A000124 -id:A000124 - cp's OEIS frontend

A123151 a(n) = A000124(n)*a(n-3) for n > 2, otherwise n!.

1, 1, 2, 7, 11, 32, 154, 319, 1184, 7084, 17864, 79328, 559636, 1643488, 8408768, 67715956, 225157856, 1294950272, 11647144432, 43005150496, 273234507392, 2702137508224, 10923308225984, 75685958547584, 813343389975424, 3560998481670784, 26641457408749568

Offset: 0

Roger L. Bagula, Oct 01 2006

G. C. Greubel, Table of n, a(n) for n = 0..500

Cf. A000124, A123150.

function a(n) // a = A123151
  if n le 2 then return Factorial(n);
  else return (n^2+n+2)*a(n-3)/2;
  end if;
end function;
[a(n): n in [0..30]]; // G. C. Greubel, Jul 17 2023

a[n_]:= a[n]= If[n<3, n!, (1/2)*(n^2+n+2)*a[n-3]];
Table[a[n], {n,0,30}]

@CachedFunction # a = A123151
def a(n): return factorial(n) if (n<3) else (n^2+n+2)*a(n-3)/2
[a(n) for n in (0..30)] # G. C. Greubel, Jul 17 2023

a(n) = n! for n < 3, otherwise a(n) = A000124(n)*a(n-3), where A000124(n) = (n^2 + n + 2)/2.

Edited by N. J. A. Sloane, Oct 04 2006

Edited by G. C. Greubel, Jul 17 2023

A166021 a(n) = 2*A000124(A003056(n-1)) if A002262(n-1)=0, otherwise a(n-1)+1.

Original entry on oeis.org

2, 4, 5, 8, 9, 10, 14, 15, 16, 17, 22, 23, 24, 25, 26, 32, 33, 34, 35, 36, 37, 44, 45, 46, 47, 48, 49, 50, 58, 59, 60, 61, 62, 63, 64, 65, 74, 75, 76, 77, 78, 79, 80, 81, 82, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121

Offset: 1

Antti Karttunen, Oct 05 2009

Complement of A136272.

A204009 a(n) is a binary vector for selecting distinct terms from A000124 that when summed give n; it uses the greedy algorithm.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 1000, 1001, 1010, 1011, 10000, 10001, 10010, 10011, 10100, 100000, 100001, 100010, 100011, 100100, 100101, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 10000000, 10000001, 10000010, 10000011, 10000100

Offset: 0

Frank M Jackson, Jan 09 2012

nonn

a(n) is a binary vector for selecting terms from A000124 that when summed give n. It uses the greedy algorithm to select from multiple solutions.

			14 can be written as 7+4+2+1, i.e., 1111, or as 11+2+1, i.e., 10011, and the latter is chosen because it uses the greedy algorithm for selection.

Wikipedia, "Complete" sequence. [Wikipedia calls a sequence "complete" (sic) if every positive integer is a sum of distinct terms. This name is extremely misleading and should be avoided. - _N. J. A. Sloane_, May 20 2023]

Cf. A000124, A072638.

complete[m_Integer] := (m(m+1)/2+1); gentable[n_Integer] := (m=n; ptable={0}; While[m!=0, (i=0; While[complete[i]<=m&&ptable[[i+1]]!=1, (AppendTo[ptable, 0]; i++)]; ptable[[i]]=1; m=m-complete[i-1])]; ptable); decimal[n_Integer] := (gentable[n]; Sum[2^(k-1)*ptable[[k]], {k, 1, Length[ptable]}]); Table[IntegerString[decimal[s], 2], {s, 0, 100}]

a(n) x A000124 = n, where x is the inner product and the binary vector is in ascending powers of 2 with infinite trailing zeros.

Edited by N. J. A. Sloane, May 20 2023

A271861 Recursive sequence based on the central polygonal numbers (A000124) and A002260.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 9, 8, 10, 12, 15, 14, 6, 16, 19, 11, 13, 18, 21, 24, 20, 28, 27, 25, 22, 30, 23, 34, 37, 36, 26, 29, 33, 17, 41, 44, 40, 39, 32, 35, 45, 31, 49, 52, 48, 55, 54, 51, 38, 46, 50, 58, 61, 57, 64, 67, 66, 56, 43, 59, 47, 68, 71, 63, 74, 77, 81

Offset: 1

Max Barrentine, Apr 15 2016

nonn

Conjectured to be a permutation of the natural numbers.

The central polygonal numbers can be constructed by starting with the natural numbers, setting A000124(0)=1 and obtaining A000124(n+1) by reversing the order of the next A000124(n) numbers after A000124(n). This procedure doesn't produce a permutation of the natural numbers for A000124 because the sequence is strictly increasing. The present sequence is constructed by the same procedure, except that a(n+1) is obtained by reversing the next a(A002260(n)) numbers.

