A238303 -id:A238303 - cp's OEIS frontend

A143593 Triangle read by rows, square of A238303 (an infinite lower triangular matrix with 1's in the first column and the rest 2's).

1, 3, 4, 5, 8, 4, 7, 12, 8, 4, 9, 16, 12, 8, 4, 11, 20, 16, 12, 8, 4, 13, 24, 20, 16, 12, 8, 4, 15, 28, 24, 20, 16, 12, 8, 4, 17, 32, 28, 24, 20, 16, 12, 8, 4, 19, 36, 32, 28, 24, 20, 16, 12, 8, 4

Offset: 1

Gary W. Adamson & Roger L. Bagula, Aug 26 2008

Row sums = A056220 starting with offset 1: (1, 7, 17, 31, 49, 71, 97,...).

			The square of the infinite lower triangular matrix (1; 1,2; 1,2,2;...) =
1;
3, 4;
5, 8, 4;
7, 12, 8, 4;
9, 16, 12, 8, 4;
11, 20, 16, 12, 8, 4;
13, 24, 20, 16, 12, 8, 4;
...

Cf. A056220, A238303.

Triangle read by rows, square of an infinite lower triangular matrix with 1's in the first column and the rest 2's. Square of (A000012 * (S(k) * 0^(n-k)), 1<=k<=n

A238339 Square number array read by ascending antidiagonals: T(1,k) = 2k + 1, and T(n,k) = (2n^(k+1)-n-1)/(n-1) otherwise.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 13, 7, 1, 1, 9, 25, 29, 9, 1, 1, 11, 41, 79, 61, 11, 1, 1, 13, 61, 169, 241, 125, 13, 1, 1, 15, 85, 311, 681, 727, 253, 15, 1, 1, 17, 113, 517, 1561, 2729, 2185, 509, 17, 1, 1, 19, 145, 799, 3109, 7811, 10921, 6559, 1021, 19, 1

Offset: 0

Philippe Deléham, Feb 24 2014

			Square array begins:
1..1...1.....1......1.......1........1........1...
1..3...5.....7......9......11.......13.......15...
1..5..13....29.....61.....125......253......509...
1..7..25....79....241.....727.....2185.....6559...
1..9..41...169....681....2729....10921....43689...
1.11..61...311...1561....7811....39061...195311...
1.13..85...517...3109...18661...111973...671845...
1.15.113...799...5601...39215...274513..1921599...
1.17.145..1169...9361...74897...599185..4793489...
1.19.181..1639..14761..132859..1195741.10761679...
1.21.221..2221..22221..222221..2222221.22222221...

Cf. A238303.

T:= proc(n, k); if n=1 then 2*k+1 else (2*n^(k+1)-n-1)/(n-1) fi end:
seq(seq(T(n-k, k), k=0..n), n=0..10); # Georg Fischer, Oct 14 2023

T(0,k) = A000012(k) = 1;
T(1,k) = A005408(k) = 2k+1;
T(2,k) = A036563(k+2);
T(3,k) = A058481(k+1);
T(4,k) = A083584(k);
T(5,k) = A137410(k);
T(6,k) = A233325(k);
T(7,k) = A233326(k);
T(8,k) = A233328(k);
T(9,k) = A211866(k+1);
T(10,k) = A165402(k+1);
T(n,0) = A000012(n) = 1;
T(n,1) = A005408(n) = 2*n+1;
T(n,2) = A001844(n) = 2*n^2 + 2*n + 1.

Definition amended by Georg Fischer, Oct 14 2023

A247349 Regular triangle obtained by procedure described in comment in the case of m=3.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 2, 1, 2, 3, 3, 2, 1, 2, 3, 2, 3, 3, 2, 1, 4, 2, 1, 4, 2, 3, 3, 3, 2, 3, 3, 3, 2, 1, 4, 3, 3, 2, 1, 4, 3, 2, 3, 3, 3, 4, 3, 2, 3, 3, 3, 3, 2, 1, 4, 3, 3, 3, 3, 2, 1, 4, 4, 3, 2, 5, 3, 2, 1, 4, 4, 3, 2, 5, 3, 3, 3, 4, 4, 4, 3, 2, 5, 3, 3, 3, 4, 3, 2, 1, 4

Offset: 1

Michel Marcus, Sep 16 2014

"Consider an array of numbers formed by a rotating queue: starting with just the number 1, to obtain the next row we move everything in the last row m steps to the left, with numbers at the front of the row cycling around and appearing at the back. We then append 1 plus the head of the last row to the new row." (from the Introduction of article by P. J. Graber).

Philip Jameson Graber, Rotation Remainders, arXiv:1409.4113 [math.NT], 2014.

Cf. A238303 (triangle obtained when m=0).

moveleft(v, m) = {va = v; for (i=1, m, nb = #va; vb = vector(nb, i, if (i

cp's OEIS Frontend

A143593 Triangle read by rows, square of A238303 (an infinite lower triangular matrix with 1's in the first column and the rest 2's).

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A238339 Square number array read by ascending antidiagonals: T(1,k) = 2k + 1, and T(n,k) = (2n^(k+1)-n-1)/(n-1) otherwise.

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A247349 Regular triangle obtained by procedure described in comment in the case of m=3.

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cp's OEIS Frontend

A143593 Triangle read by rows, square of A238303 (an infinite lower triangular matrix with 1's in the first column and the rest 2's).

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A238339 Square number array read by ascending antidiagonals: T(1,k) = 2*k + 1, and T(n,k) = (2*n^(k+1)-n-1)/(n-1) otherwise.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Formula

Extensions

A247349 Regular triangle obtained by procedure described in comment in the case of m=3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

A238339 Square number array read by ascending antidiagonals: T(1,k) = 2k + 1, and T(n,k) = (2n^(k+1)-n-1)/(n-1) otherwise.