A125714 -id:A125714 - cp's OEIS frontend

A125750 A Moessner triangle using (1, 3, 5, ...).

1, 3, 5, 10, 19, 11, 42, 89, 64, 19, 216, 498, 415, 160, 29, 1320, 3254, 3023, 1385, 335, 41, 9360, 24372, 24640, 12803, 3745, 623, 55, 75600, 206100, 223116, 127799, 42938, 8750, 1064, 71, 685440, 1943568, 2227276, 1380076, 516201, 122010, 18354, 1704

Offset: 1

Gary W. Adamson, Dec 06 2006

Right border of the triangle = A028387, left border = A007680.

			Circling the 1, 3, 6, ...(-th) terms in the sequence (1, 3, 5, 7, ...), we get A018387: (1, 5, 11, 19, 29, ...). Taking partial sums of the remaining terms, we get (3, 10, 19, 32, ...) in row 2 and we circle 3 and 19. In row 3 we circle the 10.
First few rows of the triangle are:
    1;
    3,   5;
   10,  19,  11;
   42,  89,  64,  19;
  216, 498, 415, 160,  29;
  ...

J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, pp. 63-64.

Joshua Zucker, Table of n, a(n) for n = 1..78
G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
R. Krebbers, L. Parlant, and A. Silva, Moessner's theorem: an exercise in coinductive reasoning in Coq, Theory and practice of formal methods, 309-324, Lecture Notes in Comput. Sci., 9660, Springer, 2016.
Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
Ivan Paasche, Ein neuer Beweis des Moessnerschen Satzes S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 1-5 (1953). [Two years are listed at the beginning of the journal issue.]
Ivan Paasche, Beweis des Moessnerschen Satzes mittels linearer Transformationen, Arch. Math. (Basel) 6 (1955), 194-199.
Ivan Paasche, Eine Verallgemeinerung des Moessnerschen Satzes, Compositio Math. 12 (1956), 263-270.
Hans Salié, Bemerkung zu einem Satz von A. Moessner, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 7-11 (1953). [Two years are listed at the beginning of the journal issue.]
Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Cf. A125714, A125751, A125752.

Using "Moessner's Magic" (Conway and Guy, pp. 63-64; cf. A125714), we circle the 1, 3, 6, 10, ...(-th) terms in the sequence (1, 3, 5, 7, ...) and take partial sums of the remaining terms, making row 2. Circle the terms in row 2 one place offset to the left of row 1 terms, then take partial sums. Continue with analogous operations for succeeding rows. The triangle = leftmost circled terms in each row.

More terms from Joshua Zucker, Jun 17 2007

A125751 A Moessner triangle using (1, 2, 1, 2, 1, 2, ...).

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 10, 18, 9, 2, 38, 78, 53, 15, 1, 186, 422, 344, 129, 23, 1, 1106, 2704, 2484, 1123, 268, 32, 2, 7718, 19998, 20080, 10342, 2991, 490, 42, 2, 61662, 167520, 180466, 102700, 34211, 6891, 824, 54, 1, 554330, 1567518, 1789474, 1103206

Offset: 1

Gary W. Adamson, Dec 06 2006

Circle terms n = 1, 3, 6, 10, ... in the sequence (1, 2, 1, 2, 1, 2, ...). Partial sums of the uncircled terms becomes row 2. Circle the terms in row 2 that are one place offset to the left of the circled row 1 terms. Take partial sums and continue with analogous operations. (Cf. A125714 and "The Book of Numbers", p. 64.)

Left border (1, 2, 4, 10, 38, 186, 1106, 7718, 61662, ...).

			First few rows of the triangle are:
   1;
   2,  1;
   4,  5,  2;
  10, 18,  9,  2;
  38, 78, 53, 15,  1;
  ...

J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, p. 64.

Joshua Zucker, Table of n, a(n) for n = 1..65
G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Cf. A125714, A125750, A125752.

More terms from Joshua Zucker, Jun 17 2007

Corrected the comment concerning the left border - R. J. Mathar, Sep 17 2009

A125752 Moessner triangle using the Fibonacci terms.

Original entry on oeis.org

1, 1, 2, 4, 9, 8, 26, 69, 77, 55, 261, 806, 1088, 920, 610, 4062, 14362, 22887, 22856, 17034, 10946, 98912, 395253, 728605, 847832, 721756, 502606, 317811, 3809193, 17008391, 35644614, 47557978, 46166656, 35655012, 23828383, 14930352

Offset: 1

Gary W. Adamson, Dec 06 2006

A Moessner triangle is generated with the recurrence described in A125714, starting from a first row M(1,c) filled with the Fibonacci numbers M(1,c) = A000045(c), c >= 1.
Subsequent rows n are generated from the numbers in their previous rows with the rule:
Mark/circle all elements M(n-1, A000217(t)) of the previous row n-1, t >= 1.
Define the elements M(n,.) as the partial sums of the M(n-1,.) that have not been marked:
M(n,c) = Sum_{j=1..c} M(n-1,A014132(j)), c >= 1. The T(n,m) are then defined by reading the marked/circled terms "along antidiagonals": T(n,m) = M(n+m-1, A000217(m)), n >= 1, 1 <= m <= n.

