A018834 -id:A018834 - cp's OEIS frontend

A281823 Least number k such that (k-n)^2 contains k as a substring.

1, 12, 1, 16, 108, 1, 4, 2, 116, 3, 1, 1, 1, 1, 1, 1, 4, 2, 2, 2, 1, 3, 1, 9, 4, 2, 4, 2, 5, 2, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 3, 2, 1, 3, 1, 2, 3, 3, 3, 2, 4, 3, 1, 4, 1, 1, 4, 2, 3, 2, 4, 2, 1, 2, 1, 4, 2, 6

Offset: 0

Paolo P. Lava, Jan 31 2017

			a(1) = 12 because (12 - 1)^2 = 11^2 = 121 contains 12 as a substring and it is the least number with this property.

Paolo P. Lava, Table of n, a(n) for n = 0..10000

Cf. A018834, A281822.

with(numtheory): P:= proc(q) local a,b,d,j,k,n,ok;
for n from 0 to q do for k from 1 to q do a:=ilog10(k)+1; b:=(n-k)^2; d:=ilog10((k-n)^2)-ilog10(k)+1;
ok:=0; for j from 1 to d do if k=(b mod 10^a) then ok:=1; break; else b:=trunc(b/10); fi; od;
if ok=1 then print(k); break; fi; od; od; end: P(10^6);

nk[n_]:=Module[{k=1},While[SequenceCount[IntegerDigits[(k-n)^2],IntegerDigits[ k]]==0,k++];k]; Array[lnk,90,0] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Apr 13 2021 *)

Typo in definition corrected by Harvey P. Dale, Feb 27 2017.

Entries, Maple code and b-file corrected at the suggestion of Harvey P. Dale, Feb 28 2017.

A295900 Numbers n such that n^3 contains the consecutive substring 2,3,5,7.

Original entry on oeis.org

1331, 3108, 3176, 4093, 4643, 5846, 6178, 6797, 9175, 10731, 13076, 13245, 13309, 13310, 14093, 14526, 16291, 17852, 20095, 20791, 21835, 23635, 23766, 24093, 28452, 28672, 28673, 28674, 28675, 29211, 31080, 31760, 33907, 34093, 34986, 36449, 38538, 38599, 39526

Offset: 1

K. D. Bajpai, Nov 29 2017

			1331 is in the sequence because 1331^3 = 2357947691 contains substring of prime digits "2357".
3108 is in the sequence because 3108^3 = 30022235712 contains substring of prime digits "2357".

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A018834, A029942, A085823.

Select[Range[100000], MemberQ[Partition[IntegerDigits[#^3], 4, 1], {2, 3, 5, 7}] &]

isok(n) = {c = n^3; ret = 0; while (c > 1, if ((c % 10000) == 2357, ret = 1; break); c = floor(c/10);); return (ret);} \\ Michel Marcus, Dec 15 2017

A295900_list = [n for n in range(1,10**6) if '2357' in str(n**3)] # Chai Wah Wu, Feb 09 2018

A342468 Number of multiples of n up to n^2 containing the substring n in base 10.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 8, 2, 2, 3, 2, 4, 2, 3, 2, 2, 3, 3, 2, 2, 2, 10, 2, 3, 2, 3, 4, 2, 2, 4, 2, 28, 2, 4, 3, 3, 4, 5, 2, 3, 4, 14, 2, 3, 3, 5, 5, 3, 3, 4, 4, 8, 2, 5, 2, 3, 21, 5, 7, 3, 3, 19, 2, 4, 2, 6, 6, 3

