cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A279967 Square array read by antidiagonals upwards in which each term is the sum of prior elements in the same row, column, diagonal, or antidiagonal that divide n; the array is seeded with an initial value a(1)=1.

Original entry on oeis.org

1, 1, 2, 2, 2, 7, 2, 9, 10, 15, 2, 10, 1, 13, 17, 8, 0, 13, 1, 14, 9, 8, 0, 13, 3, 30, 13, 10, 2, 16, 1, 23, 5, 7, 14, 15, 2, 8, 28, 32, 2, 23, 2, 9, 49, 12, 0, 48, 2, 11, 1, 20, 3, 18, 13, 28, 0, 4, 1, 56, 5, 8, 16, 35, 46, 4, 2, 6, 2, 10
Offset: 1

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Author

Alec Jones, Dec 24 2016

Keywords

Comments

From Hartmut F. W. Hoft, Jan 23 2017: (Start)
Shown by induction and direct (modular) computations for
column 1: Every number is even, except for the first two 1's; in addition to row 3, value 2 occurs in rows 4*k and 4*k+1, and every value in rows 4*k+2 and 4*k+3 is divisible by 4, for all k>=1.
column 2: The first four entries, 2, 2, 9 and 10, contain the only odd number; no nonzero entry in row k>3 has 9 as a factor, and value 0 occurs in rows 4*k+1 and 4*k+2, for all k>=1.
Conjecture:
a({1, 6, 8, 9, 10, 15, 26, 45, 48, 84, 96, 112, 115, 252, 336, 343}) =
{1, 7, 9,10, 15, 17, 30, 49, 48,104,117, 115, 122, 257, 343, 395} are the only numbers in the sequence with the property a(n) >= n (verified through n=500500, i.e., the triangle with 1000 antidiagonals).
This conjecture together with Bouniakowsky's conjecture that certain quadratic integer polynomials generate infinitely many primes (e.g. see A002496 for n^2+1 and A188382 for 2*n^2+n+1) implies that in every column in the triangle infinitely many prime sequence indices occur and therefore infinitely many 0's whenever the column contains no 1's. The proof is based on the fact that for a large enough prime sequence index p in whose prior column no 1 occurs then a(p)=0; therefore infinitely many 0's occur in that column. Obviously, once value 1 occurs in a column no 0 value can occur in a subsequent row.
Conjecture:
Every row in the triangle contains exactly two 1's.
(End)

Examples

			After 6 terms, the array looks like:
.
1   2   7
1   2
2
We have a(6) = 7 because a(1) = 1, a(3) = 2, a(4) = 2, and a(5) = 2 divide 6; 1 + 2 + 2 + 2 = 7.
From _Hartmut F. W. Hoft_, Jan 23 2017: (Start)
1   2   7  15  17   9  10  15  49  13   4  31  22
1   2  10  13  14  13  14   9  18  46  12  66
2   9   1   1  30   7   2   3  35  12   3
2  10  13   3   5  23  20  16  14  17
2   0  13  23   2   1   8  11   2
8   0   1  32  11   5   3   6
8  16  28   2  56  42   8
2   8  48   1   2 104
2   0   4  10   1
12   0   2  10
28   6   2
2  42
2
.
Expanded the triangle to the first 13 antidiagonals of the array, i.e. a(1) ... a(91), to show the start of the 2- and 0-value patterns in columns 1 and 2. The first 0 beyond column 2 is a(677) in row 27, column 11 of the triangle.
A188382(n)=2*n^2+n+1 for n>=0 are the alternate sequence indices for column 1 starting in row 1, 2*n^2+n+2 for n>=1 are the alternate sequence indices for column 2 starting in row 2, and 2*n^2+n+11 for n>=5 are the alternate sequence indices for column 11 starting in row 1.
The sequence indices in the triangle for row positions k>=1 in columns 1,..., 5 are given in sequences A000124(k), A152948(k+3), A152950(k+3), A145018(k+4) and A167499(k+4).
(End)
		

Crossrefs

Cf. A279966 for the related sequence which counts prior terms.
Cf. A269347 for a one-dimensional version of this sequence.
Cf. also A279211, A279212.