			The upper left corner of the array M(n,c) is
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, ...
1, 4, 9, 22, 43, 77, 166, 310, 543, 920, 1907, 3504, 6088, 10269, 17034, ...
4, 26, 69, 235, 545, 1088, 2995, 6499, 12587, 22856, 57601, 121003, 230773, ...
26, 261, 806, 3801, 10300, 22887, 80488, 201491, 432264, 847832, 2586423, ...
261, 4062, 14362, 94850, 296341, 728605, 3315028, 9488917, 22445416, ...
4062, 98912, 395253, 3710281, 13199198, 35644614, 213010460, 690899755, ...
and dropping the columns with column numbers in A014132, reading the remaining array by antidiagonals leads to the final triangle T(n,m):
    1;
    1,   2;
    4,   9,    8;
   26,  69,   77,  55;
  261, 806, 1088, 920, 610;
  ...

J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, p. 64.

Joshua Zucker, Table of n, a(n) for n = 1..55
G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
R. Krebbers, L. Parlant, and A. Silva, Moessner's theorem: an exercise in coinductive reasoning in Coq, Theory and practice of formal methods, 309-324, Lecture Notes in Comput. Sci., 9660, Springer, 2016.
Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
Ivan Paasche, Ein neuer Beweis des Moessnerschen Satzes S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 1-5 (1953). [Two years are listed at the beginning of the journal issue.]
Ivan Paasche, Beweis des Moessnerschen Satzes mittels linearer Transformationen, Arch. Math. (Basel) 6 (1955), 194-199.
Ivan Paasche, Eine Verallgemeinerung des Moessnerschen Satzes, Compositio Math. 12 (1956), 263-270.
Hans Salié, Bemerkung zu einem Satz von A. Moessner, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 7-11 (1953). [Two years are listed at the beginning of the journal issue.]
Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Cf. A125714, A125750, A125751, A081667.

T(n,n) = A081667(n-1).

More terms from Joshua Zucker, Jun 17 2007

Description of starting row corrected, comments detailed with formulas by R. J. Mathar, Sep 17 2009

A125777 Moessner triangle based on A000217.

Original entry on oeis.org

1, 3, 6, 13, 28, 21, 69, 161, 137, 55, 433, 1078, 1017, 477, 120, 3141, 8245, 8437, 4460, 1337, 231, 25873, 71008, 77620, 45058, 15415, 3220, 406, 238629, 680451, 786012, 492264, 186729, 44955, 6930, 666, 2436673, 7184170, 8699205, 5804448, 2394150

Offset: 1

Gary W. Adamson, Dec 07 2006

Begin with the triangular numbers A000217 and circle every T(k)-th term, getting the doubly triangular numbers, A002817. Per instructions shown in A125714, take partial sums of the uncircled terms in row 1, denoting this as row 2. Circle the row 2 terms which are one place to the left of row 1 terms. Take partial sums again in analogous operations for subsequent rows.

Left border = A104989: (1, 3, 13, 69, 433...). Right border = the doubly triangular numbers starting (1, 6, 21...): A002817.

			First few rows of the triangle are as follows:
    1;
    3,    6;
   13,   28,   21;
   69,  161,  137,  55;
  433, 1078, 1017, 477, 120;
  ...

J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, p. 64.

Joshua Zucker, Table of n, a(n) for n = 1..55
G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
M. Niqui and J. J. M. M. Rutten, A proof of Moessner's theorem by coinduction, High.-Order Symb. Comput. 24(3) (2011), 191-206.
Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Cf. A125714, A002817, A104989, A000217.

More terms from Joshua Zucker, Jun 17 2007

A156628 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions 0 and {(m+1)*(m+2)/2-2, m>0} and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 13, 7, 3, 1, 71, 33, 13, 4, 1, 461, 191, 71, 20, 5, 1, 3447, 1297, 461, 120, 28, 6, 1, 29093, 10063, 3447, 836, 181, 38, 7, 1, 273343, 87669, 29093, 6616, 1333, 270, 49, 8, 1, 2829325, 847015, 273343, 58576, 11029, 2150, 375, 61, 9, 1