Offset: 1

Yi-Hsuan Hsu, Mar 13 2021

Since the definition includes n, a(n) >= 1.
Called "Self-Replicating Numbers": "An n-order self-replicating number appears as a substring in exactly n multiples of itself up to its square, including itself" (Zaelin Goodman's Code Golf post).
There are exactly six 1st-order numbers (1, 2, 3, 4, 7, and 9).
Any number n always has an order a(n) >= log_10(n) (when n < 10, floor(log_10(n))=0). This is because there will always be at least one multiple where n is a substring (n itself), as well as any multiples of 10*n (n followed by any number of zeros).
Due to the above, for all integers x >= 1, the series of x-order self-replicating numbers is finite; a(n)=x for the last time at n=10^x-1.
For example, consider a(9)=1. It is the last possible order 1 because the only multiples where 9 is a substring are multiples of 10 (90, 900, ...), which are all > 9^2.

			a(5) = 3 because (5, 15, 25) contain 5 as a substring.
a(20) = 5 because (20, 120, 200, 220, 320) contain 20 as a substring.

Yi-Hsuan Hsu, Table of n, a(n) for n = 1..1000
Zaelin Goodman, Self-Replicating Numbers

Cf. A018834.

Table[Function[{d}, Count[n Range[n], ?(SequenceCount[IntegerDigits[#], d] > 0 &)]]@ IntegerDigits[n], {n, 86}] (* _Michael De Vlieger, Mar 13 2021 *)

a(n) = sum(k=1, n, #strsplit(Str(k*n), Str(n))>1); \\ Michel Marcus, Mar 14 2021

def a(n):
    k = 0
    for i in range(1,n+1):
        if str(n) in str(i*n):
            k += 1
    return k

A370004 Least k>0 such that the decimal expansion of k^2 contains k+n as a substring.

Original entry on oeis.org

1, 11, 2, 13, 104, 14, 3, 15, 108, 16, 11, 17, 4, 39, 18, 77, 760, 19, 52, 117, 5, 118, 34, 21, 120, 121, 22, 41, 123, 23, 6, 125, 12, 24, 42, 128, 504, 25, 352, 130, 16, 26, 7, 133, 377, 27, 322, 135, 136, 44, 26, 393, 24, 747, 139, 29, 8, 141, 108, 142, 30, 143, 22, 144, 380, 31, 606, 146, 1064, 147, 32

Offset: 0

Giorgos Kalogeropoulos, Feb 07 2024

This sequence is defined for all n. Proof: Given n, consider k = 10^x + n where 10^x > n^2. Since k^2 = (k+n) * 10^x + n^2, k^2 contains k+n as a substring. Furthermore, x = ceiling(log_10(1+n^2)) satisfies the inequality, therefore a(n) <= 10^ceiling(log_10(1+n^2)) + n. - Jason Yuen, Feb 26 2024

			a(3) = 13 because 13 is the least positive integer such that 13^2 = 169 contains 13 + 3 = 16 as a substring.
a(4) = 104 because 104 is the least positive integer such that 104^2 = 10816 contains 104 + 4 = 108 as a substring.

Michael S. Branicky, Table of n, a(n) for n = 0..10000

Cf. A000290, A018834.

Table[k=1;While[!StringContainsQ[ToString[k^2],ToString[k+n]],k++];k,{n,0,70}]
lks[n_]:=Module[{k=1},While[SequenceCount[IntegerDigits[k^2],IntegerDigits[k+n]]==0,k++];k]; Array[lks,80,0] (* Harvey P. Dale, May 26 2025 *)

a(n) = my(k=1); while (#strsplit(Str(k^2), Str(k+n))<2, k++); k; \\ Michel Marcus, Feb 07 2024

from itertools import count
def a(n): return next(k for k in count(1) if str(k+n) in str(k*k))
print([a(n) for n in range(71)]) # Michael S. Branicky, Feb 07 2024

A385709 Least prime p such that the decimal expansion of p^2 contains exactly n distinct primes as substrings.

Original entry on oeis.org

11, 5, 23, 61, 73, 239, 487, 523, 569, 3461, 1319, 3373, 8923, 4937, 12619, 11489, 15569, 32189, 105173, 135319, 46619, 56473, 177127, 234161, 295861, 471923, 664319, 2366387, 3183613, 1092389, 3513877, 7702319, 4632077, 10666177, 13977923, 20825939, 35821939

Offset: 1

Zhining Yang, Jul 07 2025

			a(9) = 569 because 569^2 = 323761, which contains 9 distinct primes as substring:{2,3,7,23,37,61,761,3761,23761}, and no prime less than 569 has 9 solutions.

