A132777 -id:A132777 - cp's OEIS frontend

A191217 Numbers n such that sigma(n) is congruent to 2 modulo 4.

5, 10, 13, 17, 20, 26, 29, 34, 37, 40, 41, 45, 52, 53, 58, 61, 68, 73, 74, 80, 82, 89, 90, 97, 101, 104, 106, 109, 113, 116, 117, 122, 136, 137, 146, 148, 149, 153, 157, 160, 164, 173, 178, 180, 181, 193, 194, 197, 202, 208, 212, 218, 226, 229, 232, 233, 234, 241, 244, 245, 257, 261, 269

Offset: 1

Luis H. Gallardo, May 26 2011

nonn

These numbers are exactly the numbers of the form 2^a * p^(4b+1) * m^2 where p is a prime number congruent to 1 modulo 4, a is a nonnegative integer, and m is a positive integer coprime to p. In particular, they are also sums of two squares: the sequence has the first 12 terms in common with A132777.

I corrected the above comment by adding the exponent (4b+1) to p, because otherwise it would miss terms like a(614) = 3125 = 5^5, a(1140) = 6250 = 2 * 5^5, a(4421) = 28125 = 5^5 * 3^2, etc. - Antti Karttunen, May 25 2022

			For n=2, a(2) = 10 since sigma(10) = 18 = 4*4 + 2 is congruent to 2 modulo 4

Ray Chandler, Table of n, a(n) for n = 1..10000

Similar to, but different from, A230779, which is a subsequence.
Cf. A191218, A228058, A332226 for other subsequences.
Cf. A353812 (characteristic function).

with(numtheory): gen := proc(b) local n,s,d; for n from 1 to b do s := sigma(n);
if modp(s,4)=2 then print(n); fi; od; end;

for(n=1,10^3,if(2==(sigma(n)%4),print1(n,", "))) /* Joerg Arndt, May 27 2011 */

A230486 Numbers n such that n^n is representable as the sum of two nonzero squares.

Original entry on oeis.org

5, 10, 13, 17, 20, 25, 26, 29, 30, 34, 37, 40, 41, 50, 52, 53, 58, 60, 61, 65, 68, 70, 73, 74, 78, 80, 82, 85, 89, 90, 97, 100, 101, 102, 104, 106, 109, 110, 113, 116, 120, 122, 125, 130, 136, 137, 140, 145, 146, 148, 149, 150, 156, 157, 160, 164, 169, 170

Offset: 1

Alex Ratushnyak, Oct 20 2013

nonn

If n is even, then n must have a prime factor of the form 4k+1. If n is odd, then all prime factors must be of the form 4k+1. - T. D. Noe, Oct 21 2013

The above is also a sufficient condition: the sequence consists exactly in even multiples of Pythagorean primes A002144, and products of such primes (A008846). - M. F. Hasler, Sep 02 2018

			5^5 = 55^2 + 10^2.
10^10 = 99712^2 + 7584^2.
13^13 = 17106843^2 + 3198598^2.
17^17 = 28735037644^2 + 1240110271^2.

G. H. Hardy and E. M. Wright, Theory of Numbers, Oxford, Sixth Edition, 2008, p. 395.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A000312 (n^n), A004431, A132777.
A subsequence of A000404 (numbers that are the sum of 2 nonzero squares).
Sequence A002144 (primes of the form 4k + 1) and A008846 (products of such primes) are subsequences.

t = {}; Do[f = FactorInteger[n]; p = Transpose[f][[1]]; If[EvenQ[n], If[MemberQ[Mod[p, 4], 1], AppendTo[t, n]], If[Union[Mod[p, 4]] == {1}, AppendTo[t, n]]], {n, 2, 200}]; t (* T. D. Noe, Oct 21 2013 *)

select( is_A230486(n)={(n=factor(n)[,1]%4) && if(n[1]==2, Set(n)[1]==1, Set(n)==[1])}, [1..200]) \\ M. F. Hasler, Sep 02 2018

from itertools import count, islice
from sympy import primefactors
def A230486_gen(startvalue=2): # generator of terms >= startvalue
    return filter(lambda n:all(p&3==1 for p in primefactors(n)) if n&1 else any(p&3==1 for p in primefactors(n)),count(max(startvalue,2)))
A230486_list = list(islice(A230486_gen(),20)) # Chai Wah Wu, May 15 2023

A230486 = { n | A000312(n) is in A000404 } = A004277*A002144 U A008846. - M. F. Hasler, Sep 02 2018

Extended by T. D. Noe, Oct 21 2013

A362110 a(n) is the smallest integer k such that n can be expressed as the arithmetic mean of k distinct nonzero squares, or 0 if no such k exists.

Original entry on oeis.org

1, 0, 0, 1, 2, 0, 3, 0, 1, 2, 5, 0, 2, 3, 3, 1, 2, 3, 5, 2, 4, 3, 3, 5, 1, 2, 3, 3, 2, 3, 3, 5, 5, 2, 3, 1, 2, 3, 3, 2, 2, 3, 3, 5, 2, 3, 3, 5, 1, 2, 3, 2, 2, 3, 3, 3, 3, 2, 4, 3, 2, 3, 3, 1, 2, 3, 3, 2, 4, 3, 3, 3, 2, 2, 3, 5, 4, 3, 3, 2, 1, 2, 3, 4, 2, 3, 3, 3, 2, 2, 3, 3, 4, 3, 3, 5, 2, 3, 3

Offset: 1

Yifan Xie, Apr 16 2023

nonn

			For n = 2, if k = 1, 2*1 = 2 is not a square; but, from the upper bound formula, (k + 1) * (2*k + 1) <= 12, so k <= 1. So, a satisfactory k does not exist; hence a(2) = 0.

Cf. A360530 (allows repeated squares).

Cf. A001422, A033461, A129210, A132777, A248509.

Upper bound: (a(n) + 1) * (2*a(n) + 1) <= 6*n. Proof: Because (Sum_{m=1..k} (i_m)^2)/k = n, n*k = Sum_{m=1..k} (i_m)^2. Since each i_m is distinct, n*k >= Sum_{m=1..k} m^2 = k * (k + 1) * (2*k + 1)/6, hence (k + 1) * (2*k + 1) <= 6*n.

a(A132777(n)) = 2. - Thomas Scheuerle, Apr 16 2023

Name qualified and other edits by Peter Munn, Apr 21 2023

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A191217 Numbers n such that sigma(n) is congruent to 2 modulo 4.

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A230486 Numbers n such that n^n is representable as the sum of two nonzero squares.

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A362110 a(n) is the smallest integer k such that n can be expressed as the arithmetic mean of k distinct nonzero squares, or 0 if no such k exists.

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