A329476 Main diagonal of the square array A(n,k). Let D(x) = floor(log_10(x))+1. Then A(1,1) = 1; A(n,n) = #{A(i,j) | D(A(i,j)) = D(A(n-1,n-1)), 1 <= i,j <= n-1}. A(i,n) = A(n,n) + n + 1 - i, 1 <= i < n (column); A(n,j) = A(1,n) + n + 1 - j, 1 <= j < n (row).

1, 1, 4, 9, 10, 15, 26, 39, 54, 71, 90, 100, 34, 125, 61, 154, 92, 162, 152, 189, 228, 269, 312, 357, 404, 453, 504, 557, 612, 669, 728, 789, 852, 917, 984, 1000, 124, 1073, 199, 1150, 278, 1231, 361, 1316, 448, 1405, 539, 1498, 634, 1595, 733, 1696, 836, 1801, 943

Offset

1,3

Comments

Main diagonal of the square array A(n,k). Define D(x) to be the number of digits of x in base 10. A(1,1)= 1; Then A(n,n) = #{A(i,j) | D(A(i,j)) = D(A(n-1,n-1), 1 <= i,j <= n-1}. After the new diagonal A(n,n) is computed, populate the cells above and to the left of the new diagonal: A(i,n) = A(n,n) + n + 1 - i, 1 <= i < n (column); A(n,j) = A(1,n) + n + 1 - j, 1 <= j < n (row).

Links

Table of n, a(n) for n=1..55.

Example

In the array below, A(5,5) = 10. Since it has two digits, we count the numbers in the array that have two digits up to that point. That would be 15. So A(6,6) = 15. Then we populate the 6th column up from the diagonal with 16, 17, 18, 19, 20. Then we populate the 6th row left from the diagonal with 21, 22, 23, 24, 25.
   1,  2,  6, 12, 14, 20, 32, 46, 62, 80, ...
   3,  1,  5, 11, 13, 19, 31, 45, 61, 79, ...
   8,  7,  4, 10, 12, 18, 30, 44, 60, 78, ...
  15, 14, 13,  9, 11, 17, 29, 43, 59, 77, ...
  18, 17, 16, 15, 10, 16, 28, 42, 58, 76, ...
  25, 24, 23, 22, 21, 15, 27, 41, 57, 75, ...
  38, 37, 36, 35, 34, 33, 26, 40, 56, 74, ...
  53, 52, 51, 50, 49, 48, 47, 39, 55, 73, ...
  70, 69, 68, 67, 66, 65, 64, 63, 54, 72, ...
  89, 88, 87, 86, 85, 84, 83, 82, 81, 71, ...

Crossrefs

Sequence in context: A122635 A191912 A191911 * A164098 A007936 A050943

Adjacent sequences: A329473 A329474 A329475 * A329477 A329478 A329479

Keyword

nonn,base

Author

Ali Sada, Nov 13 2019

Status

approved