A351753 -id:A351753 - cp's OEIS frontend

A357432 a(1) = 1; a(2) = 2; for n > 2, a(n) is the smallest positive number not occurring earlier such that a(n) plus the sum of all previous terms appears in the string concatenation of a(1)..a(n-1).

1, 2, 9, 17, 62, 38, 47, 115, 93, 87, 122, 30, 88, 51, 85, 4, 3, 31, 32, 21, 221, 64, 68, 302, 53, 116, 92, 268, 42, 48, 18, 78, 76, 97, 50, 153, 233, 108, 63, 20, 8, 16, 89, 12, 77, 537, 24, 377, 83, 46, 306, 28, 107, 197, 170, 126, 61, 566, 218, 82, 43, 25, 14, 148, 147, 6, 209, 145, 37, 103

Offset: 1

Scott R. Shannon, Sep 28 2022

The sequence is conjectured to be a permutation of the positive integers. In the first 20000 terms the fixed points are 393, 514, 1546, and 7854, although more likely exist.

			a(4) = 17 as a(1) + a(2) + a(3) + 17 = 1 + 2 + 9 + 17 = 29, and "29" appears in the string concatenation of a(1)..a(3) = "129".

Scott R. Shannon, Image of the first 20000 terms. The green line is a(n) = n.

Cf. A357433 (base 2), A000027, A007908, A000217, A351753, A337227.

A352245 a(0) = 1; for n >= 1, a(n) = the decimal value of the binary number of the index of where n first appears in the concatenation of all previous binary terms. If the binary value of n has not previously appeared then a(n) = 0.

Original entry on oeis.org

1, 1, 0, 1, 0, 2, 1, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 8, 0, 0, 6, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 20, 0, 8, 0, 0, 0, 0, 0, 6, 0, 2, 0, 0, 0, 0, 0, 0, 0, 56, 0, 26, 1, 0, 0, 69, 0, 0, 0, 0, 0, 0, 0, 47, 20, 0, 71, 8, 84, 0, 110, 57, 0, 0, 0, 0, 0, 0, 0, 27, 6, 79, 155, 4, 2, 0, 0, 0, 0, 0, 0, 134

Offset: 0

Scott R. Shannon, Mar 09 2022

nonn

In the first 250000 terms the longest run of consecutive 0 terms is seven, the first occurrence of which starts at a(43). It is unknown if longer runs exists. See the companion sequence A352246 for the indices where a(n) = 0.

			a(1) = 1 as the binary string concatenation up to a(0) = '1', and the binary value of 1 is '1' which appears at index 1 in 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) = 1 as the binary string concatenation up to a(2) = '110', and the binary value of 3 is '11' which appears at index 1 in the string.
a(5) = 2 as the binary string concatenation up to a(4) = '11010', and the binary value of 5 is '101' which appears at index 2 in the string.
a(17) = 8 as the binary string concatenation up to a(16) = '1101010100010001000', and the binary value of 17 is '10001' which appears at index 8 in the string.

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

Cf. A352246, A342303 (from end), A351753, A341766.

from itertools import count, islice
def agen():
    b = "1"
    yield 1
    for k in count(1):
        bk = bin(k)[2:]
        idx = b.find(bk) + 1
        yield idx
        b += bin(idx)[2:]
print(list(islice(agen(), 93))) # Michael S. Branicky, Mar 18 2022

A352246 Indices k where A352245(k) = 0.

Original entry on oeis.org

2, 4, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19, 22, 23, 24, 25, 27, 28, 29, 30, 31, 33, 35, 36, 37, 38, 39, 41, 43, 44, 45, 46, 47, 48, 49, 51, 54, 55, 57, 58, 59, 60, 61, 62, 63, 66, 70, 73, 74, 75, 76, 77, 78, 79, 86, 87, 88, 89, 90, 91, 96, 99, 101, 103, 107, 117, 118, 123, 126, 127, 140, 147, 151

Offset: 0

Scott R. Shannon, Mar 09 2022

nonn

See A352245 for further details and examples.