			Start with the natural numbers:
1, 2, 3, 4, 5, 6, 7, 8, 9...
a(A002260(1))=1, so reverse the order of the next term, leaving the sequence unchanged:
   (1)
1, (2), 3, 4, 5, 6, 7, 8, 9...
a(A002260(2))=1, so reverse the order of the next term, leaving the sequence unchanged:
      (1)
1, 2, (3), 4, 5, 6, 7, 8, 9...
a(A002260(3))=2, so reverse the order of the next 2 terms:
         (2)
1, 2, 3, (5, 4), 6, 7, 8, 9...
a(A002260(4))=1, so reverse the order of the next term, leaving the sequence unchanged:
            (1)
1, 2, 3, 5, (4), 6, 7, 8, 9...
a(A002260(5))=2, so reverse the order of the next 2 terms:
               (2)
1, 2, 3, 5, 4, (7, 6), 8, 9...
a(A002260(6))=3, so reverse the order of the next 3 terms:
                  (3)
1, 2, 3, 5, 4, 7, (9, 8, 6)...

Max Barrentine, Table of n, a(n) for n = 1..1227

Cf. A000124, A002260.

A271863 Recursive sequence based on the central polygonal numbers (A000124) and A004736.

Original entry on oeis.org

1, 2, 4, 3, 8, 6, 7, 10, 12, 5, 11, 19, 16, 14, 18, 15, 22, 25, 17, 9, 24, 13, 29, 23, 32, 28, 26, 31, 27, 39, 20, 38, 40, 33, 35, 30, 34, 49, 36, 46, 37, 21, 45, 43, 48, 44, 51, 59, 41, 56, 42, 50, 55, 53, 58, 54, 67, 62, 70, 64, 57, 65, 63, 52, 60, 69, 47

Offset: 1

Max Barrentine, Apr 15 2016

nonn

Conjectured to be a permutation of the natural numbers.

The central polygonal numbers can be constructed by starting with the natural numbers, setting A000124(0)=1 and obtaining A000124(n+1) by reversing the order of the next A000124(n) numbers after A000124(n). This procedure doesn't produce a permutation of the natural numbers for A000124 because the sequence is strictly increasing. The present sequence is constructed by the same procedure, except that a(n+1) is obtained by reversing the next a(A004736(n)) numbers.

			Start with the natural numbers:
1, 2, 3, 4, 5, 6, 7, 8...
a(A004736(1))=1, so reverse the order of the next term, leaving the sequence unchanged:
   (1)
1, (2), 3, 4, 5, 6, 7, 8...
a(A004736(2))=2, so reverse the order of the next 2 terms:
      (2)
1, 2, (4, 3), 5, 6, 7, 8, 9...
a(A004736(3))=1, so reverse the order of the next term, leaving the sequence unchanged:
         (1)
1, 2, 4, (3), 5, 6, 7, 8...
a(A004736(4))=4, so reverse the order of the next 4 terms:
            (4)
1, 2, 4, 3, (8, 7, 6, 5)...
a(A004736(5))=2, so reverse the order of the next 2 terms:
               (2)
1, 2, 4, 3, 8, (6, 7), 5...
a(A004736(6))=1, so reverse the order of the next term, leaving the sequence unchanged:
                  (1)
1, 2, 4, 3, 8, 6, (7), 5...

Max Barrentine, Table of n, a(n) for n = 1..1082

Cf. A000124, A004736.

A271865 Recursive sequence based on the central polygonal numbers (A000124) and A004738.

Original entry on oeis.org

1, 2, 4, 3, 6, 9, 7, 8, 10, 13, 5, 15, 12, 14, 16, 19, 11, 23, 20, 17, 22, 18, 24, 27, 21, 31, 35, 28, 32, 34, 26, 33, 29, 37, 25, 41, 45, 39, 47, 30, 44, 46, 42, 40, 36, 49, 43, 53, 57, 51, 58, 50, 61, 54, 52, 60, 55, 59, 38, 63, 56, 67, 71, 65, 72, 75, 70

Offset: 1

Max Barrentine, Apr 16 2016

nonn

Conjectured to be a permutation of the natural numbers.

The central polygonal numbers can be constructed by starting with the natural numbers, setting A000124(0)=1 and obtaining A000124(n+1) by reversing the order of the next A000124(n) numbers after A000124(n). This procedure doesn't produce a permutation of the natural numbers for A000124 because the sequence is strictly increasing. The present sequence is constructed by the same procedure, except that a(n+1) is obtained by reversing the next a(A004738(n)) numbers.

			Start with the natural numbers:
1, 2, 3, 4, 5, 6, 7, 8, 9...
a(A004738(1))=1, so reverse the order of the next term, leaving the sequence unchanged:
   (1)
1, (2), 3, 4, 5, 6, 7, 8, 9...
a(A004738(2))=2, so reverse the order of the next 2 terms:
      (2)
1, 2, (4, 3), 5, 6, 7, 8, 9...
a(A004738(3))=1, so reverse the order of the next term, leaving the sequence unchanged:
         (1)
1, 2, 4, (3), 5, 6, 7, 8, 9...
a(A004738(4))=2, so reverse the order of the next 2 terms:
            (2)
1, 2, 4, 3, (6, 5), 7, 8, 9...
a(A004738(5))=4, so reverse the order of the next 4 terms:
               (4)
1, 2, 4, 3, 6, (9, 8, 7, 5)...
a(A004738(6))=2, so reverse the order of the next 2 terms:
                  (2)
1, 2, 4, 3, 6, 9, (7, 8), 5...
a(A004738(7))=1, so reverse the order of the next term, leaving the sequence unchanged:
                     (1)
1, 2, 4, 3, 6, 9, 7, (8), 5...

