A235915 a(1) = 1, a(n) = a(n-1) + (digsum(a(n-1)) mod 5) + 1, digsum = A007953.

1, 3, 7, 10, 12, 16, 19, 20, 23, 24, 26, 30, 34, 37, 38, 40, 45, 50, 51, 53, 57, 60, 62, 66, 69, 70, 73, 74, 76, 80, 84, 87, 88, 90, 95, 100, 102, 106, 109, 110, 113, 114, 116, 120, 124, 127, 128, 130, 135, 140, 141

Offset: 1

Author

Ben Paul Thurston, Jan 16 2014

Keywords

• nonn
• base

Examples

			For n = 7, a(6) is 16, where the sum of the digits is 7, of which the remainder when divided by 5 is 2, then plus 1 is 3. Thus a(7) is a(6) + 3 or 19.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Ben Paul Thurston, Low Kolmorogov complexity but never repeating series?

Crossrefs

Cf. A007953.

Programs

Maple

a:= proc(n) a(n):= `if`(n=1, 1, a(n-1) +1 +irem(
            add(i, i=convert(a(n-1), base, 10)), 5)) end:
seq(a(n), n=1..100);  # Alois P. Heinz, Feb 15 2014

Mathematica

NestList[#+Mod[Total[IntegerDigits[#]],5]+1&,1,50] (* Harvey P. Dale, Nov 23 2023 *)

PARI

digsum(n)=d=eval(Vec(Str(n))); sum(i=1, #d, d[i])
a=vector(1000); a[1]=1; for(n=2, #a, a[n]=a[n-1]+digsum(a[n-1])%5+1); a \\ Colin Barker, Feb 14 2014

Python

def adddigits(i):
  s = str(i)
  t=0
  for j in s:
    t = t+int(j)
  return t
n = 1
a = [1]
for i in range(0, 100):
  r = adddigits(n)%5+1
  n = n+r
  a.append(n)
print(a)