Scott R. Shannon, Image of the zero indices for the first 100000 terms of A352245.

Cf. A352245, A342303, A351753, A341766.

from itertools import count, islice
def agen():
    b = "1"
    for k in count(1):
        bk = bin(k)[2:]
        idx = b.find(bk) + 1
        if idx == 0: yield k
        b += bin(idx)[2:]
print(list(islice(agen(), 73))) # Michael S. Branicky, Mar 18 2022

A357930 a(0) = 0; for n > 0, let S = concatenation of a(0)..a(n-1); a(n) is the number of times the digit at a(n-1) digits back from the end of S appears in S.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 7, 7, 8, 8, 7, 7, 8, 8, 7, 13, 7, 14, 7, 13, 13, 15, 15, 13, 15, 15, 6, 17, 16, 19, 8, 10, 10, 22, 10, 23

Offset: 0

Scott R. Shannon, Oct 21 2022

			a(7) = 3 as a(6) = 3 and the string concatenation of a(0)..a(6) = "0112223", and the digit 3 digits back from the end of the string concatenation is 2, and 2 has appeared three times in the string.

Michael De Vlieger, Table of n, a(n) for n = 0..9999
Michael De Vlieger, Table of n, a(n) for n = 0..10^5
Michael De Vlieger, Scalar scatterplot of a(n), n = 0..2^16, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.
Michael De Vlieger, Log-log scatterplot of a(n), n = 0..2^20, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.
Scott R. Shannon, Image of the first 5000000 terms.

Cf. A117707, A356348, A000217, A351753.

function a = A357930( max_n )
    a = 0; s = '0'; c = zeros(1,10); c(1) = 1;
    for n = 2:max_n
        k = c(s(end-a(n-1))-47); sk = num2str(k);
        c(sk-47) = c(sk-47)+1; s = [s sk]; a(n) = k;
    end
end % Thomas Scheuerle, Oct 21 2022

A355715 a(0) = 0; for n > 0, a(n) is the total number of binary bits that n has in common with all previous terms.

Original entry on oeis.org

0, 0, 2, 1, 3, 2, 7, 8, 8, 9, 16, 15, 17, 17, 18, 19, 32, 35, 39, 42, 33, 36, 40, 40, 50, 50, 57, 57, 50, 49, 53, 54, 92, 91, 94, 93, 85, 87, 89, 90, 101, 105, 106, 113, 103, 109, 108, 116, 143, 146, 144, 149, 145, 151, 146, 153, 161, 169, 161, 170, 159, 169, 158, 170, 184, 192, 187, 194, 181

Offset: 0

Scott R. Shannon, Jul 15 2022

Scott R. Shannon, Table of n, a(n) for n = 0..10000

cf. A030190, A342303, A351753, A355374.

a(1) = 0 as a(0) = 0, and 0 shares no bits in common with 1.
a(2) = 2 as a(0) = 0, a(1) = 0, and 2 = 10_2 has the 0-bit in common with both previous terms.
a(3) = 1 as a(2) = 2 = 10_2 and 3 = 11_2 shares a 1-bit in common with 2.
a(6) = 7 as a(0) = 0, a(1) = 0, a(2) = a(5) = 2 = 10_2, a(4) = 3 = 11_2 and 6 = 110_2 shares four 0-bits and three 1-bits, seven bits in all, with these previous terms.

A357880 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number such that a(n) plus the sum of all previous terms appears in the string concatenation of a(1)..a(n-1).

