A212011 -id:A212011 - cp's OEIS frontend

A212010 Triangle read by rows: T(n,k) = total number of parts in the last k shells of n.

1, 2, 3, 3, 5, 6, 6, 9, 11, 12, 8, 14, 17, 19, 20, 15, 23, 29, 32, 34, 35, 19, 34, 42, 48, 51, 53, 54, 32, 51, 66, 74, 80, 83, 85, 86, 42, 74, 93, 108, 116, 122, 125, 127, 128, 64, 106, 138, 157, 172, 180, 186, 189, 191, 192, 83, 147, 189, 221, 240

Offset: 1

Omar E. Pol, Apr 26 2012

The set of partitions of n contains n shells (see A135010). Let m and n be two positive integers such that m <= n. It appears that in any set formed by m connected shells, or m disconnected shells, or a mixture of both, the sum of all parts of the j-th column equals the total number of parts >= j in the same set (see example). More generally it appears that any of these sets has the same properties mentioned in A206563 and A207031.

It appears that the last k shells of n contain p(n-k) parts of size k, where p(n) = A000041(n). See also A182703.

			For n = 5 the illustration shows five sets containing the last k shells of 5 and below we can see that the sum of all parts of the first column equals the total number of parts in each set:
--------------------------------------------------------
.  S{5}       S{4-5}     S{3-5}     S{2-5}     S{1-5}
--------------------------------------------------------
.  The        Last       Last       Last       The
.  last       two        three      four       five
.  shell      shells     shells     shells     shells
.  of 5       of 5       of 5       of 5       of 5
--------------------------------------------------------
.
.  5          5          5          5          5
.  3+2        3+2        3+2        3+2        3+2
.    1        4+1        4+1        4+1        4+1
.      1      2+2+1      2+2+1      2+2+1      2+2+1
.      1        1+1      3+1+1      3+1+1      3+1+1
.        1        1+1      1+1+1    2+1+1+1    2+1+1+1
.          1        1+1      1+1+1    1+1+1+1  1+1+1+1+1
. ---------- ---------- ---------- ---------- ----------
.  8         14         17         19         20
.
So row 5 lists 8, 14, 17, 19, 20.
.
Triangle begins:
1;
2,    3;
3,    5,   6;
6,    9,  11,  12;
8,   14,  17,  19,  20;
15,  23,  29,  32,  34,  35;
19,  34,  42,  48,  51,  53,  54;
32,  51,  66,  74,  80,  83,  85,  86;
42,  74,  93, 108, 116, 122, 125, 127, 128;
64, 106, 138, 157, 172, 180, 186, 189, 191, 192;

Mirror of triangle A212000. Column 1 is A138137. Right border is A006128.

Cf. A135010, A138121, A181187, A182703, A206563, A207031, A207032, A212001, A212011

T(n,k) = A006128(n) - A006128(n-k).

T(n,k) = Sum_{j=n-k+1..n} A138137(j).

A212000 Triangle read by rows: T(n,k) = total number of parts in the last n-k+1 shells of n.

Original entry on oeis.org

1, 3, 2, 6, 5, 3, 12, 11, 9, 6, 20, 19, 17, 14, 8, 35, 34, 32, 29, 23, 15, 54, 53, 51, 48, 42, 34, 19, 86, 85, 83, 80, 74, 66, 51, 32, 128, 127, 125, 122, 116, 108, 93, 74, 42, 192, 191, 189, 186, 180, 172, 157, 138, 106, 64, 275, 274, 272, 269, 263, 255, 240

Offset: 1

Omar E. Pol, Apr 26 2012

It appears that the last k shells of n contain p(n-k) parts of size k, where p(n) = A000041(n). See also A182703.

			For n = 5 the illustration shows five sets containing the last n-k+1 shells of 5 and below we can see that the sum of all parts of the first column equals the total number of parts in each set:
--------------------------------------------------------
.  S{1-5}     S{2-5}     S{3-5}     S{4-5}     S{5}
--------------------------------------------------------
.  The        Last       Last       Last       The
.  five       four       three      two        last
.  shells     shells     shells     shells     shell
.  of 5       of 5       of 5       of 5       of 5
--------------------------------------------------------
.
.  5          5          5          5          5
.  3+2        3+2        3+2        3+2        3+2
.  4+1        4+1        4+1        4+1          1
.  2+2+1      2+2+1      2+2+1      2+2+1          1
.  3+1+1      3+1+1      3+1+1        1+1          1
.  2+1+1+1    2+1+1+1      1+1+1        1+1          1
.  1+1+1+1+1    1+1+1+1      1+1+1        1+1          1
. ---------- ---------- ---------- ---------- ----------
. 20         19         17         14          8
.
So row 5 lists 20, 19, 17, 14, 8.
.
Triangle begins:
1;
3,     2;
6,     5,   3;
12,   11,   9,   6;
20,   19,  17,  14,  8;
35,   34,  32,  29,  23,  15;
54,   53,  51,  48,  42,  34,  19;
86,   85,  83,  80,  74,  66,  51,  32;
128, 127, 125, 122, 116, 108,  93,  74,  42;
192, 191, 189, 186, 180, 172, 157, 138, 106, 64;

Mirror of triangle A212010. Column 1 is A006128. Right border gives A138137.

