Gonzalo Martínez - cp's OEIS frontend

A385963 a(n) is the maximum number of distinct positive integers whose sum of squares is equal to n^2.

0, 1, 1, 1, 1, 2, 1, 3, 1, 4, 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 9, 9, 9, 11, 10, 11, 11, 11, 11, 11, 13, 12, 12, 13, 14, 13, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 20, 21, 21, 21, 21, 22, 22, 22, 22, 23, 22, 24, 24, 24, 24, 24, 25

Offset: 0

Gonzalo Martínez, Jul 13 2025

nonn

An upper bound is a(n) <= r for the largest r with 1^2 + ... + r^2 <= n^2, and with equality only at n = 0,1,24, the latter being a(70) = 24 (see comments A001032).

			For n = 11, there are A030273(11) = 4 partitions of 11^2 into distinct squares: {11^2}, {2^2, 6^2, 9^2}, {1^2, 2^2, 4^2, 10^2}, {1^2, 2^2, 4^2, 6^2, 8^2}, where the largest cardinality of these sets is 5. Therefore, a(11) = 5.

David A. Corneth, Table of n, a(n) for n = 0..10000
David A. Corneth, n, a(n), one tuple of size a(n) where sum of elements of tuples is n^2

Cf. A030273, A001032.

a(n)=poldegree(polcoef(prod(k=1, n, 1 + y*x^(k^2), 1 + O(x^(n^2+1))), n^2)) \\ Andrew Howroyd, Jul 13 2025

More terms from Andrew Howroyd, Jul 13 2025

A385711 Primes whose digits are all distinct and pairwise coprime.

Original entry on oeis.org

2, 3, 5, 7, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 127, 137, 149, 157, 167, 173, 179, 197, 251, 257, 271, 317, 347, 419, 431, 457, 479, 491, 521, 523, 541, 547, 571, 587, 617, 719, 743, 751, 761, 853, 857, 859, 941, 947, 971, 1237, 1259

Offset: 1

Gonzalo Martínez, Jul 07 2025

This sequence has 252 terms, the last being 95471, which have at most 5 digits. This is because each term has at most one even digit and at most 4 odd digits, since gcd(3, 9) = 3.

All terms are in A038618, since if zero is among the digits of a prime p, then p must have at least 3 digits, where at least one of them is greater than 1, say d, and in such a case gcd(0, d) = d ! = 1.

			857 is a term since it is prime and gcd(8, 5) = gcd(5, 7) = gcd(8, 7) = 1.

Gonzalo Martínez, Table of n, a(n) for n = 1..252

Subsequence of A029743.

Cf. A038618.

Select[Prime[Range[10000]], UnsameQ @@ (d = IntegerDigits[#]) && AllTrue[Subsets[d, {2}], CoprimeQ @@ # &] &] (* Amiram Eldar, Jul 13 2025 *)

A385637 Primes whose decimal expansion consists of the concatenation of m i’s followed by m j’s, ..., iiijjj, iijj and ij, i != j, where 1 <= i, j <= 9 and m > 0.

Original entry on oeis.org

13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 44443333444333443343, 55555553333333555555333333555553333355553333555333553353

Offset: 1

Gonzalo Martínez, Jul 05 2025

Similar to A385481, but the blocks of i's and j's are concatenated from longest length to shortest, where a(n) contains terms of a length not recorded in A385481, such as length 56, 182 and 272.
a(23) has 182 digits and starts with 13 2's followed by 13 9's.
a(24) has 272 digits and starts with 16 9's followed by 16 7's.
a(25), if it exists, has m > 200 and > 40200 digits.

			For i = 4, j = 3 and m = 4, by concatenating 44443333, 444333, 4433 and 43 the prime 44443333444333443343 is obtained.

Gonzalo Martínez, Table of n, a(n) for n = 1..24

Cf. A385481, A059170, A034845.

A385536 Integers k formed by the concatenation of 3 consecutive prime numbers such that k is divisible by at least one of those 3 primes.

