A094550 -id:A094550 - cp's OEIS frontend

A001912 Numbers k such that 4*k^2 + 1 is prime.

1, 2, 3, 5, 7, 8, 10, 12, 13, 18, 20, 27, 28, 33, 37, 42, 45, 47, 55, 58, 60, 62, 63, 65, 67, 73, 75, 78, 80, 85, 88, 90, 92, 102, 103, 105, 112, 115, 118, 120, 125, 128, 130, 132, 135, 140, 142, 150, 153, 157, 163, 170, 175, 192, 193, 198, 200

Offset: 1

N. J. A. Sloane

Complement of A094550. - Hermann Stamm-Wilbrandt, Sep 16 2014
Positive integers whose square is the sum of two triangular numbers in exactly one way (A000217(k) + A000217(k+1) = k*(k+1)/2 + (k+1)*(k+2)/2 = (k+1)^2). In other words, positive integers k such that A052343(k^2) = 1. - Altug Alkan, Jul 06 2016
4*a(n)^2 + 1 = A002496(n+1). - Hal M. Switkay, Apr 03 2022

E. Kogbetliantz and A. Krikorian, Handbook of First Complex Prime Numbers, Gordon and Breach, NY, 1971, p. 1.
M. Kraitchik, Recherches sur la Théorie des Nombres. Gauthiers-Villars, Paris, Vol. 1, 1924, Vol. 2, 1929, see Vol. 1, p. 11.
C. S. Ogilvy, Tomorrow's Math. 2nd ed., Oxford Univ. Press, 1972, p. 116.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..10000
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. [Annotated scans of a few pages from Volumes 1 and 2]
E. Kogbetliantz and A. Krikorian Handbook of First Complex Prime Numbers, Gordon and Breach, NY, 1971 [Annotated scans of a few pages]
Marek Wolf, Search for primes of the form m^2+1, arXiv:0803.1456 [math.NT], 2008-2010.

Cf. A002496, A005574, A062325, A090693, A094550, A214517 (first differences).

[n: n in [1..100] | IsPrime(4*n^2+1)] // Vincenzo Librandi, Nov 21 2010

A001912 := proc(n)
    option remember;
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            if isprime(4*a^2+1) then
                return a;
            end if;
        end do:
    end if;
end proc: # R. J. Mathar, Aug 09 2012

Select[Range[200], PrimeQ[4#^2 + 1] &] (* Alonso del Arte, Dec 20 2013 *)

is(n)=isprime(4*n^2 + 1) \\ Charles R Greathouse IV, Apr 28 2015

a(n) = A005574(n+1)/2.

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

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

A110701 Friendly run sums: numbers S with two run sums (sum of positive integer runs) that share one common number, i.e., S = a + (a+1) + ... + b = b + (b+1) + ... + c with a < b < c.

Original entry on oeis.org

9, 21, 30, 42, 65, 70, 99, 105, 117, 133, 135, 154, 175, 180, 225, 231, 275, 285, 341, 342, 345, 364, 385, 414, 440, 450, 455, 481, 495, 540, 546, 567, 630, 645, 675, 693, 744, 750, 765, 825, 833, 936, 945, 990, 1035, 1045, 1140, 1161, 1170, 1176, 1178

Offset: 1

Ron Knott, Aug 04 2005

nonn

The sums are the difference of two triangular numbers A000217. The common numbers themselves are A094550. The sums, n > 0, are n = (b-a+1)(a+b)/2 = (b+c)(c-b+1)/2 where b^2 + a - a^2 = c^2 + c - b^2, is solvable in integers for 0 < a < b < c. Since the runs have something in common, they are "friends". The series of sums *without* the common number is A110702. The numbers common to the two runs are A094550.

			2 + 3 + 4 = 4 + 5 share the common number 4. The sum of both is 9 and this is the smallest such sum with a common "friend" (4), so a(1)=9.

Ron Knott Runsums
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Cf. A001227, A094550, A110702, A110703.

A110702 Numbers S with two runsums (sum of positive integer runs) where the 2 runs are separated by a single number gap (b), i.e., S = a + (a+1) + ... + (b-1) = (b+1) + ... + c with a < b-1, b < c.

