A063930 -id:A063930 - cp's OEIS frontend

A055096 Triangle read by rows, sums of 2 distinct nonzero squares: T(n,k) = k^2+n^2, (n>=2, 1 <= k <= n-1).

5, 10, 13, 17, 20, 25, 26, 29, 34, 41, 37, 40, 45, 52, 61, 50, 53, 58, 65, 74, 85, 65, 68, 73, 80, 89, 100, 113, 82, 85, 90, 97, 106, 117, 130, 145, 101, 104, 109, 116, 125, 136, 149, 164, 181, 122, 125, 130, 137, 146, 157, 170, 185, 202, 221, 145, 148, 153, 160

Offset: 2

Antti Karttunen, Apr 04 2000

Discovered by Bernard Frénicle de Bessy (1605?-1675). - Paul Curtz, Aug 18 2008
Terms that are not hypotenuses in primitive Pythagorean triangles, are replaced by 0 in A222946. - Reinhard Zumkeller, Mar 23 2013
This triangle T(n,k) gives the circumdiameters for the Pythagorean triangles with a = (n+1)^2 - k^2, b = 2*(n+1)*k and c = (n+1)^2 + k^2 (see the Floor van Lamoen entries or comments A063929, A063930, A002283, A003991). See also the formula section. Note that not all Pythagorean triangles are covered, e.g., (9,12,15) does not appear. - Wolfdieter Lang, Dec 03 2014

			The triangle T(n, k) begins:
n\k   1   2   3   4   5   6   7   8   9  10  11 ...
2:    5
3:   10  13
4:   17  20  25
5:   26  29  34  41
6:   37  40  45  52  61
7:   50  53  58  65  74  85
8:   65  68  73  80  89 100 113
9:   82  85  90  97 106 117 130 145
10: 101 104 109 116 125 136 149 164 181
11: 122 125 130 137 146 157 170 185 202 221
12: 145 148 153 160 169 180 193 208 225 244 265
...
13: 170 173 178 185 194 205 218 233 250 269 290 313,
14: 197 200 205 212 221 232 245 260 277 296 317 340 365,
15: 226 229 234 241 250 261 274 289 306 325 346 369 394 421,
16: 257 260 265 272 281 292 305 320 337 356 377 400 425 452 481,
...
Formatted and extended by _Wolfdieter Lang_, Dec 02 2014 (reformatted Jun 11 2015)
The successive terms are (1^2+2^2), (1^2+3^2), (2^2+3^2), (1^2+4^2), (2^2+4^2), (3^2+4^2), ...

Reinhard Zumkeller, Rows n = 2..121 of triangle, flattened
M. de Frénicle, Méthode pour trouver la solutions des problèmes par les exclusions, in: "Divers ouvrages de mathématiques et de physique, par Messieurs de l'Académie royale des sciences", Paris, 1693, pp 1-44.
Antti Karttunen, Larger table, showing also locations of 4k+1 primes and squares
Eric Weisstein's World of Mathematics, Congruum Problem.
Index entries for sequences related to sums of squares

Sorting gives A024507. Count of divisors: A055097, Möbius: A055132. For trinv, follow A055088.
Cf. A001844 (right edge), A002522 (left edge), A033429 (central column).
Cf. A000290, A033583, A079273, A190816, A226141, A331987.

a055096 n k = a055096_tabl !! (n-1) !! (k-1)
a055096_row n = a055096_tabl !! (n-1)
a055096_tabl = zipWith (zipWith (+)) a133819_tabl a140978_tabl
-- Reinhard Zumkeller, Mar 23 2013

[n^2+k^2: k in [1..n-1], n in [2..15]]; // G. C. Greubel, Apr 19 2023

sum2distinct_squares_array := (n) -> (((n-((trinv(n-1)*(trinv(n-1)-1))/2))^2)+((trinv(n-1)+1)^2));

