A342130 -id:A342130 - cp's OEIS frontend

A342303 a(0) = 1; for n >= 1, a(n) = the decimal value of the binary number of the total binary digits back from the end of the concatenation of all previous binary terms where the binary value of n last appeared. If the binary value of n has not previously appeared then a(n) = 0.

1, 1, 0, 3, 0, 5, 6, 4, 0, 0, 7, 14, 0, 16, 10, 15, 13, 0, 21, 0, 39, 10, 58, 8, 0, 49, 16, 81, 68, 36, 49, 72, 0, 39, 33, 25, 25, 0, 40, 16, 11, 106, 6, 7, 0, 9, 10, 26, 60, 85, 11, 70, 40, 9, 214, 30, 32, 52, 16, 0, 65, 30, 6, 226, 0, 24, 130, 161, 20, 0, 99, 0, 68, 216, 136, 0, 62, 26, 129

Offset: 0

Scott R. Shannon, Mar 07 2021

The longest run of 0 terms, corresponding to binary values that have not previously appeared, for n up to 100000 is four, starting at n = 512. Not all powers of 2 are 0 as it may have appeared as a previous value's offset, e.g., a(65536) = 412323.

			a(1) = 1 as the binary string concatenation up to a(0) = '1', and the binary value of 1 is '1' which appears 1 (1_2) digit back from the end of the string.
a(2) = 0 as the binary string concatenation up to a(1) = '11', while the binary value of 2 is '10' which does not appear in the string.
a(3) = 3 as the binary string concatenation up to a(2) = '110', and the binary value of 3 is '11' which appears 3 (11_2) digits back from the end of the string.
a(5) = 5 as the binary string concatenation up to a(4) = '110110', and the binary value of 5 is '101' which appears 5 (101_2) digits back from the end of the string.
a(10) = 7 as the binary string concatenation up to a(9) = '11011010111010000', and the binary value of 10 is '1010' which appears 7 (111_2) digits back from the end of the string.

Scott R. Shannon, Image of the first 100000 terms.

Cf. A030190, A341766, A342130, A337227, A181391,

from itertools import count, islice
def agen():
    b = "1"
    yield 1
    for k in count(1):
        bk = bin(k)[2:]
        idx = b.rfind(bk)
        if idx == -1:
            yield 0
            b += "0"
        else:
            yield len(b) - idx
            b += bin(len(b) - idx)[2:]
print(list(islice(agen(), 79))) # Michael S. Branicky, Mar 18 2022

A342127 Numbers m such that the product of m and the string m in reverse contains m as a substring.

Original entry on oeis.org

0, 1, 5, 6, 10, 47, 50, 60, 75, 78, 100, 125, 152, 457, 500, 600, 750, 1000, 1025, 1052, 1250, 1520, 5000, 5625, 6000, 7500, 10000, 10025, 10052, 10250, 10520, 12266, 12500, 15200, 23258, 43567, 50000, 56250, 60000, 62656, 75000, 82291, 90625, 98254, 100000, 100025, 100052, 100250, 100520

Offset: 1

Scott R. Shannon, Mar 01 2021

Numerous patterns exist in the terms, e.g., all numbers of the form 1*10^k, 5*10^k, 6*10^k, 75*10^k, 10^(k+2)+25, where k>=0, are in the sequence.

			6 is a term as 6*reverse(6) = 6*6 = 36 contains '6' as a substring.
47 is a term as 47*reverse(47) = 47*74 = 3478 contains '47' as a substring.
1052 is a term as 1052*reverse(1052) = 1052*2501 = 2631052 contains '1052' as a substring.

Robert Israel, Table of n, a(n) for n = 1..153

Cf. A061205, A342130 (base 2), A001477, A004086, A181721, A203565, A332795.

filter:= proc(n) local L,d,Lp,r,i;
  L:= convert(n,base,10);
  d:= nops(L);
  r:= add(L[-i]*10^(i-1),i=1..d);
  Lp:= convert(n*r,base,10);
  ormap(t -> Lp[t..t+d-1] = L, [$1..nops(Lp)+1-d])
end proc:
select(filter, [$0..120000]); # Robert Israel, Mar 24 2024

Select[Range[0,110000],SequenceCount[IntegerDigits[# IntegerReverse[#]],IntegerDigits[#]]>0&] (* Harvey P. Dale, Apr 20 2024 *)

isok(m) = #strsplit(Str(m*fromdigits(Vecrev(digits(m)))), Str(m)) > 1; \\ Michel Marcus, Mar 01 2021

def ok(n): return (s:=str(n)) in str(n*int(s[::-1]))
print([k for k in range(10**6) if ok(k)]) # Michael S. Branicky, Mar 25 2024

cp's OEIS Frontend

A342303 a(0) = 1; for n >= 1, a(n) = the decimal value of the binary number of the total binary digits back from the end of the concatenation of all previous binary terms where the binary value of n last appeared. If the binary value of n has not previously appeared then a(n) = 0.

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