Rakesh Khanna A - cp's OEIS frontend

User: Rakesh Khanna A

Rakesh Khanna A has authored 3 sequences.

A346576 Let x run through the list of numbers with no zeros (A052382); replace each digit d of x by the digit (x mod d).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 4, 3, 2, 1, 10, 0, 12, 0, 10, 2, 16, 4, 12, 10, 20, 0, 12, 20, 0, 12, 26, 3, 10, 20, 31, 0, 10, 24, 35, 0, 14, 10, 20, 32, 42, 0, 12, 21, 32, 45, 10, 20, 30, 40, 50, 0, 14, 24, 36, 10, 20, 31, 42, 50, 64, 0, 16, 27, 10

Offset: 1

Rakesh Khanna A, Jul 24 2021

Graph of the sequence generates a fractal-like image.

			If x = 247 we get 132 as 247 mod 2 = 1, 247 mod 4 = 3, and 247 mod 7 = 2. As 247 is the 205th zeroless number,  a(205) =  132.

Michel Marcus, Table of n, a(n) for n = 1..7380
Rakesh Khanna A, Graph of the sequence

Cf. A034838, A052382.

See A347323 for another version.

#include 
#define START 1
#define END 1000
int main(){
  unsigned int R,N,M,power_cntr;
  int mod1,mod2;
  for(N=START;N<=END;N++){
    R=N;
    M=0;
    power_cntr=1;
    while(R!=0){
      mod1=R%10;
      if(mod1==0) break;
      mod2=N%mod1;
      M+=mod2*power_cntr;
      power_cntr*=10;
      R=R/10;}
    if(mod1!=0) printf("%u %u\n",N,M);}
  return 0;}

f[n_] := FromDigits @ Mod[n, IntegerDigits[n]]; f /@ Select[Range[100], !MemberQ[IntegerDigits[#], 0] &] (* Amiram Eldar, Jul 26 2021 *)

a(m) = my(d=digits(m)); fromdigits(Vec(apply(x->(m % x), d)));
apply(x->a(x), select(x->vecmin(digits(x)), [1..100])) \\ Michel Marcus, Jul 24 2021

def f(k, digits): return int("".join(map(str, map(lambda x: k%x, digits))))
def aupton(terms):
    alst, k = [], 1
    while len(alst) < terms:
        s = str(k)
        if '0' not in s: alst.append(f(k, list(map(int, s))))
        k += 1
    return alst
print(aupton(73)) # Michael S. Branicky, Aug 22 2021

A335075 Positions of 1's when Kolakoski sequence is grouped into four independent numbers as 1, 2, 11, 22.

Original entry on oeis.org

1, 5, 7, 13, 17, 23, 25, 29, 31, 37, 41, 43, 49, 51, 55, 59, 61, 67, 71, 73, 79, 85, 91, 95, 97, 101, 103, 109, 113, 115, 121, 125, 131, 133, 137, 141, 143, 149, 155, 157, 161, 167, 173, 175, 179, 185, 187, 191, 193, 199, 201, 205, 209, 211, 217, 221, 223, 227, 229

Offset: 1

Rakesh Khanna A, May 22 2020

nonn

			A000002 can be grouped as 1, 22, 11, 2, 1, 22, 1, 22, 11, 2, ... The positions of occurrence of 1's (1, 5, 7, ...) are given in this sequence.

Cf. A000002.

s = {1, 2, 2}; Do[s = Append[s, Mod[n-1, 2] + 1], {n, 3, 230}, {s[[n]]}]; Position[Split[s], {1}] // Flatten (* Amiram Eldar, Jul 24 2020 *)

A334955 Number of 1's in n-th string when A000002 is expressed as 1, 2, 2, 11, 21, 221, 22112, 11221211, 21221121121, 2212211212212112, ....

Original entry on oeis.org

0, 2, 1, 1, 2, 5, 6, 7, 12, 19, 29, 43, 62, 95, 148, 216, 322, 482, 727, 1083, 1627, 2457, 3660, 5518, 8250, 12403, 18594, 27914, 41863, 62781, 94158, 141205, 211797, 317782, 476743, 714948, 1072526, 1608732, 2413223, 5428986

Offset: 3

Rakesh Khanna A, May 24 2020

nonn

Representing A000002 as a tree. Each branch of this tree is a string. Starting from n=3, each 1 in n-th string generates either 1 or 2 in (n+1)-th string and each 2 in n-th string generates either 11 or 22 in (n+1)-st string based on the previously generated term of either 2 or 1. Hence number of terms in (n+1)-st string is sum of all terms in n-th string.

Cf. A000002.

MAX_NUM = 10000 # Number of terms in Kolakoski Sequence
K = [1,2,2]
previous = 2
# Generate Kolakoski Sequence
for index1 in range(2,MAX_NUM):
    generator = K[index1]
    if (generator == 1 and previous == 1):
        K.append(2)
        previous = 2
    elif(generator == 2 and previous == 1):
        K.append(2)
        K.append(2)
        previous = 2
    elif(generator == 1 and previous == 2):
        K.append(1)
        previous = 1
    elif(generator == 2 and previous == 2):
        K.append(1)
        K.append(1)
        previous = 1
branch_sum = 1
index2 = 2
cntr1 = 1
while(index2 < MAX_NUM):
    ones_cntr = 0
    # This for loop extracts strings from Kolakoski Sequence
    start_index = index2
    end_index = index2 + branch_sum
    branch_sum = 0
    for index3 in range(start_index , end_index):
        branch_sum = branch_sum + K[index3]
        ones_cntr = ones_cntr + (K[index3]%2)
    index2 = end_index
    print(str(cntr1)+"   "+str(ones_cntr)+"\n")
    cntr1 = cntr1 + 1

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User: Rakesh Khanna A

A346576 Let x run through the list of numbers with no zeros (A052382); replace each digit d of x by the digit (x mod d).

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A335075 Positions of 1's when Kolakoski sequence is grouped into four independent numbers as 1, 2, 11, 22.

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A334955 Number of 1's in n-th string when A000002 is expressed as 1, 2, 2, 11, 21, 221, 22112, 11221211, 21221121121, 2212211212212112, ....

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