Programs

  • Mathematica
    (*  printing of the triangle is commented out of function a279967[]  *)
    pCol[{i_, j_}] := Map[{#, j}&, Range[1, i-1]]
    pDiag[{i_, j_}] := If[j>=i, Map[{#, j-i+#}&, Range[1, i-1]], Map[{i-j+#, #}&, Range[1, j-1]]]
    pRow[{i_, j_}] := Map[{i, #}&, Range[1, j-1]]
    pAdiag[{i_, j_}] := Map[{i+j-#, #}&, Range[1, j-1]]
    priorPos[{i_, j_}] := Join[pCol[{i, j}], pDiag[{i, j}], pRow[{i, j}], pAdiag[{i, j}]]
    seqPos[{i_, j_}] := (i+j-2)(i+j-1)/2+j
    antiDiag[k_] := Map[{k+1-#, #}&, Range[1, k]]
    upperTriangle[k_] := Flatten[Map[antiDiag, Range[1, k]], 1]
    a279967[k_] := Module[{ut=upperTriangle[k], ms=Table[" ", {i, 1, k}, {j, 1, k}], h, pos, val, seqL={1}}, ms[[1, 1]]=1; For[h=2, h<=Length[ut], h++, pos=ut[[h]]; val=Apply[Plus, Select[Map[ms[[Apply[Sequence, #]]]&, priorPos[pos]], #!=0 && Mod[seqPos[pos], #]==0&]]; AppendTo[seqL, val]; ms[[Apply[Sequence, pos]]]=val]; (* Print[TableForm[ms]]; *) seqL]
    a279967[13] (* values in first 13 antidiagonals *)
    (* Hartmut F. W. Hoft, Jan 23 2017 *)

A300401 Array T(n,k) = n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1) read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 4, 4, 3, 4, 7, 8, 7, 4, 5, 11, 14, 14, 11, 5, 6, 16, 22, 24, 22, 16, 6, 7, 22, 32, 37, 37, 32, 22, 7, 8, 29, 44, 53, 56, 53, 44, 29, 8, 9, 37, 58, 72, 79, 79, 72, 58, 37, 9, 10, 46, 74, 94, 106, 110, 106, 94, 74, 46, 10, 11, 56, 92, 119
Offset: 0

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Author

Keywords

Comments

Antidiagonal sums are given by 2*A055795.
Rows/columns n are binomial transform of {n, A152947(n+1), n, 0, 0, 0, ...}.
Some primes in the array are
n = 1: {2, 7, 11, 29, 37, 67, 79, 137, 191, 211, 277, 379, ...} = A055469, primes of the form k*(k + 1)/2 + 1;
n = 3: {3, 7, 37, 53, 479, 653, 1249, 1619, 2503, 3727, 4349, 5737, 7109, 8179, 9803, 11839, 12107, ...};
n = 4: {11, 37, 79, 137, 211, 821, 991, 1597, 1831, 2081, 2347, ...} = A188382, primes of the form 8*(2*k - 1)^2 + 2*(2*k - 1) + 1.

Examples

			The array T(n,k) begins
0     1    2    3    4     5     6     7     8     9    10    11  ...
1     2    4    7   11    16    22    29    37    46    56    67  ...
2     4    8   14   22    32    44    58    74    92   112   134  ...
3     7   14   24   37    53    72    94   119   147   178   212  ...
4    11   22   37   56    79   106   137   172   211   254   301  ...
5    16   32   53   79   110   146   187   233   284   340   401  ...
6    22   44   72  106   146   192   244   302   366   436   512  ...
7    29   58   94  137   187   244   308   379   457   542   634  ...
8    37   74  119  172   233   302   379   464   557   658   767  ...
9    46   92  147  211   284   366   457   557   666   784   911  ...
10   56  112  178  254   340   436   542   658   784   920  1066  ...
11   67  134  212  301   401   512   634   767   911  1066  1232  ...
12   79  158  249  352   467   594   733   884  1047  1222  1409  ...
13   92  184  289  407   538   682   839  1009  1192  1388  1597  ...
14  106  212  332  466   614   776   952  1142  1346  1564  1796  ...
15  121  242  378  529   695   876  1072  1283  1509  1750  2006  ...
16  137  274  427  596   781   982  1199  1432  1681  1946  2227  ...
17  154  308  479  667   872  1094  1333  1589  1862  2152  2459  ...
18  172  344  534  742   968  1212  1474  1754  2052  2368  2702  ...
19  191  382  592  821  1069  1336  1622  1927  2251  2594  2956  ...
20  211  422  653  904  1175  1466  1777  2108  2459  2830  3221  ...
...
The inverse binomial transforms of the columns are
0     1    2    3    4     5     6     7     8     9    10    11  ...  A001477
1     1    2    4    7    11    22    29    37    45    56    67  ...  A152947
0     1    2    3    4     5     6     7     8     9    10    11  ...  A001477
0     0    0    0    0     0     0     0     0     0     0     0  ...
0     0    0    0    0     0     0     0     0     0     0     0  ...
0     0    0    0    0     0     0     0     0     0     0     0  ...
...
		