Zhining Yang, Table of n, a(n) for n = 1..66

Cf. A018834, A046837.

b = Table[{}, 9]; Do[d = IntegerDigits[p^2];
 t = Union@Select[FromDigits /@ Flatten[Table[Partition[d, k, 1], {k, Length@d}], 1], PrimeQ]; c = Length@t;
 If[b[[c]] == {}, b[[c]] = {p, p^2, t, c}], {p, Prime@Range@120}]; b // Grid

A161783 Squares n^2 whose decimal expansion contains n as a substring.

Original entry on oeis.org

1, 25, 36, 100, 625, 2500, 3600, 5776, 10000, 62500, 141376, 250000, 360000, 390625, 577600, 1000000, 6250000, 14137600, 14379264, 25000000, 36000000, 39062500, 57760000, 87909376, 100000000, 214651801, 625000000, 1413760000, 2500000000

Offset: 1

Claudio Meller, Jun 19 2009

14379264 is in the list because 14379264 = 3792^2 and 3792 is a substring of 14379264.

			1 contains its square root (1); 25 contains its square root (5); 3600 contains -- but does not end with -- its square root (60). - _Dominick Cancilla_, Jul 20 2010

Equals A018834^2. Cf. A035383.

fQ[n_] := StringPosition[ IntegerString[n^2], IntegerString@n] != {}; lst = {}; k = 1; While[k < 50001, If[ fQ@k, AppendTo[lst, k^2]]; k++ ]; lst (* Robert G. Wilson v, Jul 23 2010 *)

Edited by N. J. A. Sloane, Jul 23 2010

A179782 Numbers n such that the decimal representation of n is contained as substring in that of the n-th pentagonal number.

Original entry on oeis.org

0, 1, 5, 7, 25, 67, 482, 551, 625, 667, 2937, 6667, 9284, 9376, 9649, 48179, 49900, 55712, 66667, 89517, 90625, 161579, 631206, 666667, 890625, 1348613, 2089517, 3863187, 4999000, 6666667, 7109376, 7477735, 8575619, 10721030, 12890625

Offset: 1

Jonathan Vos Post, Jul 27 2010

This is to pentagonal numbers A000326 as A119238 is to triangular numbers A000217 and as A018834 is to squares A000290. All numbers of the form (10^n-1)/3*2+1 are contained in this list {1, 7, 67, 667, 6667, 66667, 666667, 6666667, 66666667, ...} Alois P. Heinz. Extension: Values 8-25 by Claudio Meller, 26-37 by D. S. McNeil.

			The 5th pentagonal number, 35, which contains 5.
The 7th pentagonal number, 70, which contains 7.
The 25th pentagonal number, 925, which contains 25.
The 67th pentagonal number, 6700, which contains 67.
The 482nd pentagonal number, 348245, which contains 482.
The 667th pentagonal number, 667000, which contains 667.

Cf. A000217, A000290, A000326, A018834, A067275, A119238.

[n for n in range(10**4) if str(n) in str((3*n**2-n)//2)]

Code clarified by D. S. McNeil, Aug 08 2010

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A281823 Least number k such that (k-n)^2 contains k as a substring.

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A295900 Numbers n such that n^3 contains the consecutive substring 2,3,5,7.

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A342468 Number of multiples of n up to n^2 containing the substring n in base 10.

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A370004 Least k>0 such that the decimal expansion of k^2 contains k+n as a substring.

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A385709 Least prime p such that the decimal expansion of p^2 contains exactly n distinct primes as substrings.

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A161783 Squares n^2 whose decimal expansion contains n as a substring.

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A179782 Numbers n such that the decimal representation of n is contained as substring in that of the n-th pentagonal number.

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