80, 77, 86, 82, 77, 79, 81, 74, 85, 77, 89, 80, 70, 71, 72, 73, 74, 75, 76, 77, 78, 63, 80, 64, 82, 65, 84, 66, 86, 67, 88, 68, 90, 69, 92, 70, 47],

%e ...

%e For n = 3, m = A230624(3) = 10, and row 3 of the triangle is [7, 7, 8, 7, 5], corresponding to the identities (where x_b is the base-b expansion of x):

%e 10 = 111_2 + 3 = 7 + 3,

%e = 21_3 + 3 = 7 + 3

%e = 20_4 + 2 = 8 + 2

%e = 12_5 + 3 = 7 + 3

%e = 5_b + 5 = 5 + 5 for all b >= 6.

%Y Cf. A230624.

%K nonn,tabf,base

%O 3,1

%A _N. J. A. Sloane_, Dec 30 2021