Original entry on oeis.org

5, 15, 21, 33, 54, 54, 85, 90, 100, 117, 110, 135, 153, 161, 204, 195, 252, 261, 315, 308, 315, 315, 351, 385, 405, 418, 414, 450, 455, 476, 510, 533, 595, 609, 637, 650, 705, 693, 684, 777, 792, 870, 884, 931, 986, 999, 1071, 1105, 1121, 1110, 1125, 1210, 1230

Offset: 1

Ron Knott, Aug 04 2005

nonn

The sums are the difference of two triangular numbers A000217. The sum series where the missing number is included is A110701. The numbers in the gap between the two runs are A094550.

			1+2+3+4+5 and 7+8 have a gap between them of (6). The sum of both is 15 so 15 is in the ordered series (as a(2)).

Ron Knott Runsums
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Cf. A001227, A094550, A110701, A110703.

A110703 Numbers S with two neighboring run sums (sum of positive integer runs) S = a+(a+1)+..+b=(b+1)+(b+2)...+c, 0

Original entry on oeis.org

3, 15, 27, 30, 42, 75, 90, 105, 135, 147, 165, 243, 252, 270, 273, 315, 363, 375, 378, 420, 462, 495, 507, 612, 660, 675, 693, 735, 750, 780, 810, 855, 858, 867, 945, 1050, 1083, 1155, 1170, 1215, 1287, 1323, 1365, 1470, 1485, 1518, 1587, 1785, 1815, 1875, 1950

Offset: 1

Ron Knott, Aug 04 2005

nonn

In other words, numbers n such that a list of consecutive numbers can be split into two parts in which their sums both equal n. - A. D. Skovgaard, May 22 2017
If the two runs overlap in one number, the runs are Friends and their sums are A110701. The sums are the difference of two triangular numbers A000217.
The subsequence where there is more than one possible splitting begins 105, 945, 1365, 2457, 2625, 3990, 5145, 8505, ... - Jean-François Alcover, May 22 2017
a(n) seems to always be divisible by 3.- A. D. Skovgaard, May 22 2017. This is true. Sequence lists values of n = t(t+1)/2 - k(k+1)/2 = m(m+1)/2 - t(t+1)/2 with k < t < m. Since any triangular number must be of the form 3w or 3w+1, then there are two possibilities for n = 3w - k(k+1)/2 = m(m+1)/2 - 3w or n = 3w + 1 - k(k+1)/2 = m(m+1)/2 - 3w - 1. For first case, if k(k+1)/2 = 3u+1, there is no solution for m. Similarly for second case, if k(k+1)/2 = 3u, there is no solution for m. So always n must be divisible by 3. - Altug Alkan, May 22 2017

			3 = 1+2 = 3, so 3 is a term.
15 = 4+5+6 = 7+8 so 15 is a term.
a(6) = 75 because 75 = 3+4+5+6+7+8+9+10+11+12 = 13+14+15+16+17.

Ron Knott Runsums
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Cf. A001227, A094550, A110701, A110702.

Select[Range[1000], False =!= Reduce[# == Sum[k, {k, x, y}] == Sum[k, {k, y + 1, z}] && z >= y >= x > 0, {x, y, z}, Integers] &] (* Giovanni Resta, May 22 2017 *)

Initial 3 added by A. D. Skovgaard, May 22 2017

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A001912 Numbers k such that 4*k^2 + 1 is prime.

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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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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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A110701 Friendly run sums: numbers S with two run sums (sum of positive integer runs) that share one common number, i.e., S = a + (a+1) + ... + b = b + (b+1) + ... + c with a < b < c.

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A110702 Numbers S with two runsums (sum of positive integer runs) where the 2 runs are separated by a single number gap (b), i.e., S = a + (a+1) + ... + (b-1) = (b+1) + ... + c with a < b-1, b < c.

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A110703 Numbers S with two neighboring run sums (sum of positive integer runs) S = a+(a+1)+..+b=(b+1)+(b+2)...+c, 0

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