Max Barrentine, Table of n, a(n) for n = 1..1124

Cf. A000124, A004738.

A161254 Number of partitions of n into central polygonal numbers A000124.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 7, 10, 11, 14, 17, 21, 24, 29, 34, 41, 46, 55, 62, 73, 81, 96, 107, 124, 137, 158, 175, 199, 221, 250, 276, 310, 343, 383, 421, 469, 516, 572, 626, 693, 757, 833, 908, 1000, 1088, 1192, 1294, 1417, 1535, 1674, 1813, 1974, 2133, 2315, 2501, 2710, 2921

Offset: 0

R. H. Hardin, Jun 06 2009

nonn

			1 + x + 2*x^2 + 2*x^3 + 4*x^4 + 4*x^5 + 6*x^6 + 7*x^7 + 10*x^8 + 11*x^9 + ...
a(4) = 4 since 4 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1 is a partition in 4 ways. a(7) = 7 since 7 = 4 + 2 + 1 = 4 + 1 + 1 + 1 = 2 + 2 + 2 + 1 = 2 + 2 + 1 + 1 + 1 = 2 + 1 + 1 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1 + 1 is a partition in 7 ways. - _Michael Somos_, May 29 2012

R. H. Hardin, Table of n, a(n) for n = 0..1000

Cf. A000124.

G.f.: 1 / (Product_{k>0} (1 - x^( (k^2 - k)/2 + 1))). - Michael Somos, May 29 2012

A161255 Number of partitions of n into central polygonal numbers A000124 where every part appears at least 2 times.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 5, 3, 6, 5, 9, 6, 12, 9, 16, 12, 20, 16, 26, 21, 33, 27, 41, 35, 50, 43, 62, 54, 75, 66, 91, 82, 108, 99, 131, 120, 154, 144, 183, 172, 214, 204, 254, 240, 294, 284, 344, 331, 398, 387, 462, 449, 531, 521, 612, 600, 702, 690, 803, 792, 915, 905, 1043

Offset: 1

R. H. Hardin, Jun 06 2009

nonn

R. H. Hardin, Table of n, a(n) for n=1..1000

Cf. A000124.

A161256 Number of partitions of n into central polygonal numbers A000124 where every part appears at least 3 times.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 2, 3, 3, 5, 4, 5, 6, 8, 7, 9, 9, 12, 12, 14, 14, 19, 18, 22, 23, 28, 27, 32, 34, 40, 42, 47, 48, 57, 58, 66, 69, 78, 81, 89, 94, 106, 111, 120, 128, 143, 148, 161, 170, 187, 196, 214, 225, 244, 257, 277, 294, 319, 335, 357, 377, 409, 428, 460, 483, 519, 546, 582

Offset: 1

R. H. Hardin Jun 06 2009

nonn

R. H. Hardin, Table of n, a(n) for n=1..1000

A161257 Number of partitions of n into central polygonal numbers A000124 where every part appears at least 4 times.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 4, 3, 6, 4, 6, 5, 9, 7, 9, 8, 13, 10, 13, 11, 19, 14, 19, 16, 26, 21, 28, 25, 36, 29, 39, 34, 49, 41, 55, 48, 66, 56, 71, 65, 86, 78, 94, 87, 111, 99, 121, 113, 143, 128, 155, 146, 180, 165, 196, 189, 226, 211, 246, 237, 281, 267, 310, 300, 351, 334

Offset: 1

R. H. Hardin Jun 06 2009

nonn

R. H. Hardin, Table of n, a(n) for n=1..1000

cp's OEIS Frontend

A123151 a(n) = A000124(n)*a(n-3) for n > 2, otherwise n!.

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A166021 a(n) = 2*A000124(A003056(n-1)) if A002262(n-1)=0, otherwise a(n-1)+1.

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A204009 a(n) is a binary vector for selecting distinct terms from A000124 that when summed give n; it uses the greedy algorithm.

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Examples

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Programs

Mathematica

Formula

Extensions

A271861 Recursive sequence based on the central polygonal numbers (A000124) and A002260.

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A271863 Recursive sequence based on the central polygonal numbers (A000124) and A004736.

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A271865 Recursive sequence based on the central polygonal numbers (A000124) and A004738.

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A161254 Number of partitions of n into central polygonal numbers A000124.

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Examples

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Formula

A161255 Number of partitions of n into central polygonal numbers A000124 where every part appears at least 2 times.

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A161256 Number of partitions of n into central polygonal numbers A000124 where every part appears at least 3 times.

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A161257 Number of partitions of n into central polygonal numbers A000124 where every part appears at least 4 times.

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