Original entry on oeis.org

235, 357, 71113, 374143, 616771, 167173179, 143914471451, 909071909089909091, 104651162787110465116278991046511627907

Offset: 1

Gonzalo Martínez, Jul 02 2025

Subsequence of A132903.
If p is A255669, as it divides the concatenation of the next two primes, then p divides the concatenation of p with the next two primes. Thus, the first 4 terms of A255669 give rise to a(2), a(3), a(5) and a(6). In this sequence the number formed by concatenating 3 consecutive primes is allowed to be divisible by at least one of those 3, which generates more possibilities.
a(9) has p > 10^11 and thus >= 36 digits. - Michael S. Branicky, Jul 02 2025
a(9) has p <=  A258182(11) - 36 = 1046511627871.
a(10) has p <= A258182(22) - 226.

			235 is a term, since it is the concatenation of the consecutive primes 2, 3, 5 and is divisible by 5.
2, 3 and 5: 235 = 5*47
3, 5, and 7: 357 = 3*119 = 7*51
7, 11 and 13: 71113 = 7*10159
37, 41 and 43: 374143 = 43*8701
61, 67 and 71: 616771 = 61*10111
167, 173 and 179: 167173179 = 167*1001037
1439, 1447 and 1451: 143914471451 = 1447*99457133
909071, 909089 and 909091: 909071909089909091 = 909091*999979000001

Cf. A132903, A255669, A258182.

tcat:= proc(a,b,c)
   c + (b + 10^(1+ilog10(b))*a)*10^(1+ilog10(c))
end proc:
R:= NULL: count:= 0:
q:= 2: r:= 3:
while count < 8 do
  p:= q; q:= r; r:= nextprime(r);
  x:= tcat(p,q,r);
  if igcd(x,p*q*r)>1 then
    R:= R,x; count:= count+1;
  fi
od:
R; # Robert Israel, Jul 06 2025

cat[s_] := FromDigits[Flatten[IntegerDigits[s]]]; q[s_] := AnyTrue[s, Divisible[cat[s], #] &]; cat /@ Select[Partition[Prime[Range[72000]], 3, 1], q] (* Amiram Eldar, Jul 03 2025 *)

a(9) from Michael S. Branicky, Jul 03 2025

A385515 Repdigit numbers whose square does not contain the repeated digit.

Original entry on oeis.org

2, 3, 4, 7, 8, 9, 22, 33, 44, 77, 88, 333, 444, 3333, 33333, 44444, 88888, 333333, 3333333, 33333333, 333333333, 3333333333, 33333333333, 333333333333, 3333333333333, 33333333333333, 333333333333333, 3333333333333333, 33333333333333333, 333333333333333333

Offset: 1

Gonzalo Martínez, Jul 01 2025

For n >= 18, all terms are of the form 33...3; that is, elements of A002277.
A002277(m) is a term, for m > 0. Proof: 3 is in a(n) because 3^2 = 9. If 33...3 is composed of k 3's, with k > 1, it is satisfied that 33...3^2 = 11...1088...89; i.e., (k - 1) 1's followed by a 0, then (k - 1) 8's and a 9, so that 3 is not among the digits of its square.
Let's see that there are no other terms of the form 33...3 besides 2, 4, 7, 8, 9, 22, 44, 77, 88, 444, 44444, 88888. In this sequence there are no repdigits of the form 11...1, 55...5, 66...6, since their squares end in 1, 5 and 6 respectively. On the other hand, 9 is the only number of the form 999...9, since if it has 2 or more 9's its square starts with 9. Suppose that dd...d contains 6 or more digits. We already saw that the cases d = 1, 5, 6 and 9 are discarded. Let us analyze what happens for d = 2, 4, 7 and 8:
For d = 2, we have that 22...2^2 == 284 (mod 10^3).
For d = 4, we have that 44...4^2 == 469136 (mod 10^6).
For d = 7, we have that 77...7^2 == 729 (mod 10^3).
For d = 8, we have that 88...8^2 == 876544 (mod 10^6).
Thus, we conclude that a(n) only consists of digits 3 for n >= 18. And, in fact, a(n) consists of (n - 12) 3's.

			22 is a term since 22^2 = 484 does not contain the digit 2.

Cf. A002277, A010785, A029783.

Intersection of A010785 and A029783.

