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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A351381 Table read by downward antidiagonals: T(n,k) = n*(k+1)^2.

Original entry on oeis.org

4, 9, 8, 16, 18, 12, 25, 32, 27, 16, 36, 50, 48, 36, 20, 49, 72, 75, 64, 45, 24, 64, 98, 108, 100, 80, 54, 28, 81, 128, 147, 144, 125, 96, 63, 32, 100, 162, 192, 196, 180, 150, 112, 72, 36, 121, 200, 243, 256, 245, 216, 175, 128, 81, 40, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44
Offset: 1

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Author

Bernard Schott, Mar 28 2022

Keywords

Comments

When m and k are both positive integers and k | m, with m/k = n, then T(n,k) = S(m,k) = (m+k) + (m-k) + (m*k) + (m/k) = S(n*k,k) = n*(k+1)^2, problem proposed by Yakov Perelman.
All terms are nonsquarefree (A013929).

Examples

			Table begins:
  n \ k |   1      2      3      4      5      6      7      8      9     10
  ----------------------------------------------------------------------------
     1  |   4      9     16      25    36     49     64     81    100    121
     2  |   8     18     32      50    72     98    128    162    200    242
     3  |  12     27     48      75   108    147    192    243    300    363
     4  |  16     36     64     100   144    196    256    324    400    484
     5  |  20     45     80     125   180    245    320    405    500    605
     6  |  24     54     96     150   216    294    384    486    600    726
     7  |  28     63    112     175   252    343    448    567    700    847
     8  |  32     72    128     200   288    392    512    648    800    968
     9  |  36     81    144     225   324    441    576    729    900   1089
    10  |  40     90    160     250   360    490    640    810   1000   1210
  ............................................................................
T(3,4) = 75 = 3*(4+1)^2 corresponds to S(3*4,4) = S(12,4) = (12+4) + (12-4) + (12*4) + 12/4 = 75.
S(10,5) = (10+5) + (10-5) + (10*5) + (10/5) = T(10/5,5) = T(2,5) = 72.
		

References

  • I. Perelman, L'Algèbre Récréative, Chapitre IV, Les équations de Diophante, Deux nombres et quatre opérations, Editions en langues étrangères, Moscou, 1959, pp. 101-102.
  • Ya. I. Perelman, Algebra Can Be Fun, Chapter IV, Diophantine Equations, Two numbers and four operations, Mir Publishers Moscow, 1979, pp. 131-132.

Crossrefs

Cf. A013929.
Cf. A000290 \ {0,1} (row 1), A001105 \ {0,2} (row 2), A033428 \ {0,3} (row 3), A016742 \ {0,4} (row 4), A033429 \ {0,5} (row 5), A033581 \ {0,6} (row 6).
Cf. A008586 \ {0} (column 1), A008591 \ {0} (column 2), A008598 \ {0} (column 3), A008607 \ {0} (column 4), A044102 \ {0} (column 5).
Cf. A045991 \ {0} (diagonal).

Programs

  • Mathematica
    T[n_, k_] := n*(k + 1)^2; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Mar 29 2022 *)

Formula

T(n,k) = n*(k+1)^2.
T(n,n) = (n+1)^3 - (n+1)^2 = A045991(n+1) for n >= 1.
G.f.: x*(1 + y)/((1 - x)^2*(1 - y)^3). - Stefano Spezia, Mar 31 2022