A094552 -id:A094552 - cp's OEIS frontend

A094550 Numbers n such that there are integers a < b with a+(a+1)+...+(n-1) = (n+1)+(n+2)+...+b.

4, 6, 9, 11, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 30, 31, 32, 34, 35, 36, 38, 39, 40, 41, 43, 44, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 59, 61, 64, 66, 68, 69, 70, 71, 72, 74, 76, 77, 79, 81, 82, 83, 84, 86, 87, 89, 91, 93, 94, 95, 96, 97, 98, 99, 100, 101, 104

Offset: 1

T. D. Noe, May 10 2004

Liljestrom shows that n is in this sequence if and only if 4n^2+1 is composite.
Complement of A001912.
From Hermann Stamm-Wilbrandt, Sep 16 2014: (Start)
For n > 1, A047209 is a subset of this sequence [ 4*n^2+1 is divisible by 5 if n is (1 or 4) mod 5].
A092464 is a subset of this sequence [4*n^2+1 is divisible by 13 if n is (4 or 9) mod 13].
The above are for divisibility by 5, 13; notation (1,4,5), (4,9,13). Divisibility by p for a and p-a; notation (a,p-a,p). These are the next tuples: (2,15,17), (6,23,29), (3,34,37), (16,25,41), ... . The corresponding sequences are a subset of this sequence [ 2,15,19,32,36,49,... for (2,15,17) ]. These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences [solution to 4*a^2+1=0 (mod p)].
For n>1, A000290 (squares) is a subset of this sequence (4,9,16,25,...) [ 4*(m^2)^2+1 is divisible by  m^2+(m+1)^2, tuple (m^2, (m+1)^2, m^2+(m+1)^2) ].
(End)

			6 is in this sequence because 1+2+3+4+5 = 7+8.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
R. J. Liljestrom and Richard Zucker, Numerical Fulcrums (PowerPoint Format)

Cf. A094551, A094552, A094553.

Cf. A001912.

[n: n in [1..100] |not IsPrime(4*n^2 + 1)]; // Vincenzo Librandi, Sep 27 2012

lst={}; Do[i1=n-1; i2=n+1; s1=i1; s2=i2; While[i1>1 && s1!=s2, If[s1T. D. Noe, Nov 12 2010 *)

A094551 Numbers k such that there are integers a < b with a+(a+1)+...+(k-1) = k+(k+1)+...+b.

Original entry on oeis.org

3, 7, 8, 9, 13, 15, 18, 19, 20, 21, 23, 26, 27, 28, 31, 33, 37, 38, 43, 44, 45, 46, 48, 49, 51, 53, 55, 56, 57, 58, 59, 60, 62, 63, 68, 69, 72, 73, 75, 77, 78, 79, 80, 83, 85, 87, 88, 91, 92, 93, 94, 97, 98, 99, 102, 103, 108, 110, 111, 113, 115, 117, 118, 121, 123, 124, 128

Offset: 1

T. D. Noe, May 10 2004

nonn

From Robert Israel, Oct 28 2024: (Start)
Numbers k such that 2 * (2*k-1)^2 is the sum of two distinct squares.
Numbers k such that 2*k-1 has at least one prime factor in A002144. (End)

			7 is in this sequence because 4+5+6 = 7+8.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A002144, A094550, A094552, A094553.

filter:= proc(k)
member(1, numtheory:-factorset(2*k-1) mod 4)
end proc:
select(filter, [$1..1000]); # Robert Israel, Oct 28 2024

lst={}; Do[i1=n-1; i2=n; s1=i1; s2=i2; While[i1>1 && s1!=s2, If[s1

A094553 Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = n^2+(n+1)^2+...+b^2.

Original entry on oeis.org

5, 13, 25, 35, 39, 41, 51, 61, 85, 111, 113, 143, 145, 160, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 856, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2251, 2381, 2471, 2521, 2611, 2665, 2813, 2965, 3031

Offset: 1

T. D. Noe, May 10 2004

nonn

A094551 generalized to squares. Compare to A094552, which has far fewer solutions. For many values of n (5, 13, 25, 41, 61, 85,...), the value of b-a increases by 2 for each successive n. These n are the same as A001844. In other words, when n=i^2+(i+1)^2, then a=n-i-1 and b=n+i-1. The other values of n (35, 39, 51, 111, 143, 160, 856,...), A094523, have comparatively large values of b-a.

			13 is in this sequence because 10^2+11^2+12^2 = 13^2+14^2.

Cf. A001844 (sum of two consecutive squares), A094550, A094551, A094552, A094523.

lst={}; Do[i1=n-1; i2=n; s1=i1^2; s2=i2^2; While[i1>1 && s1!=s2, If[s1

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A094550 Numbers n such that there are integers a < b with a+(a+1)+...+(n-1) = (n+1)+(n+2)+...+b.

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A094551 Numbers k such that there are integers a < b with a+(a+1)+...+(k-1) = k+(k+1)+...+b.

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A094553 Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = n^2+(n+1)^2+...+b^2.

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