Offset: 0

Paul D. Hanna, Feb 17 2009

			To generate the array, start with all 1's in row 0; from then on,
obtain row n+1 from row n by first removing terms in row n at
positions 0 and {(m+1)*(m+2)/2-2,m>0} and then taking partial sums.
This square array A begins:
(1), (1), 1, 1, (1), 1, 1, 1, (1), 1, 1, 1, 1, (1), 1, 1, 1, 1, 1, ...;
(1), (2), 3, 4, (5), 6, 7, 8, (9), 10, 11, 12, 13, (14), 15, 16, ...;
(3), (7), 13, 20, (28), 38, 49, 61, (74), 89, 105, 122, 140, (159),...;
(13), (33), 71, 120, (181), 270, 375, 497, (637), 817, 1019, 1244, ...;
(71), (191), 461, 836, (1333), 2150, 3169, 4413, (5906), 8001, ...;
(461), (1297), 3447, 6616, (11029), 19030, 29483, 42775, (59324),...;
(3447), (10063), 29093, 58576, (101351), 185674, 300329, 451277, ...;
(29093), (87669), 273343, 573672, (1024949), 1982310, 3330651, ...;
(273343), (847015), 2829325, 6159976, (11320359), 23009602, 39998897, ...;
where terms in parenthesis at positions {0,1,4,8,13,..} in a row
are removed before taking partial sums to obtain the next row.
...
RELATION TO SPECIAL TRIANGLE.
Triangle A104980 begins:
1;
1, 1;
3, 2, 1;
13, 7, 3, 1;
71, 33, 13, 4, 1;
461, 191, 71, 21, 5, 1;
3447, 1297, 461, 133, 31, 6, 1;
29093, 10063, 3447, 977, 225, 43, 7, 1; ...
in which column 0 and column 1 are found in square array A.
...
Matrix square of A104980 = triangle A104988 which begins:
1;
2, 1;
8, 4, 1;
42, 20, 6, 1;
266, 120, 38, 8, 1;
1954, 836, 270, 62, 10, 1;
16270, 6616, 2150, 516, 92, 12, 1;
151218, 58576, 19030, 4688, 882, 128, 14, 1; ...
where column 1 and column 2 are also found in square array A.

Cf. columns: A003319, A104981, A156629, related triangles: A104980, A104988.

Cf. related tables: A136212, A136213, A125714, A135876, A127054, A125781, A136217.

{T (n, k)=local (A=0, b=2, c=1, d=0); if (n==0, A=1, until (d>k, if (c==b* (b+1)/2-2, b+=1, A+=T (n-1, c); d+=1); c+=1)); A}

Column 0 = Column 0 of triangle A104980 = A003319.
Column 1 = Column 1 of triangle A104980 = A104981.
Column 3 = column 1 of A104988 (matrix square of A104980).
Column 5 = column 2 of A104988 (matrix square of A104980).

A125312 Moessner triangle based on primes.

Original entry on oeis.org

2, 3, 5, 10, 21, 13, 48, 105, 80, 29, 264, 628, 553, 232, 47, 1730, 4378, 4235, 2059, 543, 73, 13024, 34620, 36078, 19553, 6063, 1095, 107, 110542, 306362, 339554, 200769, 70350, 15166, 2000, 151, 1044900, 3003012, 3507070, 2228398, 861305, 212514

Offset: 1

Gary W. Adamson, Dec 10 2006

Row sums are 2, 8, 44, 262, 1724, 13024, ... Conjecture: log row n-th sum tends to (2n-3) + some unknown fractional part. E.g., log 1724 = 7.45... while log 13024 = 9.43... Right border = A011756.

			First few rows of the triangle are:
     2;
     3,    5;
    10,   21,   13;
    48,  105,   80,   29;
   164,  628,  553,  232,  47;
  1736, 4378, 4235, 2059, 543, 73;
  ...

J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996, p. 64.

Joshua Zucker, Table of n, a(n) for n = 1..55
G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
R. Krebbers, L. Parlant, and A. Silva, Moessner's theorem: an exercise in coinductive reasoning in Coq, Theory and practice of formal methods, 309-324, Lecture Notes in Comput. Sci., 9660, Springer, 2016.
Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
Ivan Paasche, Ein neuer Beweis des Moessnerschen Satzes S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 1-5 (1953). [Two years are listed at the beginning of the journal issue.]
Ivan Paasche, Beweis des Moessnerschen Satzes mittels linearer Transformationen, Arch. Math. (Basel) 6 (1955), 194-199.
Ivan Paasche, Eine Verallgemeinerung des Moessnerschen Satzes, Compositio Math. 12 (1956), 263-270.
Hans Salié, Bemerkung zu einem Satz von A. Moessner, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 7-11 (1953). [Two years are listed at the beginning of the journal issue.]
Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Cf. A125714, A125777, A011756.

Begin with the primes and circle every (n*(n+1)/2)-th prime: 1, 5, 13, 29, 47, ... = A011756. Following the instructions in A125714, take partial sums of the uncircled terms, making this row 2. Circle the terms in row 2 one place to the left of row 1 terms. Take partial sums of the uncircled terms, continuing with analogous procedures for subsequent rows.

Corrected and extended by Joshua Zucker, Jun 17 2007

cp's OEIS Frontend

A125750 A Moessner triangle using (1, 3, 5, ...).

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A125751 A Moessner triangle using (1, 2, 1, 2, 1, 2, ...).

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A125752 Moessner triangle using the Fibonacci terms.

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A125777 Moessner triangle based on A000217.

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A156628 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions 0 and {(m+1)*(m+2)/2-2, m>0} and then taking partial sums, starting with all 1's in row 0.

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A125312 Moessner triangle based on primes.

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