References

  • Miklós Bóna, Introduction to Enumerative Combinatorics, McGraw-Hill, 2007.
  • L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Reidel Publishing Company, 1974.
  • R. P. Stanley, Enumerative Combinatorics, second edition, Cambridge University Press, 2011.

Crossrefs

Programs

  • Maple
    T := (n, k) -> n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1);
    for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;
  • Mathematica
    T[n_, k_] := n (Binomial[k, 2] + 1) + k (Binomial[n, 2] + 1);
    Table[T[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 07 2018 *)
  • Maxima
    T(n, k) := n*(binomial(k, 2) + 1) + k*(binomial(n, 2) + 1)$
    for n:0 thru 20 do
      print(makelist(T(n, k), k, 0, 20));
    
  • PARI
    T(n, k) = n*(binomial(k,2) + 1) + k*(binomial(n,2) + 1);
    tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, Mar 12 2018

Formula

T(n,k) = T(k,n) = n*A152947(k+1) + k*A152947(n+1).
T(n,0) = A001477(n).
T(n,1) = A000124(n).
T(n,2) = A014206(n).
T(n,3) = A273465(3*n+2).
T(n,4) = A084849(n+1).
T(n,n) = A179000(n-1,n), n >= 1.
T(2*n,2*n) = 8*A081436(n-1), n >= 1.
T(2*n+1,2*n+1) = 2*A006000(2*n+1).
T(n,n+1) = A188377(n+3).
T(n,n+2) = A188377(n+2), n >= 1.
Sum_{k=0..n} T(k,n-k) = 2*(binomial(n, 4) + binomial(n, 2)).
G.f.: -((2*x*y - y - x)*(2*x*y - y - x + 1))/(((x - 1)*(y - 1))^3).
E.g.f.: (1/2)*(x + y)*(x*y + 2)*exp(x + y).

A306362 Prime numbers in A317298.

Original entry on oeis.org

3, 11, 37, 79, 137, 211, 821, 991, 1597, 1831, 2081, 2347, 2927, 3571, 3917, 4657, 5051, 6329, 8779, 9871, 11027, 14197, 14879, 17021, 20101, 21737, 26107, 27967, 28921, 33931, 34981, 39341, 40471, 41617, 50087, 51361, 59341, 60727, 62129, 66431, 69379, 70877
Offset: 1

Views

Author

Stefano Spezia, Feb 10 2019

Keywords

Comments

Conjecture: Except the first term a(1) = 3, all the other terms do not end with 3.
It is easy to prove that the numbers A317298(n) end with 3 only when n ends with 1. In this case A317298(10*n+1) = (10*n + 1)*(20*n + 3), which is composite for n > 0. Therefore the conjecture is true. - Bruno Berselli, Feb 11 2019
Essentially (apart from the 3) the same as A188382, because for even n, A317298(n=2k) has the form 1+2*k+8*k^2 and for odd n A317298(n) is a multiple of n and not prime. - R. J. Mathar, Feb 14 2019

Crossrefs

Programs

  • Mathematica
    Select[Table[(1/2)*(1 + (-1)^n + 2*n + 4*n^2),{n,1,300}], PrimeQ]
  • PARI
    for(n=0, 300, if(ispseudoprime(t=(1/2)*(1 + (-1)^n + 2*n + 4*n^2)), print1(t", ")));
Showing 1-3 of 3 results.