Cf. A135010, A138121, A181187, A182703, A206563, A207031, A207032, A212001, A212011

T(n,k) = A006128(n) - A006128(k-1).

T(n,k) = Sum_{j=k..n} A138137(j).

A212001 Triangle read by rows: T(n,k) = sum of all parts of the last n-k+1 shells of n.

Original entry on oeis.org

1, 4, 3, 9, 8, 5, 20, 19, 16, 11, 35, 34, 31, 26, 15, 66, 65, 62, 57, 46, 31, 105, 104, 101, 96, 85, 70, 39, 176, 175, 172, 167, 156, 141, 110, 71, 270, 269, 266, 261, 250, 235, 204, 165, 94, 420, 419, 416, 411, 400, 385, 354, 315, 244, 150, 616, 615

Offset: 1

Omar E. Pol, Apr 26 2012

The set of partitions of n contains n shells (see A135010). It appears that the last k shells of n contain p(n-k) parts of size k, where p(n) = A000041(n). See also A182703.

			For n = 5 the illustration shows five sets containing the last n-k+1 shells of 5 and below the sum of all parts of each set:
--------------------------------------------------------
.  S{1-5}     S{2-5}     S{3-5}     S{4-5}     S{5}
--------------------------------------------------------
.  The        Last       Last       Last       The
.  five       four       three      two        last
.  shells     shells     shells     shells     shell
.  of 5       of 5       of 5       of 5       of 5
--------------------------------------------------------
.
.  5          5          5          5          5
.  3+2        3+2        3+2        3+2        3+2
.  4+1        4+1        4+1        4+1          1
.  2+2+1      2+2+1      2+2+1      2+2+1          1
.  3+1+1      3+1+1      3+1+1        1+1          1
.  2+1+1+1    2+1+1+1      1+1+1        1+1          1
.  1+1+1+1+1    1+1+1+1      1+1+1        1+1          1
. ---------- ---------- ---------- ---------- ----------
.     35         34         31         26         15
.
So row 5 lists 35, 34, 31, 26, 15.
.
Triangle begins:
1;
4,     3;
9,     8,   5;
20,   19,  16,  11;
35,   34,  31,  26,  15;
66,   65,  62,  57,  46,  31;
105, 104, 101,  96,  85,  70,  39;
176, 175, 172, 167, 156, 141, 110,  71;
270, 269, 266, 261, 250, 235, 204, 165,  94;
420, 419, 416, 411, 400, 385, 354, 315, 244, 150;

Mirror of triangle A212011. Column 1 is A066186. Right border is A138879.

Cf. A135010, A138121, A182703, A206563, A211980, A212000, A212010.

T(n,k) = A066186(n) - A066186(k-1).

T(n,k) = Sum_{j=k..n} A138879(j).

A211990 Triangle read by rows: T(n,k) = total number of regions in the last k shells of n.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 2, 3, 4, 5, 2, 4, 5, 6, 7, 4, 6, 8, 9, 10, 11, 4, 8, 10, 12, 13, 14, 15, 7, 11, 15, 17, 19, 20, 21, 22, 8, 15, 19, 23, 25, 27, 28, 29, 30, 12, 20, 27, 31, 35, 37, 39, 40, 41, 42, 14, 26, 34, 41, 45, 49, 51, 53, 54, 55, 56, 21, 35, 47

Offset: 1

Omar E. Pol, Apr 27 2012

For the definition of "region of n" see A206437. For the definition of "last section of n" see A135010.

Apparently differs from A027300 at the right border.

			For n = 5 and k = 2 we have that the 4th shell of 5 contains two regions: [2] and [4,2,1,1,1]. Then we can see that the 5th shell of 5 contains two regions: [3] and [5,2,1,1,1,1,1]. Therefore the total number of regions in the last two shells of 5 is T(5,2) = 2+2 = 4 (see illustration in the link section).
Triangle begins:
1;
1,   2;
1,   2,  3;
2,   3,  4,  5;
2,   4,  5,  6,  7;
4,   6,  8,  9, 10, 11;
4,   8, 10, 12, 13, 14, 15;
7,  11, 15, 17, 19, 20, 21, 22;
8,  15, 19, 23, 25, 27, 28, 29, 30;
12, 20, 27, 31, 35, 37, 39, 40, 41, 42;
14, 26, 34, 41, 45, 49, 51, 53, 54, 55, 56;
21, 35, 47, 55, 62, 66, 70, 72, 74, 75, 76, 77;

Omar E. Pol, Illustration of the seven regions of 5

Mirror of triangle A211980. Column 1 is A187219. Right border gives A000041, n >= 1.

Cf. A027300, A135010, A138121, A182703, A186114, A206437, A212010, A212011.

T(n,k) = A000041(n) - A000041(n-k), if 1

T(n,k) = A000041(n), if k = n.

T(n,k) = Sum_{j=1..k} A187219(n-j+1,1).

Showing 1-4 of 4 results.

cp's OEIS Frontend

A212010 Triangle read by rows: T(n,k) = total number of parts in the last k shells of n.

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A212000 Triangle read by rows: T(n,k) = total number of parts in the last n-k+1 shells of n.

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A212001 Triangle read by rows: T(n,k) = sum of all parts of the last n-k+1 shells of n.

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A211990 Triangle read by rows: T(n,k) = total number of regions in the last k shells of n.

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