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
Examples
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), ...
Links
- 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
Crossrefs
Programs
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Haskell
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
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Magma
[n^2+k^2: k in [1..n-1], n in [2..15]]; // G. C. Greubel, Apr 19 2023
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Maple
sum2distinct_squares_array := (n) -> (((n-((trinv(n-1)*(trinv(n-1)-1))/2))^2)+((trinv(n-1)+1)^2));
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Mathematica
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 *) -
SageMath
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
Formula
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).
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)
Extensions
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
Comments