A289868 Consider the digit reverse of a number x. Take the sum of these digits and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to x.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 17, 21, 25, 42, 63, 84, 143, 286, 2355, 5821, 6618, 11709, 12482, 33747, 39571, 129109, 466957, 1162248, 1565166, 1968084, 3636638, 3853951, 4898376, 13443745, 13933175, 17118698, 22421197, 24153462, 147440984, 209989875, 245742153

Offset: 0

Author

Paolo P. Lava, Jul 14 2017

Keywords

• nonn
• base

Comments

Numbers of iterations for the listed terms are 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 5, 4, 5, 5, 5, 6, 6, 9, 10, 10, 11, 11, 12, 12, 14, 15, 17, 17, 17, 18, 18, 18, 20, 20, 20, 21, 21, 23, 23, 24.

From David A. Corneth, Jul 20 2017: (Start)

If n is in the sequence and its highest digit is m then n * (10\m) is in the sequence for m * (10\m) < 10.

Let T(q, k, n) = b(n) from the following recursion: for 0 <= i <= q-1, b(i) = 1 if i = k, else, b(i) = 0. Then b(n) = Sum(j=1..n, b(n-j)). If some m has q digits d1,..,dk,..,dq with d1 nonzero then after n iterations, we have Sum(j=1..q, T(q, k, n)*d(q-k+1)). This enables an iterative approach to finding solutions with q digits. (End)

Examples

			Digit reverse of 17 is 71 and 7 + 1 = 8, 1 + 8 = 9, 8 + 9 = 17;
Digit reverse of 286 is 682 and 6 + 8 + 2 = 16, 8 + 2 + 16 = 26, 2 + 16 + 26 = 44, 16 + 26 + 44 = 86, 26 + 44 + 86 = 156, 44 + 86 + 156 = 286.
		

Crossrefs

Cf. A000045, A000073, A001590, A004086, A248134, A269309, A269307, A269311.

Programs

Maple

P:=proc(q) local a,b,k,n; for n from 0 to q do a:=convert(n,base,10); b:=convert(a,`+`); while b
				

Mathematica

Select[Range[10^6], Function[n, Total@ NestWhile[Append[Drop[#, 1], Total@ #] &, Reverse@ IntegerDigits@ n, Total@ # < n &] == n]] (* Michael De Vlieger, Jul 20 2017 *)

PARI

is(n) = {my(d = Vecrev(digits(n))); while(vecsum(d)David A. Corneth, Jul 20 2017