Select[Union@ Flatten@ Table[k (10^n - 1)/9, {k, 0, 9}, {n, 18}] ,ContainsNone[IntegerDigits[#^2],IntegerDigits[#]]&] (* James C. McMahon, Jul 07 2025 *)

a(n) = A002277(n - 12), for n >= 18.

A385481 Primes whose decimal expansion consists of the concatenation of ij, iijj, iiijjj,..., and m i’s followed by m j’s, i != j, where 1<= i, j <= 9 and m > 0.

Original entry on oeis.org

13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 31331133311133331111, 37337733377733337777, 43443344433344443333, 49449944499944449999

Offset: 1

Gonzalo Martínez, Jun 30 2025

a(25) has 812 digits and starts with 292299222999...999, where the last concatenated strings have 28 2's followed by 28 9's.
Since each term is prime, then j = 1, 3, 7, and 9 and i + j == 1, 2 (mod 3).
m == 1 (mod 3). Proof. If k is a term whose last concatenated string has m i's followed by m j's, then it has 2(1 + 2 + ... + m) = m(m + 1) digits, whose sum is (i + j)*m(m + 1)/2, so that if m == 0, 2 (mod 3), then the sum of digits of k is a multiple of 3 and so k is not prime.
Are there infinite primes of this form?
From Michael S. Branicky, Jul 01 2025: (Start)
A probabilistic argument suggests the sequence is finite.
a(26), if it exists, has m > 321 and > 103362 digits. (End)

			for i = 3, j = 1 and m = 4, by concatenating 31, 3311, 333111, 33331111 the prime 31331133311133331111 is obtained.

Gonzalo Martínez, Table of n, a(n) for n = 1..25

Cf. A059170, A034845.

from gmpy2 import is_prime, mpz
from itertools import count, islice, product
def agen(): yield from (p for m in count(1) for i in "123456789" for j in "1379" if i != j and is_prime(p:=int(mpz("".join(i*k+j*k for k in range(1, m+1))))))
print(list(islice(agen(), 24))) # Michael S. Branicky, Jun 30 2025

A385331 a(1) = 1 and a(n) is the smallest prime factor of n-th numerator of partial sum for Liouville's constant (A145571), for n > 1.

Original entry on oeis.org

1, 11, 3, 4447, 7823, 3, 7, 13, 3, 61, 31, 3, 11, 7, 3, 9281, 19163, 3, 17, 53861, 3, 599, 397, 3

Offset: 1

Gonzalo Martínez, Jun 25 2025

a(n) = A145571(n) itself when that numerator is prime: are there any others after a(2) = 11?

			For n=3, the prime decomposition A145571(3) = 110001 = 3*37*991 so that a(3) = 3.

Cf. A012245, A145571.

a(n) = if(n>1, forprime(p=3, oo, if(sum(k=1, n, Mod(10, p)^(n!-k!))==0, return(p))), 1); \\ Jinyuan Wang, Jun 25 2025

a(3k) = 3 for k >= 1, since A145571(n) is an odd number whose sum of digits is n so at n = 3k it is divisible by 3.

a(10)-a(24) from Jinyuan Wang, Jun 25 2025

A385332 Integers k such that the set {k, k^2, ..., k^9} contains at least 8 zeroless numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 11, 17, 68, 121, 786

Offset: 1

Gonzalo Martínez, Jun 25 2025

1, 2, 3, 6 and 68 are the only integers less than 3.3*10^16 such that {k, k^2,..., k^8} are all zeroless (See A124649), but of them 1^9, 2^9 and 3^9 are the only zeroless ones. If we weaken the condition and ask that 8 of the 9 numbers in the set {k, k^2,..., k^9} are zeroless, then more numbers appear that satisfy this property.

Is a(10) = 786 the largest term?

			5 is a term, since the first nine powers of 5 are: 5, 25, 125, 625, 3125, 15625, 78125, 390625 and 1953125, where 8 of the 9 are zeroless.

Cf. A124648, A124649.

zless:= n -> not has(convert(n,base,10),0):
filter:= proc(n) local i; nops(select(zless, [seq(n^i,i=1..9)]))>=8 end proc:
select(filter, [$1..1000]); # Robert Israel, Jun 26 2025

Select[Range[10^5],Total[Boole/@Positive/@Min/@IntegerDigits/@(#^Range[9])]>7&] (* James C. McMahon, Jul 01 2025 *)

A385175 Cubes using at most three distinct digits, not ending in 0.

