A094553 -id:A094553 - 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

A094552 Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = (n+1)^2+(n+2)^2+...+b^2.

Original entry on oeis.org

52, 100, 137, 513, 565, 1247, 8195, 13041, 18921, 35344, 40223, 65918, 68906, 121759, 132720, 213831, 215221, 235469, 265654, 506049, 520654, 585046, 598337, 817454, 993142, 1339560, 1579353, 2331619, 2843086, 3594812

Offset: 1

T. D. Noe, May 10 2004

nonn

A094550 generalized to squares. Note that equality is attained only for very long sums of squares.

a(31) > 4*10^6. [From Donovan Johnson, Apr 20 2010]

			52 is in this sequence because 7^2+8^2+...+51^2 = 53^2+54^2+...+65^2.

Cf. A094550, A094551, A094553.

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

a(14)-a(30) from Donovan Johnson, Apr 20 2010

A094523 Numbers n not of the form i^2+(i+1)^2 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

35, 39, 51, 111, 143, 160, 856, 2251, 2471, 2611, 3031, 3840, 3893, 4291, 5223, 5385, 5730, 7490, 7828, 9488, 21576, 27650, 30396, 31683, 38936, 41580, 48793, 56871, 60456, 64240, 64805, 66115, 85485, 90013

Offset: 1

T. D. Noe, May 10 2004

nonn

When n=i^2+(i+1)^2, then a=n-i-1 and b=n+i-1 is a solution. See A094553.

			35 is in this sequence because 18^2+19^2+...+34^2 = 35^2+36^2+...+42^2 and 35 is not the sum of two consecutive squares.

Cf. A001844 (sum of two consecutive squares), A094553.

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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A094552 Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = (n+1)^2+(n+2)^2+...+b^2.

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A094523 Numbers n not of the form i^2+(i+1)^2 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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