T[n_, k_]:= (n+1)^2 + k^2; Table[T[n, k], {n,15}, {k,n}]//Flatten (* Jean-François Alcover, Mar 16 2015, after Reinhard Zumkeller *)

def A055096(n,k): return n^2 + k^2
flatten([[A055096(n,k) for k in range(1,n)] for n in range(2,16)]) # G. C. Greubel, Apr 19 2023

a(n) = sum2distinct_squares_array(n).
T(n, 1) = A002522(n).
T(n, n-1) = A001844(n-1).
T(2*n-2, n-1) = A033429(n-1).
T(n,k) = A133819(n,k) + A140978(n,k) = (n+1)^2 + k^2, 1 <= k <= n. - Reinhard Zumkeller, Mar 23 2013
T(n, k) = a*b*c/(2*sqrt(s*(s-1)*(s-b)*(s-c))) with s =(a + b + c)/2 and the substitution a = (n+1)^2 - k^2, b = 2*(n+1)*k and c = (n+1)^2 + k^2 (the circumdiameter for the considered Pythagorean triangles). - Wolfdieter Lang, Dec 03 2014
From Bob Selcoe, Mar 21 2015:  (Start)
T(n,k) = 1 + (n-k+1)^2 + Sum_{j=0..k-2} (4*j + 2*(n-k+3)).
T(n,k) = 1 + (n+k-1)^2 - Sum_{j=0..k-2} (2*(n+k-3) - 4*j).
Therefore: 4*(n-k+1) + Sum_{j=0..k-2} (2*(n-k+3) + 4*j) = 4*n(k-1) - Sum_{j=0..k-2} (2*(n+k-3) - 4*j). (End)
From G. C. Greubel, Apr 19 2023: (Start)
T(2*n-3, n-1) = A033429(n-1).
T(2*n-4, n-2) = A079273(n-1).
T(2*n-2, n) = A190816(n).
T(3*n-4, n-1) = 10*A000290(n-1) = A033583(n-1).
Sum_{k=1..n-1} T(n, k) = A331987(n-1).
Sum_{k=1..floor(n/2)} T(n-k, k) = A226141(n-1). (End)

Edited: in T(n, k) formula by Reinhard Zumkeller k < n replaced by k <= n. - Wolfdieter Lang, Dec 02 2014

Made definition more precise, changed offset to 2. - N. J. A. Sloane, Mar 30 2015

A098847 a(n) = n*(n + 12).

Original entry on oeis.org

0, 13, 28, 45, 64, 85, 108, 133, 160, 189, 220, 253, 288, 325, 364, 405, 448, 493, 540, 589, 640, 693, 748, 805, 864, 925, 988, 1053, 1120, 1189, 1260, 1333, 1408, 1485, 1564, 1645, 1728, 1813, 1900, 1989, 2080, 2173, 2268, 2365, 2464, 2565, 2668, 2773

Offset: 0

Eugene McDonnell (eemcd(AT)mac.com), Nov 04 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Wikipedia, Hydrogen spectral series, Humphreys series.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A098832, A063930.

a(n-6), n>=7, sixth column (used for the n=6 series of the hydrogen atom) of triangle A120070.

Table[ n(n + 12), {n, 0, 50}] (* Robert G. Wilson v, Jul 14 2005 *)
LinearRecurrence[{3,-3,1},{0,13,28},50] (* Harvey P. Dale, May 24 2012 *)

a(n)=n*(n+12) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = (n+6)^2 - 6^2 = n*(n + 12), n>=0.
G.f.: x*(13 - 11*x)/(1-x)^3.
a(n) = 2*n + a(n-1) + 11 (with a(0)=0). - Vincenzo Librandi, Nov 17 2010
a(0)=0, a(1)=13, a(2)=28, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 24 2012
Sum_{n>=1} 1/a(n) = 86021/332640 = 0.258600... via Sum_{n>=0} 1/((n+x)(n+y)) = (psi(x)-psi(y))/(x-y). - R. J. Mathar, Jul 14 2012
E.g.f.: x*(13 + x)*exp(x). - G. C. Greubel, Jul 29 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 18107/332640. - Amiram Eldar, Jan 15 2021

More terms from Robert G. Wilson v, Jul 14 2005

A001283 Triangle read by rows, in which row n consists of n(n+m) for m = 1 .. n-1.

