A261236 a(n) = A261234(n) - A261237(n).

0, 1, 3, 7, 16, 40, 104, 279, 758, 2071, 5678, 15609, 43035, 119139, 331616, 928572, 2614743, 7396880, 20999683, 59784414, 170615755, 488073987, 1399625614, 4023315793, 11590737827, 33452982391

Offset: 0

Author

Antti Karttunen, Aug 16 2015

Keywords

• nonn
• base

Comments

a(n) = How many numbers whose base-3 representation begins with digit "1" are encountered before (3^n)-1 is reached when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k by k - (sum of digits in base-3 representation of k).

Examples

			For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), from which we subtract 4, to get 12 ("110" in base-3), from which we subtract 2 to get 10 ("101" in base-3), from which we subtract 2 to get 8 ("22" in base-3), which is the end point of iteration. Of the numbers encountered, 16, 12 and 10 have base-3 representations beginning with digit "1", thus a(2) = 3.
		

Crossrefs

Cf. A261234, A261237, A261230.

Programs

C

/* Use the C-program given in A261234. */

Mathematica

Table[Length@ # - First@ FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, #] &@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &], {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)

Scheme

(define (A261236 n) (- (A261234 n) (A261237 n)))

Formula

a(n) = A261234(n) - A261237(n).

Extensions

Terms a(24) & a(25) from Antti Karttunen, Jun 27 2016