Original entry on oeis.org

1, 8, 27, 64, 125, 216, 343, 512, 729, 1331, 2744, 3375, 46656, 238328, 778688, 1030301, 5177717, 7077888, 9393931, 700227072, 1003003001, 44474744007, 1000300030001, 1000030000300001, 1331399339931331, 3163316636166336, 1000003000003000001, 1000000300000030000001, 1000000030000000300000001

Offset: 1

Gonzalo Martínez, Jun 20 2025

This sequence has infinitely many terms since (10^m + 1)^3 is a term for all m >= 0.

Conjecture: a(26) = 3163316636166336 is the largest term with nonzero digits (See comments of A030294 and the data of A155146, where a(26) = A155146(47)^3).

			8, 343, and 46656 belong to this list because they are cubes that use 1, 2, and 3 distinct digits, respectively.

Cf. A000578, A030292, A155146, A018884, A202940.

Select[Range[10^6]^3,Length[Union[IntegerDigits[#]]]<4&&IntegerDigits[#][[-1]]!=0&] (* James C. McMahon, Jun 30 2025 *)
fQ[n_] := Mod[n, 10] > 0 && Length@ Union@ IntegerDigits[n^3] < 4; k = 1; lst = {}; While[k < 1000002, If[ fQ@k, AppendTo[lst, k]]; k++]; lst^3 (* Robert G. Wilson v, Jul 10 2025 *)

a(n) = A202940(n)^3.

a(28) from Robert G. Wilson v, Jul 10 2025

a(29) from David A. Corneth, Jul 10 2025

A385171 Perfect powers m^k whose decimal expansion begins with k and ends with m, where m and k are greater than 1.

Original entry on oeis.org

25, 59049, 78125, 13060694016, 17179869184, 19073486328125, 30514648531249, 53613724194557, 59120987373568, 65944160601201, 116490258898219, 324965351768751, 512908935546875, 21936950640377856, 371308922853718751, 578261433548013568, 913517247483640899

Offset: 1

Gonzalo Martínez, Jun 20 2025

Such as automorphic numbers (A003226), which are those m such that m^2 ends with m, if m^k is in this sequence, then it is a k-morphic number which also begins with k. Thus, m^k contains both m and k as substrings at its ends.
If m is in A003226 and m^2 starts with 2, then m^2 is in this sequence. For example, A003226(3)^2 = 5^2 = 25 and A003226(119)^2.
If m is in A033819 and m^3 starts with 3, then m^3 is in this sequence. For example, A033819(39)^3 = 31249^3 = 30514648531249.
This sequence has infinitely many terms since (10^m - 1)^9 is a term for all m >= 2, which starts with (m - 1) 9's and ends with m 9's.

			6^13 = 13060694016 is a term since it starts with 13 and ends with 6.

Cf. A001597, A003226, A033819, A378857, A051248, A074250.

cp's OEIS Frontend

User: Gonzalo Martínez

A385963 a(n) is the maximum number of distinct positive integers whose sum of squares is equal to n^2.

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A385711 Primes whose digits are all distinct and pairwise coprime.

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A385637 Primes whose decimal expansion consists of the concatenation of m i’s followed by m j’s, ..., iiijjj, iijj and ij, i != j, where 1 <= i, j <= 9 and m > 0.

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A385536 Integers k formed by the concatenation of 3 consecutive prime numbers such that k is divisible by at least one of those 3 primes.

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A385515 Repdigit numbers whose square does not contain the repeated digit.

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A385481 Primes whose decimal expansion consists of the concatenation of ij, iijj, iiijjj,..., and m i’s followed by m j’s, i != j, where 1<= i, j <= 9 and m > 0.

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A385331 a(1) = 1 and a(n) is the smallest prime factor of n-th numerator of partial sum for Liouville's constant (A145571), for n > 1.

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A385332 Integers k such that the set {k, k^2, ..., k^9} contains at least 8 zeroless numbers.

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