Original entry on oeis.org

1, 1, 9, 8, 79, 21, 79, 19, 574, 1, 87, 40, 2, 36, 30, 211, 593, 83, 83, 30, 128, 64, 184, 501, 148, 9, 280, 329, 203, 5, 185, 161, 3, 314, 391, 119, 150, 24, 556, 197, 195, 64, 105, 108, 8, 777, 207, 16, 302, 52, 147, 2, 111, 298, 53, 67, 66, 20, 105, 99, 37, 15, 85, 51, 183, 39, 45, 8, 14

Offset: 1

Scott R. Shannon, Oct 18 2022

It is conjectured that all numbers eventually appear. In the first 100000 terms the only fixed point is 210; it is likely no more exist.

			a(6) = 21 as a(1) + ... + a(5) + 21 = 98 + 21 = 119, and "119" appears in the string concatenation of a(1)..a(5) = "119879".

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Scott R. Shannon, Image of the first 100000 terms. The green line is a(n) = n.

Cf. A357432, A357433, A000027, A007908, A000217, A351753, A337227.

nn = 120; a[1] = a[2] = 1; s = 2; w = "11"; Do[k = 1; While[! StringContainsQ[w, ToString[k + s]], k++]; a[n] = k; s += k; w = StringJoin[w, ToString[k]], {n, 3, nn}]; Array[a, nn] (* Michael De Vlieger, Oct 20 2022 *)

A356537 Numbers k whose binary expansion is a substring of the binary expansion of binomial(k,2).

Original entry on oeis.org

3, 5, 9, 11, 17, 33, 44, 50, 58, 65, 129, 257, 396, 452, 513, 581, 864, 971, 1025, 1139, 1843, 1881, 1914, 2049, 2541, 2676, 2929, 3130, 4097, 4596, 5254, 6621, 7010, 7111, 8193, 10771, 11140, 12655, 16385, 17090, 19135, 19371, 19580, 20985, 27117, 27845, 32769, 35272, 44278, 46779, 56069

Offset: 1

Scott R. Shannon, Aug 11 2022

All numbers of the form 2^m+1, m>=1, are in the sequence. There are 152 terms below 100 million.

			9 is a term as 9 = 1001_2 and binomial(9,2) = 9!/(2!7!) = 36 = 100100_2 and "100100" contains "1001" as a substring.

Cf. A000217, A187752, A030190, A351753.

kmax=56100; a={}; For[k=1, k<=kmax, k++, If[StringContainsQ[ToString[FromDigits[IntegerDigits[Binomial[k, 2], 2]]], ToString[FromDigits[IntegerDigits[k,2]]]], AppendTo[a, k]]]; a (* Stefano Spezia, Aug 11 2022 *)

str(k) = Str(fromdigits(binary(k)));
isok(k) = #strsplit(str(binomial(k,2)), str(k)) > 1; \\ Michel Marcus, Aug 11 2022

from math import comb
def ok(n): return n > 0 and str(bin(n)[2:]) in str(bin(comb(n, 2))[2:])
print([k for k in range(10**5) if ok(k)]) # Michael S. Branicky, Aug 11 2022

cp's OEIS Frontend

A357432 a(1) = 1; a(2) = 2; for n > 2, a(n) is the smallest positive number not occurring earlier such that a(n) plus the sum of all previous terms appears in the string concatenation of a(1)..a(n-1).

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A352245 a(0) = 1; for n >= 1, a(n) = the decimal value of the binary number of the index of where n first appears in the concatenation of all previous binary terms. If the binary value of n has not previously appeared then a(n) = 0.

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Python

A352246 Indices k where A352245(k) = 0.

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A357930 a(0) = 0; for n > 0, let S = concatenation of a(0)..a(n-1); a(n) is the number of times the digit at a(n-1) digits back from the end of S appears in S.

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MATLAB

A355715 a(0) = 0; for n > 0, a(n) is the total number of binary bits that n has in common with all previous terms.

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A357880 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number such that a(n) plus the sum of all previous terms appears in the string concatenation of a(1)..a(n-1).

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Mathematica

A356537 Numbers k whose binary expansion is a substring of the binary expansion of binomial(k,2).

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Python