Original entry on oeis.org

6, 12, 15, 20, 24, 28, 30, 35, 40, 45, 42, 48, 54, 60, 66, 56, 63, 70, 77, 84, 91, 72, 80, 88, 96, 104, 112, 120, 90, 99, 108, 117, 126, 135, 144, 153, 110, 120, 130, 140, 150, 160, 170, 180, 190, 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 156, 168, 180

Offset: 2

N. J. A. Sloane

With a different offset: triangle read by rows: t(n, m) = T(n+1, m) = (n+1)(n+m+1) = radius of C-excircle of Pythagorean triangle with sides a=(n+1)^2-m^2, b=2*(n+1)*m and c=(n+1)^2+m^2. - Floor van Lamoen, Aug 21 2001

			The triangle T(n, m) begins:
n\m   1   2   3   4   5   6   7   8   9  10  11  12  13  14 ...
2:    6
3:   12  15
4:   20  24  28
5:   30  35  40  45
6:   42  48  54  60  66
7:   56  63  70  77  84  91
8:   72  80  88  96 104 112 120
9:   90  99 108 117 126 135 144 153
10: 110 120 130 140 150 160 170 180 190
11: 132 143 154 165 176 187 198 209 220 231
12: 156 168 180 192 204 216 228 240 252 264 276
13: 182 195 208 221 234 247 260 273 286 299 312 325
14: 210 224 238 252 266 280 294 308 322 336 350 364 378
15: 240 255 270 285 300 315 330 345 360 375 390 405 420 435
...
[Reformatted and extended by _Wolfdieter Lang_, Dec 02 2014]
----------------------------------------------------------------

T. D. Noe, Rows n = 2..100, flattened

Cf. A003991, A063929, A063930.

Row sums are in A085788. Central column is A033581.

Flatten[Table[n*(n+m), {n, 2, 10}, {m, n-1}]] (* T. D. Noe, Jun 27 2012 *)

T(n, m) = n*(n+m), n-1 >= m >= 1.

Edited comment by Wolfdieter Lang, Dec 02 2014

A063929 Radius of A-excircle of Pythagorean triangle with a = (n+1)^2 - m^2, b = 2(n+1)m and c = (n+1)^2 + m^2.

Original entry on oeis.org

2, 6, 3, 12, 8, 4, 20, 15, 10, 5, 30, 24, 18, 12, 6, 42, 35, 28, 21, 14, 7, 56, 48, 40, 32, 24, 16, 8, 72, 63, 54, 45, 36, 27, 18, 9, 90, 80, 70, 60, 50, 40, 30, 20, 10, 110, 99, 88, 77, 66, 55, 44, 33, 22, 11, 132, 120, 108, 96, 84, 72, 60, 48, 36, 24, 12, 156, 143, 130, 117

Offset: 1

Floor van Lamoen, Aug 21 2001

From Wolfdieter Lang, Dec 03 2014: (Start)
For excircles and their radii see the Eric W. Weisstein links. Here the circle radius with center J_A is considered.
Note that not all Pythagorean triangles are covered, e.g., the nonprimitive one (9, 12, 15) does not appear. However, the nonprimitive one (8, 6, 10) does appear as (n, m) = (2, 1). (End)
This triangle T appears also in the problem of finding all positive integer solutions for a and b of the general Fibonacci sequence F(a,b;k+1) = a*F(a,b;k) + b*F(a,b;k-1) (with some inputs F(a,b;0) and F(a,b;1)) such that the limit r = r(a,b) = F(a,b;k+1)/F(a,b;k) for k -> infinity becomes a positive integer r = (a + sqrt(a^2 + 4*b))/2. Namely, for any a = m >= 1 there are infinitely many b solutions b = T(n,m) = (n+1)*(n+1-m) for n >= m. The limit is r(a,b) = n+1 for a = m = 1..n, which is A003057 read as a triangle with offset 1. This entry was motivated by A249973 and A249974 by Kerry Mitchell concerned with real values of r. - Wolfdieter Lang, Jan 11 2015

