A352610 a(0) = 0; a(n+1) = Sum_{d a distinct decimal digit in a(n)} binomial(c(d)-1,2), where c(d) is the number of occurrences of d in a(j), 0 <= j <= n; see comments.

0, 0, 1, 0, 3, 0, 6, 0, 10, 16, 4, 0, 21, 6, 3, 1, 10, 43, 4, 3, 6, 6, 10, 57, 0, 45, 7, 1, 28, 1, 36, 25, 6, 21, 51, 61, 94, 10, 133, 112, 130, 230, 129, 175, 184, 206, 155, 231, 312, 353, 106, 426, 131, 416, 445, 81, 381, 517, 486, 143, 651, 642, 249, 172, 629

Offset: 0

Author

David James Sycamore, Mar 16 2022

Keywords

• nonn
• base

Comments

In other words, a(n+1) is the sum, taken over each distinct d in a(n), of the number of pairs (d,d) counted from different occurrences of d in a(j), 0 <= j <= n.

Every term is the sum of one or more triangular numbers (A000217). If no digit in a(n) has occurred before, a(n+1) = 0 (there are 7 such terms). Some numbers (0,1,3,4,6,10,...) occur multiple times, while others (2,5,8,11,...) never occur.

a(28844) = 2417583609 is the first pandigital term. Similar to A279818.

Examples

			a(1) = 0 because there are no terms prior to a(0)=0;
a(2) = 1 because 0 has occurred just twice, and 0,0 can be counted just once.
To calculate a(39) given that a(38) = 133: c(1) = 14 so (1,1) can be counted 13*14/2=91 times, c(3) = 7 so (3,3) can be counted 6*7/2 = 21 times. Therefore a(39) = 91 + 21 = 112.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• Michael De Vlieger, Log log scatterplot of a(n), n = 0..10^6, representing a(n) = 0 instead as a(n) = 1/2 for visibility.

Crossrefs

Cf. A000217, A279818.

Programs

Mathematica

Block[{a, c, d, nn}, nn = 65; d[] = -1; a[1] = d[0] = 0; Do[c = DigitCount[a[n - 1]]; MapIndexed[AddTo[d[First[#2]], #1] &, c]; a[n] = Total@ Map[# (# + 1)/2 &@ d[#] &, Position[c, ?(# > 0 &)][[All, 1]] ], {n, 2, nn}]; Array[a, nn] ] (* Michael De Vlieger, Mar 23 2022 *)