			The triangle T(n, m) begins:
n\m   1   2   3   4   5   6   7   8   9 10 11 12 13 14 15 ...
1:    2
2:    6   3
3:   12   8   4
4:   20  15  10  5
5:   30  24  18  12   6
6:   42  35  28  21  14   7
7:   56  48  40  32  24  16   8
8:   72  63  54  45  36  27  18   9
9:   90  80  70  60  50  40  30  20  10
10: 110  99  88  77  66  55  44  33  22 11
11: 132 120 108  96  84  72  60  48  36 24 12
12: 156 143 130 117 104  91  78  65  52 39 26 13
13: 182 168 154 140 126 112  98  84  70 56 42 28 14
14: 210 195 180 165 150 135 120 105  90 75 60 45 30 15
15: 240 224 208 192 176 160 144 128 112 96 80 64 48 32  1
... Formatted and extended by _Wolfdieter Lang_, Dec 02 2014
--------------------------------------------------------------
Example of general (a,b)-Fibonacci sequence positive integer limits r(a,b) (see the Jan 11 2015 comment above):
T(3, 2) = 8, that is a = m = 2 has a solution b = T(3, 2) = 8 with r = r(2,8) = n+1 = 4 = (2 + sqrt(4 + 4*8))/2. The other two solutions with r = 4 appear for b = T(3, m) with m = a = 1 and 3. In general, row n has n times the value n+1 for r, namely r(a=m,b=T(n,m)) = n+1, for m = 1..n. - _Wolfdieter Lang_, Jan 11 2015

Eric Weisstein's World of Mathematics, Excircles
Eric Weisstein's World of Mathematics, Exradius

Cf. A003991 (incircle radius), A063930 (B-excircle radius), A001283 (C-excircle radius), A055096 (circumcircle diameter).

T(n, m) = (n+1)(n-m+1), n >= m >= 1.

T(n, m) = rho_A = sqrt(s*(s-b)*(s-c)/(s-a)) with the semiperimeter s = (a + b + c)/2 and the substituted a, b, c values as given in the name. - Wolfdieter Lang, Dec 02 2014

Edited: Crossreferences commented and A055096 added by Wolfdieter Lang, Dec 02 2014

A119412 a(n) = n*(n+11).

Original entry on oeis.org

0, 12, 26, 42, 60, 80, 102, 126, 152, 180, 210, 242, 276, 312, 350, 390, 432, 476, 522, 570, 620, 672, 726, 782, 840, 900, 962, 1026, 1092, 1160, 1230, 1302, 1376, 1452, 1530, 1610, 1692, 1776, 1862, 1950, 2040, 2132, 2226, 2322, 2420, 2520

Offset: 0

Zerinvary Lajos, Jul 26 2006

Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A056115, A063930.

s=0;lst={s};Do[s+=n++ +12;AppendTo[lst, s], {n, 0, 7!, 2}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 19 2008 *)
Table[n(n+11),{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, May 19 2011 *)
LinearRecurrence[{3,-3,1},{0,12,26},50] (* Harvey P. Dale, Jun 11 2016 *)

a(n)=n*(n+11) \\ Charles R Greathouse IV, Jan 21 2015

a(n) = 2*A056115(n). - Zerinvary Lajos, Feb 12 2007
a(n) = 2*a(n-1) - a(n-2) + 2 with a(0)=0, a(1)=12. - Vincenzo Librandi, Aug 01 2010
G.f.: 2*x*(-6+5*x)/(x-1)^3. - R. J. Mathar, Jul 14 2012
Sum_{n>=1} 1/a(n) = 83711/304920 via Sum_{n>=0} 1/((n+x)(n+y)) = (psi(x)-psi(y))/(x-y). - R. J. Mathar, Jul 14 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/11 - 20417/304920. - Amiram Eldar, Jan 15 2021
From Elmo R. Oliveira, Dec 12 2024: (Start)
E.g.f.: exp(x)*x*(12 + x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

Definition simplified and the most obfuscating programs removed by R. J. Mathar, Jul 31 2010

Offset corrected (from 11 to 0) by Vincenzo Librandi, Aug 01 2010

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A055096 Triangle read by rows, sums of 2 distinct nonzero squares: T(n,k) = k^2+n^2, (n>=2, 1 <= k <= n-1).

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A098847 a(n) = n*(n + 12).

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A001283 Triangle read by rows, in which row n consists of n(n+m) for m = 1 .. n-1.

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A063929 Radius of A-excircle of Pythagorean triangle with a = (n+1)^2 - m^2, b = 2*(n+1)*m and c = (n+1)^2 + m^2.

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A119412 a(n) = n*(n+11).

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A063929 Radius of A-excircle of Pythagorean triangle with a = (n+1)^2 - m^2, b = 2(n+1)m and c = (n+1)^2 + m^2.